I like problems that can be solved in multiple ways, which can train us in seeing the world from different perspectives. Late in November we dealt with a pair of such questions involving angles in star-like figures.

Here is the first question, which was actually the second of two submitted at about the same time:

This one is hard question, no angles given.

We are supposed to find the sum of angles “A”, “B”, “C”, “D” and “E”.

Doctor Rick answered:

Hi, Daniel, thanks for writing to the Math Doctors.

Your problems are interesting. I can think of several ways you might approach each of them. I’d like to start with this one because I think it is

a good introduction to one of the ideas I have in mind.That idea is a little different from the usual way we think of geometry.

Imagine that you are a bug crawling along the five-pointed star. Start on the horizontal line between A and B, just to make it more specific. You can crawl all the way around the paths (going straight through each intersection) and eventually end up back where you started. In doing so, you will have made five turns.

Now, I have two questions for you. First,

by what angle do you turnat each corner (in terms of the angles A, B, C, D, and E)? Second,how many complete turnshave you made by the time you get back to your starting point? (If you were crawling around a circle or a square or a pentagon, you would have made one full turn, or 360 degrees.)If you can answer those questions correctly, then you can find A+B+C+D+E.

Here is one turn:

For comparison, here is what we would do with a pentagon:

At each corner the bug turns, ultimately making exactly one complete turn in a circle. So the sum of all the turns is 360°. So what is the sum of the internal angles at those corners? It’s the sum of the supplements of five angles whose sum is 360°; so it’s \(5\cdot 180°-360°=540°\). More formally, we can write the equation $$(180-A)+(180-B)+(180-C)+(180-D)+(180-E)=360\\5(180)-(A+B+C+D+E)=360\\A+B+C+D+E=5(180)-360=540°$$

Here is

a second, more traditional approach. Consider, for example, the triangle that has A and E as two of its angles. You canfind the third angle in terms of A and E. Do the same with other triangles like this one, anduse what you know about the interior angles of ordinary polygons(as opposed to star polygons like this).Those are enough ideas to get you started, I think, since I presume you and your friend like challenges. You could apply similar ideas to your other problem.

Daniel tried both ways:

Ohhhh

Thank you very much Doctor Rick.

Just when I thought about

the crawling rat thingI noticed I came back to the same point but in different direction.Which means

something like 180 degrees.Also using

the interior angles conceptI could notice how a single triangle in the star could have the sum of two angles of another triangle which makes a + b + c + d + e.In a single triangle! and one triangle have sum of 180° so the star is

likely to have total sum of 180° degreesin it.I hope I didn’t get you wrong

The rat/bug approach is unclear here; Daniel has probably misunderstood something.

The triangle approach is not fully explained here, but upon reading carefully what he says and thinking about what he might be doing, I realize that he may have discovered a very nice way, and found the right answer. And in fact, looking ahead, we’ll find that he will explain it nicely later, so I’ll wait until then to make a picture of it. Can you see what he did?

Doctor Rick said,

If you would rather be a rat than a bug, that’s up to you.

But I don’t understand what you mean by “I came back to the same point but in different direction.” Maybe you mean that you

started at A and returned to A, but you weren’t yet facing right again, because youhadn’t yet made the fifth turn. I suggested starting in themiddleof segment AB (taking these as names of vertices) so it would be clear that you had to complete all five turns.

It isn’t clear whether this is in fact what Daniel did; but let’s see what he would find if he did (sticking with the bug):

It started at A (green) and stopped when it got back to A, without making a final turn; since we don’t know the angles, we can’t know how far it turned.

Also, you didn’t really say how this led to an answer of 180°. I have to do some calculation before I’m done — notice that the turn at B, for example, is

notby the angle measure B, but thesupplementof that angle, as shown in the attached figure: the rat was facing right, and it turned through the angle I labeled β’ (beta prime) in order to crawl toward C. So, could you say more about how you got your answer by this approach?

Let’s finish this solution:

From the initial direction (green), the bug has rotated two full turns (720°) to the final position (red). Each turn is the supplement of the interior angle at that corner (e.g. \(180^\circ – B\)), so we can write an equation: $$(180^\circ – A)+(180^\circ – B)+(180^\circ – C)+(180^\circ – D)+(180^\circ – E)=720^\circ\\900-(A+B+C+D+E)=720\\A+B+C+D+E=180^\circ$$

So the sum of the five angles is indeed 180°.

Again, I don’t follow your description of your thinking on the

second method. How can you get a+b+c+d+e in a single triangle? You put a smiley there, and your words “likely to have” (and “something like” in the first approach) suggest that you know your thinking is too vague.

We’ll see what that meant soon. First, we need a better way to name the triangles:

In my figure, I named five points of intersection so that we can talk more precisely. For instance, what I was saying before was that α and ε (I used Greek letters for the angles) are two of the interior angles of triangle AGE, so that you can express the measure of angle AGE (which is the same as angle FGH in the pentagon) in terms of α and ε. You may be thinking of something different from what I did, which is fine, but you will have to express your thoughts more precisely in order to convince me that your method works. Would you like to try?

Daniel responded,

Hey there Dr. Rick

I am very-very sorry for my late reply.

Anyhow, what I meant was if we look at the star as something formed from different triangles then use the exterior angles theorem (see the first image, here) we can notice how some exterior angles of some triangles exist in the closest triangle (see the second image, further down)

This Exterior Angle Theorem says that one *exterior* angle is equal to the sum of the other *interior* angles, and is very powerful.

For example:

The triangle that contains angle “A” and “B” (the one I colored red in my second image, here) have exterior angle in the triangle that contains angle “E” (the triangle I colored green). Also the triangle that contains angle “C” and “D” (colored blue) have an exterior angle in the triangle that contains “E”.

Now in exterior angles theorem *the exterior angle of a triangle is equal to the sum of the opposite interior angles* so in the triangle that contains “E” if we name the missing angles “X” and “Y” (see the second image in the attachment) so X would be equal to a+b and Y equal to c+d

So this way we got E + Y +X = 180° because the sum triangles interior angles is 180°

Y = C+D

X = A+B

Which means A+B +C+D+E = 180°

Because the sum of triangle interior degrees is doomed to be 180°

Sorry again for my late reply and sorry for the bad English.

Please let me know if there is something wrong with my way.

I really appreciate your will to help me!

This is excellent!

Doctor Rick answered:

Hi, Daniel. There is no need to apologize for delay in responding on a math problem. Your solution is very good — in fact, it is

more elegant than either solution I had in mind. Did you come up with it on your own?Using my figure (ignoring the extension of AB and angle β’), we can write a more formal proof of your result as follows:

By the Exterior Angle Theorem on triangle ABI,

m(∠HIE) = m(∠A) + m(∠B). . . . . . . . . [1]

By the Exterior Angle Theorem on triangle CDH,

m(∠EHI) = m(∠C) + m(∠D). . . . . . . . . [2]

By the Interior Angle Sum Theorem on triangle EHI,

m(∠HIE) + m(∠EHI) + m(∠E) = 180°. . [3]

Substituting from [1] and [2] into [3] gives

m(∠A) + m(∠B) + m(∠C) + m(∠D) + m(∠E) = 180°

You can’t get much simpler than this! — But to come up with a simple idea can take a good deal of thought. Good work!

Here is a colored version of his picture to go with the proof:

The angle in the green triangle at I is α + β; the angle at H is γ + δ; so the green triangle shows that α + β + γ + δ + ε = 180°.

I suspect the method Doctor Rick had in mind, using the sum of interior angles of polygons, looked like this:

The Exterior Angle Theorem tells us that \(C’=180^\circ-A-E\), and similarly for the other interior angles of the pentagon. A theorem tells us that those angles add up to \((n-2)180^\circ\), so we can write the equation $$C’+A’+D’+B’+E’=\\(180-A-E)+(180-D-C)+(180-B-A)+(180-E-D)+(180-C-B)=3(180)\\900-2(A+B+C+D+E)=540\\A+B+C+D+E=180$$

Or, we could use the fact that the sum of the exterior angles of the pentagon is 360°:

This time, \(C’=A+E\) for the yellow triangle, and likewise for the others, so $$C’+A’+D’+B’+E’=(A+E)+(D+C)+(B+A)+(E+D)+(C+B)=360\\2(A+B+C+D+E)=360\\A+B+C+D+E=180$$

Now, if you were just asked for the sum of the angles in a multiple-choice context that implied the answer is independent of the specific angles, you might just adjust the figure so that it is inscribed in a circle, and assume (this can be rigorously justified, or you can just know this based on our other methods) that the sum of angles is unchanged. In that case, you can use the theorem about angles inscribed in a circle, that the inscribed angle is half the central angle of the subtending arc:

Here angle A is half of arc A, and so on; so the sum of the angles is half the sum of the arcs, which is 360°. And we’re done.

Two problems had been submitted at once. Here is the other, which Doctor Rick left unanswered to work on the easier one:

I got this question from a friend …

We are supposed to find the sum value of angle “a”, “b”, “c”, “d”, “e”, “f”, “g”, “h”

Please help us it is very urgent.

While Doctor Rick worked with the easier one, I wanted to suggest some ideas to supplement his:

Hi, Daniel.

I see that Doctor Rick is working with you on the pentagram question; I want to suggest some things to you that will apply ultimately to both problems. Here are pages I found at

Ask Dr. Maththat are relevant:

Measures of Interior and Exterior Angles of PolygonsThe first suggests a variant on the “bug crawl” approach; the other two do essentially the same thing, in terms of the “

winding number“, which is the number of times you wind around the center as you move around a figure. Any of these may give you additional perspectives on both problems.For the problem in this question, you may be able to see

exactly what the angles are (as drawn)just by thinking about the figure, and adding them up! Apart from that, you can break the figure up intofour triangles and a square, label the unnamed angles in each, and write equations stating what the sum of angles in each triangle and the square are.

This last suggestion is similar to what I did above when I supposed that the sum would be the same as in the simpler case of an inscribed star (or even a regular pentagram). This yields lots of parallel lines, and you can see that the eight angles would all be 45°, making the sum 360°:

(This has been rotated and relettered so a bug would visit the vertices in order, as in the other problem.)

Doctor Rick added to this as well:

Hi, Daniel, Doctor Rick here. Let me point out a few things on this problem too.

First, as I understand it, you can’t assume anything about, for instance, whether lines are parallel, just because they

lookparallel. However, what Doctor Peterson said about just finding the angles and adding them up,assumingthat the quadrilateral is a square (and maybe even that the triangles are isosceles), will give you the answerfor that particular case — and the problem seems toimply, that the sum of the angles is independent of any details not given, like angles and lengths, so the particular answer is likely to be the general answer too.That’s sort of cheating yourself, if you have a mathematical mind that really wants to

knowthat the sum is invariant. But sometimes it’s helpful to find the answer first, and then try to prove it.And that proof, as Doctor Peterson suggests, can be accomplished by

either methodI offered for the other problem. The work will be very similar to what I did there — which I haven’t exactly told you yet, I’ve just given hints. For this purpose I’ve attached a more general figure, marked up similarly to my figure for the other problem. I hope we can carry both problems through to your complete satisfaction!

We didn’t hear back on this problem, because the other was where all the discussion happened. So let’s take a look at it.

First, here is the bug approach:

Following the turns, as I did on the right, we have three full turns (winding number 3), or \(3(360) = 1080^\circ\). This is the sum of 8 external angles; so the sum of the internal angles is $$8(180)-1080=1440-1080=360^\circ$$

An alternative way to count the turns is to observe that the four triangles are all circumnavigated clockwise, while the quadrilateral in the middle is covered counterclockwise, for a net motion of 3 clockwise:

If we break up into triangles and a quadrilateral, we find that the sum of angles A through H equals the sum of external angles I through L, whose sum is 360°:

That, I think, is about as elegant as we can get.

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Our first question is from 2001:

Induction Proof with Inequalities I've been trying to solve a problem and just really don't know if my solution is correct. I have a really hard time doing these induction problems when inequalities are involved. I was hoping you could help me solve this. The question is this: Prove by induction that (1 + x)^n >= (1 + nx), where n is a non-negative integer.

Jay is right: inequality proofs are definitely trickier than others, particularly than series proofs, which tend to be fairly routine apart from the big algebraic step.

As written, the problem appears to mean what one of our problems last time meant: that the inequality is true for all real numbers *x*, for any non-negative integer *n*. If we check one case, *n* = 2, that looks right (the red curve is \(y=(1+x)^2\), and the blue line is \(y=1+2x\)):

But if we try *n* = 3, something is wrong; the red curve isn’t *always* above the blue line:

So we’ll have to clarify the problem.

Doctor Ian answered, demonstrating how one might think through such a problem from the start:

Hi Jay,Let's _assume_ that it holds for n = k.That is, for some k, it's true that (1+x)^k >= (1 + kx)What happens for the next value of n, i.e., n = (k+1)? (1+x)^(k+1) >= (1 + (k+1)x) >= (1 + kx + x) >= (1 + kx) + x Soincreasing the value of n by 1 adds x to the right sideof the inequality. What does it do to the left side? (1+x)^(k+1) >= (1 + kx) + x (1+x)^k * (1+x) >= Soincreasing the value of n by 1 multiplies the left sideof the equation by (1+x).

Note that he didn’t yet check the base case; and that he isn’t yet proving anything, but rather experimenting with the goal, by stating it (a hoped-for inequality) and thinking about what has to happen for it to be true, by rewriting each side to extract from it the expression found in the *n* = *k* case.

Now there are two things left to do. The first is to show that (or explain the conditions under which) something multiplied by (1+x) is greater than the same thing plus x: alpha * (1+x) >= alpha + x Once you've done that, you need to show that the inequality holds for the smallest value of n (in this case, n = 1), (1+x)^1 >= (1 + 1x) which should be pretty easy to do.

(Actually, the base case should be *n* = 0, as it only has to be *non-negative*, not positive; and the “alphas” on the two sides may not actually be equal, just in a known order.)

All of this is still informal – the sort of thing mathematicians do when they are thinking about a problem, sketching ideas for the actual work; they sweep this away when they clean up their proof for publication. Textbooks generally don’t show you this mess, and therefore can give you the impression that proofs are supposed to pop into your mind full-grown, and that if they don’t, you aren’t fit for math. Never believe that!

But he’s left the details for Jay to do, and only hinted at that issue I pointed out at the start, by saying, “show that (*or explain the conditions under which*) …”: As happens in real life, we will have to refine our claim before we can prove it!

He closed with a summary of how induction works:

The general idea is to show thatif it works for n = k, then it also works for n = k+1; and then to show thatit works for n = 1. (Since it works for n = 1, it must also work for n = 2; and since it works for n = 2, it must also work for n = 3; and so on.) And usually it works just the way it did here: You substitute k+1 for k in both sides, and thentry to figure out what _change_ occurs as a result of the substitution, which you do by trying to recover the original form for k, with some extra stuff hanging off it: k k+1 --------- --------------- (1+x)^k -> (1+x)^k * (1+x) (1+kx) -> (1+kx) + x ^ ^ | | original extra form stuff Then you can forget about comparing the original forms, and concentrate on comparing the extra stuff, which is usually easier.

Doctor Rob (having written simultaneously) joined in 9 minutes later, bringing up the details Doctor Ian omitted:

Thanks for writing to Ask Dr. Math, Jay.You will need a condition on x for this to work. It is false for x = -11 and n = 3, for example. One condition under which this is true for all nonnegative integers n is x >= -1.

How could this be determined? Our graphs above suggest it might be true for \(x \ge -3\); I made graphs for higher values of *n* and it looked as if \(x \ge -2\) might work. We could just make a guess like that and see what happens as we proceed with the proof.

Thebasisof the induction is n = 0, which you can verify directly is true. Nowassumeit is true for some value of n. Now if (1+x) is nonnegative, you can multiply both sides by (1+x) to get the left side in the correct form. Expand the right-hand side, and rearrange it into the form (1+x)^(n+1) >= 1 + (n+1)*x + n*x^2. Observe that if you leave out the n*x^2 term, you get something equal or smaller, because n >= 0 and x^2 >= 0 for all real x. I leave the rest to you.

The base case is $$(1+x)^0\ge 1+0x\\1\ge 1$$

Now, assuming the induction hypothesis, that $$(1+x)^k\ge 1+kx$$ we want to prove the goal, $$(1+x)^{k+1}\ge 1+(k+1)x$$ as Doctor Ian said. Again, we can rewrite this goal (still unproved) as $$(1+x)^k\cdot(1+x)\ge 1+x+kx$$

So let’s try to derive this from the hypothesis. As noted, we can multiply both sides of an inequality by the same number as long as that number is *positive*. So we’d like to assume that \(1+x>0\), which we can do as long as \(x>-1\). (I think we’ve found where Doctor Rob’s restriction came from! We don’t know that this restriction can’t be loosened so that somewhat smaller x can be allowed; but we aren’t claiming to have found the “sharpest” condition for our theorem.)

So now we can say that if $$(1+x)^k\ge 1+kx$$ then $$(1+x)^{k+1}\ge (1+kx)(1+x)=1+x+kx+kx^2$$ But \(kx^2\ge 0\) for any *x*, so we can subtract it: $$\cdots \ge 1+x+kx$$ which was our goal.

Next, consider this question from 2003, involving an exponential inequality:

Proof by Induction Use induction to prove: If n >= 6 then n! >= n(2^n) This is unlike all other induction problems. I get lost when I do the induction step. Base case: 6! >= 6(2^6) 720 >= 384 Induction Step: (n + 1)! >= (n + 1) (2^(n + 1)) if n! >= n*2^n

Let’s graph this one, too, just to check that \(n\ge 6\) is the appropriate condition; again, the LHS is in red (discrete dots because the factorial is only defined for integers), and the RHS is in blue:

(The dotted line is a curve that passes through all the factorial points.)

We can see that the factorials don’t start exceeding the blue curve until after *n* = 6.

Doctor Luis answered:

Hi Sarah, Good work establishing thebase caseand theinduction step. I'll start you up by showing you writing down my thoughts on how I would solve this problem. If at any point while reading this, you see the solution, stop and work it out on your own, and then come back and keep reading. Given n! >= n(2^n), we have to figure out a way to make that look like (n + 1)! >= (n + 1)(2^(n + 1)).

Again, we have a hypothesis and a goal. What is changing on each side? We can start with a basic fact about factorials:

The first thing we see is that(n + 1)! = (n + 1) * n!, and this is >= n * n!, which gives us the n! for which we made the assumption. This suggests the following: n! >= n * (2^n) (start from assumption) n * n! >= n * n * (2^n) (multiply by n) (n + 1)! >= n^2 * 2^n

This is a lot like the first inequality problem, isn’t it? We multiply by something positive on both sides, in order to make one side look like our goal; but in this case, he chose to multiply by *n*, and then increase it to *n *+ 1 on the left, which increases the expression and maintains the inequality. (This is not the only way to approach it!)

Now he plays with the condition \(n\ge 6\), trying to change \(n^22^n\) to \((n+1)2^{n+1}\):

Now we're getting somewhere, but there's an extra n on the right side. Maybe we can do something with the inequality n >= 6. n >= 6 n^2 >= 6n = 2 * (3n) = 2 * (n + n + n) Hmm. That 2 may come in useful in getting us from 2^n to 2^(n + 1). Now, how does n + 1 compare with n + n + n? Eureka! n + n + n >= n + 1. You can see this from n >= 6, 2n >= 12 + 1, 3n > 1 + n. Combining all these inequalities, you can prove the induction step.

Here is his proof, starting with the induction hypothesis:

$$n! \ge n\cdot 2^n\\(n+1)n! \ge n\cdot n! \ge n^2\cdot 2^n\ge 6n\cdot 2^n\ge 3n\cdot 2\cdot 2^n\ge (n+1)2^{n+1}$$

I see a simpler way to get the desired result, based on the common occurrence of \((n+1)\), and the fact that if \(n\ge 6\), then \(n\ge 2\):

$$n! \ge n\cdot 2^n\\(n+1)n! \ge (n+1)n\cdot 2^n\\(n+1)! \ge (n+1)n\cdot 2^n\ge (n+1)2\cdot 2^n=(n+1)2^{n+1}$$ Interestingly, this step works for values less than 6; but because the base case would fail in such cases, we can’t start there.

Let’s not forget that base case: $$6! \ge 6\cdot 2^6\\720\ge 384$$

Turning to series, this question is from 1997:

Triangular Numbers in a Proof I am trying to prove 1^3 = 1^2 1^3 + 2^3 = (1+2)^2 1^3 + 2^3 + 3^3 = (1+2+3)^2 1^3 + 2^3 + 3^3 + ... + n^3 = (1+2+3+...+n)^2. I have tried to find a proof by induction, but didn't get very far. I also tried working withtriangular numberssince the right side is the triangular numbers, but I could not show that the left and right sides were equal. I need some hints, or maybe the trick that is needed to verify the general case. I also found some patterns which turned out to lead to a dead end.

The first three lines are particular cases (the first being the base case), and can be proved by calculation. The last line is what is to be proved in general, and looks tricky, considering that there is a summation on both sides!

The **triangular numbers** are the sums \(1+2+3+…+n = \frac{n(n+1)}{2}\), and that formula (which we proved last week) can be useful.

Doctor Steven answered, starting with that formula and expanding it:

The right hand side is equal to [(n*(n+1))/2]^2 so you are trying to prove that the sums of the cubes of numbers is equal to (n^4 + 2n^3 + n^2)/4. Induction should work fairly well for this proof.

We’ll consider later whether that expansion was necessary; but it was easy: $$\left(\frac{n(n+1)}{2}\right)^2=\frac{n^2(n+1)^2}{2^2}=\frac{n^2(n^2+2n+1)}{4}=\frac{n^4+2n^3+n^2)}{4}$$

So now we want to prove by induction that, for any positive integer *n*, $$1^3+2^3+3^3+\cdots +n^3=\frac{n^4+2n^3+n^2}{4}$$

Start with yourbase caseof 1: (1^4 + 2*1^3 + 1^2)/4 = 1^3 = 1.Assume it's true for k: (k^4 + 2k^3 + k^2)/4 = 1^3 + 2^3 + .... + k^3. Look at k + 1: ((k+1)^4 + 2*(k+1)^3 + (k+1)^2)/4 = (k^4 + 4k^3 + 6k^2 + 4k + 1 + 2k^3 + 6k^2 + 6k + 2 + k^2 + 2k + 1)/4= (k^4 + 6k^3 + 13k^2 + 12k + 4)/4 = (k^4 + 2k^3 + k^2)/4 + (4k^3 + 12k^2 + 12k + 4)/4 = (1^3 + 2^3 + 3^3 + . . . + k^3) + (k^3 + 3k^2 + 3k + 1) = (1^3 + 2^3 + 3^3 + . . . + k^3) + (k + 1)^3 = 1^3 + 2^3 + 3^3 + . . . + k^3 + (k+1)^3. We win!

This is slightly odd, in putting the series on the right and starting on the other side; clearly this requires keeping the goal (looking like a series) in mind; but all he’s done (if I can describe expanding fourth powers as if it were easy) is to expand and simplify the formula for the next case, then split it into the formula for the assumed case plus some extras – much like Doctor Ian’s suggestion in our first example.

But I like the conclusion: “We win!” Proof is like a game, and we can celebrate at the end. That’s essentially what the traditional closing of “Q.E.D” means.

In 2001, we got another request for the same problem; Doctor Floor’s answer was added to the same page:

Hi, Madison, It is easy to check that this is correct for n = 1. Suppose that we have proven it is correct that 1^3+2^3+3^3+...+n^3 = (1+2+3+...+n)^2, which we will use as induction hypothesis. We will use the well-known identity for triangular numbers 1+2+3+...+n = n(n+1)/2

You may have noticed that some of us use a separate variable, such as *k*, in the induction hypothesis, while others stick with the same *n* used in the statement of the problem. This is a matter of taste, and also of how comfortable you are with reusing a variable within a restricted part of a proof. For beginners, it is probably better to use the *k*.

Then we can derive the following: 1^3+2^3+3^3+...+n^3+(n+1)^3 = (1+2+3+...+n)^2 + (n+1)^3 = (1+2+3+...+n)^2 + n(n+1)^2 +(n+1)^2 = (1+2+3+...+n)^2 + n(n+1)*(n+1) + (n+1)^2 = (1+2+3+...+n)^2 + 2(1+2+3+...+n)*(n+1) + (n+1)^2 = (1+2+3+...+n + n+1)^2 And the proof by induction is complete.

Here he never actually used the formula for the triangular numbers, much less expanded it. But he did some impressive juggling of expressions, keeping his eye, as always, on the goal. I wouldn’t be surprised if he had worked backward from the goal a little bit, in order to see better what to aim for, something like this: $$(1+2+3+…+n + (n+1))^2=((1+2+3+…+n) + (n+1))^2\\=(1+2+3+…+n)^2+2(1+2+3+…+n)(n+1)+(n+1)^2$$ using the fact that \((a+b)^2 = a^2+2ab+b^2\). Seeing the repeated \((n+1)\)’s would suggest how to pull them apart when going forward.

We’ll close with another series problem, this time not a polynomial:

Proof by Induction What is the expression for: n ----- \ / 2 to the power of r ----- r=0 How can we prove it?

We haven’t seen this “sigma notation” for series yet; this means “the sum, for *r* taking integer values from 0 through *n*, of \(2^r\)”. In print, it looks like $$\sum_{r=0}^n 2^r=2^0+2^1+2^2+\cdots+2^n$$

Doctor Jeremiah answered, first suggesting that we can find a (hopeful) formula by looking for a pattern:

Hi Tony, Write out the sum: 2^0 + 2^1 + 2^2 + 2^3 + 2^4 + ... 1 + 2 + 4 + 8 + 16 + ... The sum of the first 2 terms equals 3 and the 3rd term is 4 The sum of the first 3 terms equals 7 and the 4th term is 8 The sum of the first 4 terms equals 15 and the 5th term is 16 See the pattern? If you want to prove it I think induction is the best way.

We see that $$2^0 = 1=2^1-1\\2^0+2^1=1+2 = 3 = 2^2-1\\2^0+2^1+2^2=1+2+4 = 7 = 2^3-1\\2^0+2^1+2^2+2^3=1+2+4+8 = 15 = 2^4-1$$ So it looks like the formula may be $$\sum_{r=0}^n 2^r=2^{n+1}-1$$

After referring to one of the answers we looked at last week as an example of induction, he continued:

What we want to prove is that the sum of the first n terms is the n+1 term's value minus one. Written mathematically we are trying to prove: n ----- \ / 2^r = 2^(n+1)-1 ----- r=0 Induction hasthree steps: 1) Prove it's true for one value. 2) Prove it's true for the next value. The way we do step 2 isassume it's true for some arbitrary value(in this case k). We know it's true for one arbitrary value of k because of the base case. So if we canprove it's true for k+1, then we have proven it for all arbitrary values.

Three steps, or two? It’s all the same thing; what he wrote as step 2 is commonly thought of as what he will call (below) steps 2 (stating the induction hypothesis) and 3 (proving it for the next case).

Step 1:Base casen=0 n ----- \ / 2^r = 2^0 = 1 = 2^1 - 1 TRUE ----- r=0 Step 2:Assumeit's true for n=k k ----- \ / 2^r = 2^(k+1)-1 ASSUMED ----- r=0 Step 3:Try to prove for n=k+1k+1 ----- \ / 2^r = 2^((k+1)+1)-1 ----- r=0 k ----- \ / 2^r + 2^(k+1) = 2^(k+2)-1 ----- r=0

Here, as we’ve seen before, he is stating the goal; we need to show that these two sides are equal, using the hypothesis.

At this point we cansubstitute our assumption. If we prove this is true, that validates the assumption and all is fine. 2^(k+1)-1 + 2^(k+1) = 2^(k+2)-1 2*2^(k+1)-1 = 2^(k+2)-1 2^((k+1)+1)-1 = 2^(k+2)-1 2^(k+2)-1 = 2^(k+2)-1 TRUE The assumption is true and the proof therefore is true.

He changed the LHS while rewriting the RHS each time. Let’s write it out as one chain of equalities:

We assume $$\sum_{r=0}^k 2^r=2^{k+1}-1$$ and want to show that $$\sum_{r=0}^{k+1} 2^r=2^{k+2}-1$$ And here is the proof: $$\sum_{r=0}^{k+1} 2^r=\sum_{r=0}^{k} 2^r+2^{k+1}=(2^{k+1}-1)+2^{k+1}=2\cdot 2^{k+1}-1=2^{k+2}-1$$

Next week, we’ll move on to the subject of the Fibonacci series, which is a playground for inductive proofs!

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The first question is from 1998:

The Sum of the First N Squares I don't know where to begin. I have toformulate a formulafor thesum of the first N squares(i.e. 1^2 + 2^2 + 3^2 + ... + N^2)by mathematical induction.

Doctor Sonya answered, first clarifying the problem:

Hi, Rob: I'm not quite sure what exactly you are supposed to do in this problem. To "formulate" means tofinda formula. You never use mathematical induction to find a formula, only toprovewhether or not a formula you've found is actually true. Therefore I'll assume that you want tofind a formulafor the sum of the first n squares,and then provethat the formula is rightusing mathematical induction.

This is an important distinction to understand: Induction is used to prove that a formula you may have just guessed, is indeed correct. Induction, in fact, often seems unsatisfying because it doesn’t give even a hint as to how the thing being proved could have been discovered. So this assignment is not well worded.

Next, a quick overview of what we saw last week:

First, let me give youa brief account of how "mathematical induction" works. a)When do we use "mathematical induction"?Well, we sometimes encounter such problems where the behavior of a series of numbers or expressions SEEMS to change in a fixed manner. Here, I emphasize "SEEMS" because we do not know for sure if a certain formula works or not. For example, after looking at some numbers, I mightguessthat the sum of the first n squares can be found by the formula: (1/2)(n^2 + 14). However, I cannot say that this formula works foreveryn, because I really didn't try it with EVERY n, just with some of them. In such cases that involve natural numbers (1,2,3...), mathematical induction is a way to find a proof without having to spend eternity plugging values of n into the formula.

That “guess”, by the way, is intentionally wrong, and you can easily see that by trying a number of two. Disproving a guess is often very easy; proving it takes a lot more work!

b)What are the steps of a typical "mathematical induction"?1) To show that when n = 1, the formula is true. 2) Assuming that the formula is true when n = k. 3) Then show that when n = k+1, the formula is also true. According to the previous two steps, we can say that for all n greater than or equal to 1, the formula has been proven true.

Now, you may find the above explanation too abstract. So, let's just take your question as an example. We want to find the formula for 1^2+2^2+3^2+...+n^2. Now, I'm going to assume that you have already thought a lot about this problem, and you have a pretty good guess that the sum of the first n squares is: n (n + 1) (2*n + 1) ------------------ 6 (N.B. 2*n means "2 times n") However, we still don't know if this formula is true for all n, because we never tried it with n = 1,235,822 or n = 677,331 or millions of others. That's where proof by induction comes in:it will allow us to say that this formula it true for ALL n without having to test them all.

You would never come up with that formula just by random guessing, would you? For some ways you might discover this formula, see

Sum of Consecutive Squares Formula for Sum of First N squares

Now, we're ready for the three steps. 1. Whenn = 1, the sum of the first n squares is 1^2 = 1. Using the formula we've guessed at, we can plug in n = 1 and get: 1(1+1)(2*1+1)/6 = 1 So, when n = 1, the formula is true.

This step is usually pretty easy.

2. Now we'llassumethat when n = k (for some k that's greater than or equal to 1), the formula is also true. What this means is that we're just going to declare that our formula is true for k. 1^2 + 2^2 + 3^2 + ... + k^2 = k(k+1)(2*k+1)/6

This step takes no real work; it’s just a matter of replacing the general variable *n* with a specific *k* so we can make an assumption. (You can carry out the work without renaming the variable, but you might forget that you haven’t yet proved the general claim!)

3. Now we're going to try to prove,using the assumption from step 2, that the formula is true for n = k + 1. The series we need to compute is 1^2 + 2^2 + 3^2 + ... + k^2 + (k+1)^2. Since we already made an assumption about the sum of the first n squares (step 2), we canreplace the part before (k+1)^2by k*(k+1)*(2*k+1), and get the following:1^2 + 2^2 + 3^2 + ... + k^2+ (k+1)^2 =k(k+1)(2*k+1)/6+ (k+1)^2 Now we work on the right part of the equation to try to make it into our formula for n = k+1. I would like to leave this part of the work to you. Just remember that our formula says that the sum of the first k+1 squares is: (k+1)*((k+1)+1)*(2*(k+1)+1)/6)

Since Rob has had over 20 years to work on it, we can go ahead and do it now. I like to start this step by writing out what our **goal** is: to show that $$\frac{k(k+1)(2k+1}{6} + (k+1)^2 = \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$$ It will help if we simplify the RHS (right-hand side) a bit: $$\frac{k(k+1)(2k+1}{6} + (k+1)^2 = \frac{(k+1)(k+2)(2k+3)}{6}$$

Keep in mind that we don’t **know** this yet; it is what we want to **prove**!

How can we do that? We can either transform the LHS (left-hand side) into the RHS step by step; or we can just simplify both sides and show that they are equal to the same thing. Looking at the simplified version above, I see \((k + 1)\) in every term, so I’m going to keep those while combining the two terms on the LHS: $$\frac{k(k+1)(2k+1)}{6} + (k+1)^2 = \\ \frac{k(k+1)(2k+1)+6(k+1)^2}{6} = \\ \frac{(k+1)\left[k(2k+1)+6(k+1)\right]}{6} = \\ \frac{(k+1)\left[2k^2+7k+6\right]}{6} = \\ \frac{(k+1)(k+2)(2k+3)}{6}$$

My last step was to factor \(2k^2+7k+6\), but that was easy because I kept my eye on the goal and just had to *check* that it was equal to what I expected, \((k+2)(2k+3)\)!

4. Now, say what you've proved.

We’ve shown that the formula holds for *n* = 1, and that if it holds for any *k*, then it also holds for *k* + 1. Therefore, we have proved that for any positive integer *n*, $$1^2+2^2+3^2+\cdots +n^2 = \frac{n(n+1)(2n+1}{6}$$

Next, we have a question from 1996, where a pattern has been observed, and we want to show that it is “real”:

Sum of n Odd Numbers Dear Dr. Math: I am in the 9th grade, and taking trig. I was thinking about patterns in numbers one day andI came up with a peculiar law. It gives you every perfect, real, integral, square! (1,4,9,16,25.....) This is how it works:Start with 1. Now think of the set of odd numbers.Add the first odd numberthat isn't 1.(3). what do you get? 4....a perfect square! nowadd the next odd numberto 4. What do you get? (4+5=9) A perfect square. Keep adding the odd numbers to the perfect squares. (9+7=16...16+9=25...25+11=36...36+13=49...etc. etc. Now, to me, this is a simple pattern. I'm no genius, so someone must've thought of this law. Right? Your mission - find out who? I'd appreciate any help you could give.

If we keep adding consecutive odd numbers, we get a perfect square; this implies a formula for the sum of odd numbers. It is actually so well known that no name is attached to it; it is a special case of the general concept of an arithmetic series.

Doctor Luis answered:

You're very perceptive! To have noticed this simple pattern. Indeed, this pattern works as follows: 1 + 3 + 5 + ... + (2n-1) = n^2 That is, thesum of all odd numbers, up to the odd number (2n-1) is n^2. To illustrate this, think of the following example: What is 1 + 3 + 5 + ... + 997 + 999 ? Well, 999 is of the form 2(500)-1, so n, in this case, is 500, so the sum of all odd numbers (from 1) up to 999 is 500^2 or 250,000.

The first step in proving a claim is to state it as clearly as possible. Here, we might have looked at the examples and seen, say, that we get the square of 7 when the last number added was 13, which is 2(7) – 1.

The above theoremcan be proven quite easily by a method called induction, which is a very powerful technique used in mathematics to prove statements about the natural numbers. Since by now I probably have you interested, I'll explain a tad more about induction, and prove a basic relation involving, again, the natural numbers: Induction (a variation of it, at least) involvesthree stepsin proving a statement. The first one is called theBasis Step, and it just involves the assertion that the given statement is true for the natural number 1. The second step involves the assumption that the statement is true for all natural numbers less than or equal to some integer k, and, not surprisingly, is called theHypothesis Step. The third step is theInductive Step, and it involves proving that: if the statement is true for the integer k, then it is true for the integer k+1. This step usually comprises the bulk of inductive proofs.

As always, a good example clarifies a general concept. You’ll observe that Doctor Luis will, as we like to do, offering a *different* example to work through, so that our anonymous asker can enjoy doing his own. He will prove the simplest arithmetic series, one that is very well known (called the triangular number formula):

Just in case you got confused whilst trying to read what I wrote :), I'll prove the following relation (by induction, of course):1 + 2 + 3 + 4 + 5 + ... + n = n(n+1)/2Proof: Let S(n) be the sum of all integers from 1 to n, inclusive. So the statement to prove (I'll label it 'a') is that S(n) = n(n+1)/2 (a)

As we saw last week, it can be helpful to label the *n*th statement symbolically; here, the label is not for the *n*th **proposition**, but for the *n*th **sum**. This will make it easier to write about the next one.

Now that I know what it is I'm trying to prove, I'll follow the steps I told you about above.[Basis step]S(1) is clearly 1 (by definition). Let's see if statement (a) is true: S(1) = 1(1+1)/2 = 2/2 = 1 (so the Basis Step holds)

Some readers might be curious about the fact that S(1) means “the sum of natural numbers from 1 through 1 is 1”! The notation “\(1 + 2 + 3 + 4 + 5 + … + n\)” doesn’t mean that *n* is greater than 5, but is illustrative of “the sum of the first *n* natural numbers”. Mathematicians get used to treating 1 just like any other number, and not worrying about grammatical oddities (“the first *one* natural numbers”??). But the base case is generally special like that.

[Hypothesis]Assume that the statement S(n) = n(n+1)/2 is truefor the natural numbers n = 1,2,...,k.

Until now we have assumed that our proposition is true for *n* = *k*, not for **all numbers up to k**. What Doctor Luis is stating here is technically called “

**Weak** induction says, “If it worked last time, it will work this time;” **strong** induction says, “If it’s *always* worked so far, it will work this time.”

Weak induction is represented well by the **domino analogy**, where each is knocked over by the one before it; strong induction is represented well by the **stair analogy**, where each step is supported by *all* the steps below it.

The two forms are equivalent: Anything that can be proved by strong induction can also be proved by weak induction; it just may take extra work. We’ll see a couple applications of strong induction when we look at the Fibonacci sequence, though there are also many other problems where it is useful.

[Inductive Step]This is the tricky part of the proof. Here, you need to prove that if the statement in the hypothesis is true for the integer k then it is true for the integer k+1. I'll proceed as follows: By hypothesis, S(k) = k(k+1)/2 Now, I'll add (k+1) to both sides, and obtain S(k) + (k+1) = k(k+1)/2 + (k+1) = k(k+1)/2 + 2(k+1)/2 = [k(k+1) + 2(k+1)]/2 = [(k+1)(k+2)]/2

This is the same kind of reasoning we used for the sum of squares, and in fact for any summation formula; the only difference is that the goal wasn’t stated first, so it may have been harder to see why we are doing what we do, and that we have, in fact, finished the work! Once again, we wanted a factored form, so the common factor of \((k+1)\) was useful.

Ok. So far, we have S(k) + (k+1) = (k+1)(k+2)/2 Remember that S(n) is, by definition, the sum of all integers from 1 to n. Now, look at the left side of the equation. Therefore S(k) + (k+1) is simply S(k+1) ! Thus, S(k+1) = (k+1)(k+2)/2 which is what you obtain if you substitute n by (k+1) in statement (a)! What I have just proven is the Inductive Step, and this completes the proof.

This is the part I like to do ahead of time, to set up my target.

I have tried my best to explain the proof thoroughly so that you may be able to prove for yourself the interesting pattern you discovered. Hope this helped :)

I, too, will leave it to you to prove the formula for the sum of odd numbers. You’ll find that the algebra is far easier than what we just did!

This question is also from 1999:

Proof by Mathematical Induction I must prove the following statement by mathematical induction: For any integer n greater than or equal to 1,x^n - y^n is divisible by x-ywhere x and y are any integers with x not equal to y. I am confused as to how to approach this problem. Reading the examples in my textbook have not helped explain divisibility. Can you shed some light on this and get me going in the right direction?

What is to be proved here could be thought of as factoring of polynomials, as we know for example that \(x^3-y^3=(x-y)(x^2+xy+y^2)\); but it is stated of numbers, as for example if \(x=5\) and \(y=3\), then \(x^4-y^4=625-81=544\) is divisible by \(x-y=5-3=2\).

Doctor Marykim answered, starting with a proof of divisibility by a fixed number:

Hi James, Since you are not familiar with divisibility proofs by induction, I will begin witha simple example. The main point to note with divisibility induction is that the objective is to get a factor of the divisor out of the expression. As you know, induction is a three-step proof:Prove 4^n + 14 is divisible by 6Step 1. When n = 1: 4 + 14 = 18 = 6 * 3 Therefore true for n = 1, thebasisfor induction.

It is assumed that *n* is to be any positive integer. The base case is just to show that \(4^1+14=18\) is divisible by 6, and we showed that by exhibiting it as the product of 6 and an integer.

Step 2. Assume true for n = k i.e.: 4^k + 14 = 6 * A, where A is a positive integer. Rearranging this we get: 4^k = 6A - 14 Now consider n = k+1: 4^(k+1) + 14 = 4*(4^k) + 14 = 4(6A-14) + 14 (from our assumption) = 24A - 56 + 14 = 24A - 42 = 6(4A-7) Therefore, if true for n = k, then true for n = k+1

Doctor Marykim is taking the 3 steps a little differently than others, taking the second to include the inductive step proper, and step 3 to be the statement of the conclusion. What she has done here is to use the assumption, in the form \(4^k=6A-14\), to show that the next case, \(4^{k+1}+14\), is also a multiple of 6 by rewriting it and factoring out 6.

Step 3. However, the statement was proved true for n = 1 (Step 1) So the statement is true for n = 1+1 = 2 (step 2) Thus the statement is true for n = 2+1, 3+1, ... (iterating step 2) Thereforethe statement is true for all positive integral n

This explains how induction works, as a chain of implications.

That was provided as an example, focusing on what it takes to prove divisibility. Now James gets to try it himself:

Nowback to your original question: Prove (x^n) - (y^n) is divisible by x - y Step 1. This is a trivial step, so I'll leave it up to you. Step 2. Assume true for n = k, i.e.: x^k - y^k = (x-y)A, where A is a positive integer. If you follow the same steps as in the example above, you should be able toget a common factor of x-y in your final expression. Step 3. This will be exactly the same as in the other example.

Let’s do it. Keep in mind that *x* and *y* are just arbitrary integers; induction involves *n*, which can be any positive integer.

**Base case:** For *n* = 1, we are claiming that \(x^1-y^1=x-y\) is divisible by \(x-y\), which is obvious, as any number is divisible by itself. But to practice how we prove such things, the proof is $$x-y=1(x-y) \text{ for any }x\text{ and }y $$

**Induction hypothesis:** We *assume* that \(x^k – y^k = A(x-y)\) for some integer *A*, for all integers *x* and *y*. Our *goal* is to show that \(x^{k+1} – y^{k+1} = B(x-y)\) for some integer *B*.

**Inductive step:** We need to factor out \((x-y)\) from \(x^{k+1} – y^{k+1} = x\cdot x^k-y\cdot y^k\). One useful trick for factoring (equivalent to long division) is to insert a pair of equal terms in the middle, chosen so that we can factor by grouping: $$x\cdot x^k-y\cdot y^k = \\x\cdot x^k\boldsymbol{-y\cdot x^k+y\cdot x^k}-y\cdot y^k =\\(x-y)x^k+y(x^k-y^k) = \\(x-y)x^k+yA(x-y) = \\(x-y)(x^k+yA)$$ That does the job, since \(B = (x^k+yA)\) is an integer.

**Conclusion:** We have shown that \(x^n – y^n\) is divisible by \((x-y)\) for all integers *x* and *y*, when \(n=1\), and for any subsequent value of *n*, and therefore for all positive integers *n*.

Limits can be challenging. They can be even more challenging when they require L’Hôpital’s rule or more advanced methods (Maclaurin series), and then are turned inside-out by asking not for the limit itself, but for parameters that will result in a specified limit, or what values of the limit are possible. Two of us helped with such problems in November.

Here is the easier of two questions Vignesh sent us 20 minutes apart:

Hello

I have tried to solve the problem but at its first step it has become a complicated equation. Help me to solve it if I am in correct way or else guide me the perfect way to solve it.

NOTE: Multicorrect question.

The simplification is indeed going in an unpleasant direction. (Or is it? Hold that thought.)

Doctor Fenton answered, suggesting the use of Maclaurin series (the approach, as we’ll see, that Vignesh had already used in the other problem), *without* first combining the fractions:

Hi Vignesh,

I would recommend expanding the two terms in the parentheses into

Maclaurin seriesand combining them into a single series for the difference, which then becomes the numerator of a fraction with denominator x^{3}.Then the only way that fraction can have a finite limit as x→0 is for

the constant term and the coefficients of x and xall to be^{2}in the series0. That will give you equations which determine a and b. With values for a and b, you can determine the limit.

Vignesh replied, clearly very familiar with the technique:

Hello Doctor Fenton

With your suggestion I expanded the two terms in the parentheses and got the value of ‘a’ and ‘b’ which made me to complete #57, also I have done #58 but no option matched with it. Once go through the sum, if I had done a mistake please tell me.

Series expansions can be a powerful method for dealing with limits like this. Let’s look carefully at what he is doing.

The first fraction expands using the following Maclaurin series, for which Vignesh is using a formula, while I’ll use it as given in Wikipedia:

$$(1+x)^{-\frac{1}{2}}=1-\frac{1}{2}x+\frac{3}{8}x^2-\frac{5}{16}x^3+\cdots$$

This is a special case of the binomial series. The second fraction can be expanded using the following basic form, which can be found by mere long division (or by working backward from the geometric series):

$$\frac{1}{1-x}=1+x+x^2+x^3+\cdots$$

Applying this to \(\displaystyle\frac{1+ax}{1+bx}=1+(a-b)x\frac{1}{1+bx}\), by replacing \(x\) with \(-bx\), we get

$$1+(a-b)x\left[1+(-bx)+(-bx)^2+(-bx)^3+\cdots\right] =\\ 1+(a-b)x\left(1-bx+b^2x^2-b^3x^3+\cdots\right) =\\ 1+(a-b)x-(a-b)bx^2+(a-b)b^2x^3-(a-b)b^3x^4+\cdots$$

Subtracting those two series and combining like terms, we get $$\left[1-\frac{1}{2}x+\frac{3}{8}x^2-\frac{5}{16}x^3+\cdots\right]-\left[1+(a-b)x-(a-b)bx^2+(a-b)b^2x^3+\cdots\right] =\\ \left(-\frac{1}{2}-a+b\right)x+\left(\frac{3}{8}+(a-b)b\right)x^2+\left(-\frac{5}{16}-(a-b)b^2\right)x^3+\cdots$$

Dividing this by \(x^3\), we are left with $$\left(-\frac{1}{2}-a+b\right)x^{-2}+\left(\frac{3}{8}+b(a-b)\right)x^{-1}+\left(-\frac{5}{16}-(a-b)b^2\right)+\cdots$$

The omitted terms are all multiplied by positive powers of *x*, which will go to 0 so they can be ignored. The terms with negative powers of *x* go to infinity, so their coefficients must be 0; and the constant term will be the limit. This gives us two equations in *a* and *b*, $$-\frac{1}{2}-a+b=0\\\frac{3}{8}+b(a-b)=0$$ Solving the first for \(a=b-\frac{1}{2}\) and putting that into the second equation, we get $$\frac{3}{8}+b(-\frac{1}{2})=0$$ so that \(b=\frac{3}{4}\), from which \(a=\frac{1}{4}\). Vignesh is correct here, and therefore is right that \(a+b=1\).

The limit, then, is $$-\frac{5}{16}-(a-b)b^2=-\frac{5}{16}-(\frac{1}{4}-\frac{3}{4})\left(\frac{3}{4}\right)^2=-\frac{1}{32}$$

If you compare my work on the series with Vignesh’s, you will notice a sign error, which explains the wrong answer. (One sign error in all this work is impressive!)

Doctor Fenton pointed out the one error:

I have a different sign for the b

^{2}(b-a) term in the coefficient of x^{3}. That makes the limit -1/32 instead of -19/32.

Vignesh made the correction:

I made a mistake, instead of taking negative sign I took positive sign. Now I solved it, once check it.

Now the limit is \(-\frac{1}{32}\), and the requested ratio is 24, which is on the list.

Doctor Fenton:

That’s what I got. Good work!

After I finished writing this post, I decided to check this answer by plugging in the values for *a* and *b* and finding the limit directly. I found that I could do so merely by using the complement … which is just what Vignesh had done in his initial work. Could we actually finish that work? Yes!

Here is the last line of his original attempt (with the denominator backed up one step to keep it a little simpler): $$\frac{(1+bx)^2-(1+ax)^2(1+x)}{x^3\sqrt{1+x}(1+bx)(1+bx+(1+ax)\sqrt{1+x})}$$

Expanding and simplifying the numerator, we get $$\frac{x(-a^2x^2+b^2x-a^2x-2ax+2b-2a-1)}{x^3\sqrt{1+x}(1+bx)(1+bx+(1+ax)\sqrt{1+x})}$$

The factors in the denominator apart from the first approach 2; so we need the numerator to cancel with the \(x^3\). We will have a limit, therefore, if \(-a^2x^2+b^2x-a^2x-2ax+2b-2a-1\) is a constant multiple of \(x^2\). That will be true only if the coefficients of the linear and constant terms are zero: that is, if $$b^2-a^2-2a=0\\2b-2a-1=0$$

What do you think is the solution to that? Yes, \(a=\frac{1}{4}\) and \(b=\frac{3}{4}\).

And what is the limit? $$\lim_{x\to 0}\frac{x(-a^2x^2)}{x^3\sqrt{1+x}(1+bx)(1+bx+(1+ax)\sqrt{1+x})}=\\\lim_{x\to 0}\frac{-a^2}{\sqrt{1+x}(1+bx)(1+bx+(1+ax)\sqrt{1+x})}=-\frac{1}{32}$$

So series were not really needed! But that method was enlightening, and leads us to the other problem …

Vignesh actually sent the next problem 20 minutes before the other; it turns out to be considerably trickier:

Hello

I have started solving the problem and I think that I have done 90% solution but at last the equation had become quite complicated so that I am unable to finish it. Please help me.

NOTE: Multicorrect question.

Vignesh has used Maclaurin series (before Doctor Fenton had suggested it). Here are the basic series expansions he used this time:

$$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots$$

$$\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots$$

$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$$

In each case, something has to be substituted for *x*:

$$\sin(x^2)=(x^2)-\frac{(x^2)^3}{3!}+\frac{(x^2)^5}{5!}+\cdots=x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!}+\cdots$$

$$\cos(bx)=1-\frac{(bx)^2}{2!}+\frac{(bx)^4}{4!}+\cdots=1-\frac{b^2x^2}{2!}+\frac{b^4x^4}{4!}+\cdots$$

$$e^{ax}=1+(ax)+\frac{(ax)^2}{2!}+\frac{(ax)^3}{3!}+\cdots=1+ax+\frac{a^2x^2}{2!}+\frac{a^3x^3}{3!}+\cdots$$

Only enough terms have to be written so that, after cancelling powers of *x*, remaining terms will go to zero. Using terms through \(x^4\) as Vignesh has done (but being a little more consistent), the numerator becomes $$\sin(x^2)+2\cos(bx)-ax^4-2=\\ \left[x^2+\cdots\right]+2\left[1-\frac{b^2x^2}{2!}+\frac{b^4x^4}{4!}+\cdots\right]-ax^4-2=\\x^2+2-b^2x^2+\frac{b^4x^4}{12}-ax^4-2+\cdots=\\ (1-b^2)x^2+\left(\frac{b^4}{12}-a\right)x^4+\cdots$$

and the denominator becomes $$e^{ax}-1-ax-2x^2-\frac{a^3x^3}{6} =\\ \left[1+ax+\frac{a^2x^2}{2!}+\frac{a^3x^3}{3!}+\frac{a^4x^4}{4!}+\cdots\right]-1-ax-2x^2-\frac{a^3x^3}{6} = \\ \frac{a^2x^2}{2}-2x^2+\frac{a^4x^4}{24}+\cdots = \\ x^2\left(\frac{a^2}{2}-2+\frac{a^4}{24}x^2\right)+\cdots$$

Canceling \(x^2\) leaves us with the fraction $$\frac{(1-b^2)+\left(\frac{b^4}{12}-a\right)x^2+\cdots}{\frac{a^2}{2}-2+\frac{a^4}{24}x^2+\cdots}$$

When *x* goes to zero, we are left with the limit $$\frac{(1-b^2)}{\frac{a^2}{2}-2}$$ You will observe a small difference in what I have done here. But at this point I hadn’t yet duplicated his work in that way; instead, I had taken a different approach in order to check his work independently. I’ll show that in a moment.

I answered this one, while Doctor Fenton was working on the other problem:

As I understand it, “I” means the set of integers (commonly called Z), so that a and b must be

integers(positive or negative).When you say the problem is “multicorrect”, I presume you mean that

more than one answer can be correct, so you need to choose all choices in the list that are correct.The word “the” in “

thepossible values” seems to imply that it is asking forall possible valuesof a + b, and likewise for #60? But if we take #60 this way, then it would be saying that #59 is impossible, as 25/8 would not be a possible limit. It must instead be asking merelywhich of the four valuesare possible limits.I applied L’Hopital’s rule and obtained your equation (2(1-b^2)/(a^2-4) = 25/8. I didn’t find that b^4 = 12a.

Now we have a

Diophantine equationwhich is equivalent to 25a^2 + 16b^2 = 116. This can be solved rather easily; but the solutions turn out to force you to take a further step.

L’Hôpital’s rule is closely related to the use of series expansions; it amounts to using only the first term or so. The limit we are to find is

$$\lim_{x\to 0}\frac{\sin(x^2)+2\cos(bx)-ax^4-2}{e^{ax}-1-ax-2x^2-\frac{a^3x^3}{6}}$$

This has the form 0/0, so we take derivatives of the numerator and the denominator, and get

$$\lim_{x\to 0}\frac{2x\cos(x^2)-2b\sin(bx)-4ax^3}{ae^{ax}-a-4x-\frac{a^3x^2}{2}}$$

This still has the form 0/0, so we differentiate again, and get

$$\lim_{x\to 0}\frac{2\cos(x^2)-4x^2\sin(x^2)-2b^2\cos(bx)-12ax^2}{a^2e^{ax}-4-a^3x}$$

Setting *x* to 0, this becomes $$\frac{2-0-2b^2-0}{a^2-4-0} = \frac{2(1-b^2)}{a^2-4}$$ which is what Vignesh got. So this has to equal \(\frac{25}{8}\). And we can cross-multiply the equation

$$\frac{2(1-b^2)}{a^2-4} = \frac{25}{8}$$

to get

$$2(1-b^2)\cdot 8=25\cdot (a^2-4)$$

which simplifies to $$16-16b^2=25a^2-100\\ 25a^2+16b^2=116$$

But what about Vignesh’s \(\frac{b^4}{12}-a = 0\)? That comes from his second term in the numerator, which he is thinking has to be zero in order to have a limit; that would be appropriate if that term had degree *less than* 2. But since, after canceling \(x^2\), that will still be multiplied by \(x^2\), it does not have to be zero. By using L’Hôpital’s rule instead of series expansion, I never even got to see that, so I wasn’t tempted to misread it! So in fact, we don’t have a separate constraint imposed by the existence of a limit.

Vignesh replied, carrying out my suggestion to solve the Diophantine equation (that is, find all integer solutions):

I solved #59 with obtained Diophantine equation once check it. What step I have to do to solve #60?

This is good. I think I myself just saw immediately that \(a^2=4, b^2=1\) would work, and no other positive solutions for the squares are possible. We now have four solutions, each of which yields one of the four options given for the sum \(a + b\), so all of them appear to be correct. But …

I pointed out that there was more to do:

I mentioned

a further stepyou need to take; that starts withchecking. What happens when you put your values for a and b into your equation (2(1-b^2)/(a^2-4) = 25/8 ?I haven’t yet looked at #60, but will do so now.

Vignesh replied,

Yes, when I put values of a and b in the equation I am getting ‘0’.

Does it mean the values of a and b are wrong?

Yes, we get \(\frac{0}{0} = \frac{25}{8}\), which doesn’t really work.

I answered,

That means it is still

indeterminate, sowe can’t be sure yet what the limit will be.I think you need to apply L’Hopital yet again to see under what further conditions, if any, the limit will be correct. (I started that, but didn’t yet get to the conclusion.)

Or there may be a better way.

That was the end of the discussion, for some reason, and I didn’t get to go further at the time. Let’s see if we can finish the problem now.

Equivalent to applying L’Hôpital’s rule a third time as I suggested, Vignesh could just use more terms of the series expansion. Let’s do it both ways, starting with mine:

Differentiating a third time yields

$$\lim_{x\to 0}\frac{-12x\sin(x^2)-8x^3\cos(x^2)+2b^3\sin(bx)-24ax}{a^3e^{ax}-a^3}$$

which again has the form 0/0. (Note that we are assuming *a* and *b* are solutions of the equation above, so that we could apply L’Hôpital again!) So we differentiate a fourth time:

$$\lim_{x\to 0}\frac{-12\sin(x^2)-48x^2\cos(x^2)+16x^4\sin(x^2)+2b^4\cos(bx)-24a}{a^4e^{ax}}$$

This time we don’t get zeros; the limit becomes \(\displaystyle\frac{2b^4-24a}{a^4}\). We want this to equal \(\frac{25}{8}\), so we try our four possibilities, \((2, 1), (2, -1), (-2, -1), (-2, -1)\). We find that \(\frac{2b^4-24a}{a^4} = \frac{-23}{8},\frac{-23}{8},\frac{25}{8},\frac{25}{8}\) respectively. So the solution is \(a=-2,b=\pm 1\), and the answer to the problem is that \(a+b=-1\text{ or }-3\). Only two of the four apparent possibilities are real.

Using series, the fact that I had to apply L’Hôpital twice more suggests we need two more terms for each expansion (through \(x^4\)), which happens to be what I did above (where Vignesh had gone that far only in the numerator): $$\frac{(1-b^2)+\left(\frac{b^4}{12}-a\right)x^2+\cdots}{\frac{a^2}{2}-2+\frac{a^4}{24}x^2+\cdots}$$

Letting \(a=\pm2\) and \(b=\pm1\), this becomes (with signs corresponding to \(a=+2\) and \(a=-2\) respectively) $$\lim_{x\to 0}\frac{0+\left(\frac{1}{12}\mp 2\right)x^2+\cdots}{0+\frac{2}{3}x^2+\cdots} = \frac{\frac{1}{12}\mp 2}{\frac{2}{3}} = \frac{1}{8}\mp 3 = -\frac{23}{8}\text{ or }\frac{25}{8}$$

So we need to take the negative sign for *a*, and the solution as before is \(a=-2,b=\pm 1\).

How about problem 60, which asked which of the values \(\frac{1}{2}\), \(\frac{1}{8}\), \(-\frac{4}{3}\), and \(-\frac{23}{8}\) are possible? Well, we’ve seen that last one, haven’t we? That’s the limit when \(a = +2\). (We found that while we were trying to make the limit \(\frac{25}{8}\), but it *was* a valid limit – one of two possible when the limit was still indeterminate after the first part of our work.)

But we need to go back to the expression we found for the limit:

$$L = \frac{2(1-b^2)}{a^2-4}$$

What are the possible values of *L* when the parameters are integers? Fortunately, we don’t need to find all of them (and probably couldn’t); we just need to try determining values of *a* and *b* that will work for each choice we are given for *L*.

Let’s start with \(L=\frac{1}{2}\). In solving the Diophantine equation $$\frac{2(1-b^2)}{a^2-4} = \frac{1}{2}$$ we find that it simplifies to $$a^2+4b^2=8$$ The only solutions are those we have seen before, \((2, 1), (2, -1), (-2, -1), (-2, -1)\), none of which yield \(L=\frac{1}{2}\). So this choice is out. Similarly, \(L=\frac{1}{8}\) also fails.

But when we solve $$\frac{2(1-b^2)}{a^2-4} = -\frac{4}{3}$$ we get the equation $$2a^2-3b^2=5$$ which in addition to the familiar solution, is also true for \(a=\pm 4,b=\pm 3\), or \(a=\pm 16,b=\pm 13\). So this, too, is a possible limit.

The answer is that, of the four choices offered, \(L=-\frac{4}{3}\) or \(L=-\frac{23}{8}\).

I think this problem was one of the more interesting ones I’ve seen, and provides some useful experience comparing the series and L’Hôpital approaches to a limit.

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We’ll start with this basic question from 1998:

Explaining Mathematical Induction I have to do a report and the math article involves mathematical induction. I have not yet learned this and wanted to knowwhat mathematical induction is.

Doctor Barrus answered:

Hi, Kate. Mathematical induction is one way mathematicians prove things. What it is, basically, is this: Let's say I wanted to prove something about numbers [positive integers]. Step 1: First I would show that this statement istrue for the number 1. Step 2: Next, I would show thatifthe statement is true for one number,then it's true for the next number.

This is often described as *three* steps: (1) proving the **base case**, (2) stating the **inductive hypothesis**, and (3) the **inductive step**, proving the next case from that assumption.

These two steps complete a proof. It's kind of like dominoes. If you wantto knock down all the dominoesyou have, they all have to be lined up in the right way (so that they'llknock each other over), and you have to be able toknock the first domino down(so that all the others will fall). Well, showing that your statement istrue for 1is like making sure you can knock down the first domino. Proving that if your statement istrue for one number, then it's true for the next, is like making sure all the dominoes are lined up correctly, meaning that when one falls down, it will knock down the next one. When you put these things together - everything's lined up and you can knock down the first one - you've proved that you can knock down all the dominoes, or that your statement is true for all positive whole numbers. I hope that's understandable. But basically, mathematical induction is just a way of proving things about numbers.

So, we have an infinite set of claimed facts all lined up so that each one implies the next one is true. If we start the process by proving the first one is true, then all of them must be true, because each implies the next.

It’s easier to understand the idea with an example. Let’s look at a very simple one from 1997:

Proof by Induction Prove by induction on n that |A^n|=|A|^n. I'm not sure how to do this problem at all.

Doctor Sydney answered:

I'm glad you wrote. Induction can be confusing at first, but once you get the hang of it, it will be okay. You will probably remember how to do proofs by induction if you have a good understanding of how the proofs work. Let's quickly review this. Proofs by induction are commonly used when you want to prove a statement that depends on some variable (usually named n) forall positive integer valuesof that variable. For instance, in your problem you want to prove the above equality for all positive integer values of n. In other words, you want to show that the above equality holds forn = 1, n = 2, n = 3, and so on.

Induction is not applicable when the variable is a real number. But it’s possible for *n* to start at 0 or any other particular integer.

How does the inductive proof do this?First, it demonstrates that the statement holds for the "base case" - usually the n = 1 case.Next, when doing an inductive proof, youassume that the statement holds for the kth case. What is the kth case, you may be wondering? It isanycase. We write k because we want k to be able to represent any positive integer.Next, youprove that the (k+1)st case holds assuming that the kth case holds. Then the proof is done! Why? Well, think about it. In the first part we showed that the statement we are trying to prove holds for n = 1, right? Then, in the second part of the proof, we demonstrate that if the kth case holds, so does the (k+1)st case. Combining these two pieces of information, we see that since the 1st case holds (as proven in the first part of the proof), the 2nd case must hold too (according to the second part of the proof); well, if the 2nd case holds (as just demonstrated), so too must the third case hold, right? And so on. Thus we have shown that the statement holds forany n.

You’ll note that Doctor Sydney has the three steps I mentioned, and also that the inductive hypothesis step (stating that the *k*th case is true) is not merely a restatement of the problem, because the problem to be proved is that a statement about *n* is true for **all** *n*, while this is about any **specific** number *k*.

This will all become much more clear if we do an example. Let's look at your example.To show that |A|^n = |A^n| for all nby induction, we follow the steps outlined above. 1. Show that the statement holds for thebase casen = 1. When n = 1, the equation is |A|^1 = |A^1|, right? Does this equation hold? Yes, because anything to the first power is itself, so |A|^1 = |A| and |A^1| = |A|. Hence, |A|^1 = |A^1|. Thus, we have shown that the equality holds for n = 1.

It wasn’t explicitly stated what “all *n*” means; we could take it to mean any **non-negative** integer *n*, in which case the base case is *n* = 0. That, too, is easily proved: \(|A|^0 = 1\) and \(|A^0| = 1\)

2. Now, weassumethat the equality holds for the kth case. That is, we assume that|A|^k = |A^k|. 3. We now want toprovethat the (k+1)st case holds. We want to prove that|A|^(k+1) = |A^(k+1)|(**). Let's use what we know. In inductive proofs, proving that the (k+1)st case holds almost always relies on the fact that we have assumed that the kth case holds. So, let'srewrite the equation for the (k+1)st casein a way that will allow us to use information from the kth case. We know that |A|^(k+1) = |A|^k * |A|, right? Similarly, we have that |A^(k+1)| = |A^k * A| = |A^k| * |A|. Substituting into (**), we see that we want to prove that: |A|^k * |A| = |A^k| * |A| I bet you can prove that the equation above holds. I'll leave the last steps to you. Once you have proved that the above equation holds, you will have proven that the (k+1)st case holds, and thus, the proof will be over.

The proof is, of course, simple: Since we have assumed for this step that \(|A|^k = |A^k|\), we can just replace \(|A|^k\) with \(|A^k|\).

Again, the reason the proof works is as follows: First, we prove that the equation holds for n = 1. Next, we prove that if it holds for the kth case then it must also hold for the (k+1)st case. Since the equation holds for n = 1, it must also hold for n = 2 (the second part of the proof shows that if the kth case holds then so must the (k+1)st. Here, k = 1). But if the equation holds for n = 2, it must also hold for n = 3, and so on. Does that make sense to you?

The proof has set up these dominoes, and given the first one a push:

This proof, of course, was just for practice; we don’t need induction to prove such an obvious fact. But most such proofs are considerably more interesting, as we’ll see next week.

A good way to understand a concept better is to try to defend it against attack. Here is a question from 1998, presumably reflecting a statement the student was assigned to respond to:

Mathematical Induction "Proof by induction does not prove anything, because in the inductive step, one makes the assumption that P(k) is true. Since one has to do this for each k,one has assumed what was to be provedin the first place, and so the proof is invalid." What is wrong with the above argument?

In a proof, you can’t just assume that you are right. So why is it okay to assume the *k*th case is true?

Doctor Allan answered:

Here is what goes wrong in the argument: When you generally prove something using induction, you firstprovethat the theorem is true for (typically) n = 1. This is called the basis of the proof. Afterwards you prove thatifit is true for P(k)thenit is true for P(k+1). As the argument states, it is true that you assume P(k) in order to prove P(k+1), butyou can do that because you have proventhe statement for P(1). So the argument forgets the basis of the proof: that you prove P(1) to start with.

We can’t just assume that any one case is true; but we are not doing that. We are proving one case, and then proving the other cases follow from it.

Let me give you an example: Say I want to prove that n = n+1 for all n, and I assume P(k) as in the argument. Then k = k+1. I need to prove that k+1 = k+2 (the statement P(k+1)). But k+1 = k + 1 = k+1 + 1 = k+2 (here the second equality comes from the fact that we have assumed P(k). This is nonsense, and it only works becauseI have forgotten to prove the basis: that 1 = 1+1 = 2, and of course I cannot do that.

In this example, we have proved the last (inductive) step, but never proved a base case, so the inductive step has no ground to stand on. It may be true that B follows from A, but if A is not true, then we have accomplished nothing. But if A is true, then we will know that B is true.

So as you seeit is quite easy to prove weird things using inductionif you forget the basis of the proof, and that is exactly the case with the argument you mention. If you prove P(1) and thereafter prove 'if P(k) then P(k+1)' then proof by induction is a completely sound method of proof.

There are several classic fallacious “proofs” using induction. One, which is referred to in the *Ask Dr. Math* FAQ page on False Proofs, claims to prove that all people in Canada are the same age. This one is subtle, because rather than being due to failure of the base case, it comes from the chain being broken elsewhere.

Here is a long question from 2012 making a stronger argument against induction, which I’ll take in pieces:

Inductive Misunderstanding I have a question about mathematical induction. I'm sure my reasoning is out of line somewhere, but I can't seem to get past thinking that it is not useful in the sense of being a proof. Would you be so kind as to show me the hole in my thinking? :) I have had mathematical inductionexplained to me two separate ways(one of which must be erroneous). Given A1, Ar, and Ar+1: a.) One mustASSUME that Ar holds true. If Ar is true, and the base case -- A1 -- is also true, and we illustrate that if Ar holds true for Ar+1, we may assume that Ar is proven true and that it holds true for all integers. b.) One mustKNOW that Ar holds true. If Ar is true, and the base case -- A1 -- is also true, and we illustrate that Ar holds true for Ar+1, we may assume that Ar is proven true and that it holds true for all integers.

Neither of these is worded very well. In his notation, \(A_r\) is the general proposition to be proved for any value of variable *r*; \(A_1\) is the base case, where the variable *r* is 1, \(A_{r+1}\) is the next case. The statement “\(A_r\) holds true for \(A_{r+1}\)” suggests that Brandon is a little confused about the notation. Subsequently he will use “*n*” and “*a*” in place of *r* without explanation. So a major goal will be to clarify what the symbols mean.

It might help to look at an example. Let's suppose we have the example of an arithmetic series. Let us follow method b. Ar = n(n + 1)/2 Proving for 1 (base case): 1(1 + 1)/2 = 1 Ar+1 = (a + 1)(a + 2)/2 (I cheated, but if you do the math, it works). Usingmethod b, we have, at least to my understanding, proven Ar to be true. However,it was known that Ar was true beforehand, so what is the sense in proving it again? That suggests that mathematical induction (at least in the form of method b) isuseless. It proves only what has already been proven to be true!

What Brandon means, I believe, is that we are trying to prove that for any positive integer *n*, the sum \(1 + 2 + 3 + … + n = \frac{n(n+1)}{2}\), so the *r*th case would claim that \(1 + 2 + 3 + … + r = \frac{r(r+1)}{2}\), and the *r*+1st case would claim that \(1 + 2 + 3 + … + r + (r+1) = \frac{(r+1)(r+2)}{2}\).

Brandon is using \(A_r\) to represent both a **statement** that can be true, and an **expression** that has a numerical value. He is also failing to distinguish between *all n* and a *specific r*. These confusions are probably both common among students!

His issue is that if we *know* already that \(A_r\) is true, then there is nothing to prove. His other idea is that we merely *assume* that \(A_r\) is true, which is no better:

Conversely, if one thinks aboutmethod a, it is equally as useless. Given method a, let's suppose the following: Ar = 1 such that Ar is any integer, including an integer larger than 1, such as 2. Remember,we're assuming Ar to be true, so we can theoretically do this. 1 = 1 (base case) Ar + 1 = 1 + 1 Recall that Ar = 2, and therefore, Ar+1 also is equal to one. That insinuates thatwe can assume anything at all to be true, and we can prove it, regardless of whether or not it is in alignment with reality. That leaves us only with method b as the valid use of mathematical induction, and we mentioned before that it is useless. I just can't seem to see wrap my mind around how mathematical induction could be useful, nor do I see the error in my thinking.

Doctor Rick answered, first clarifying the notation, carefully describing what induction means:

Hi, Brandon, Both descriptions of mathematical induction you present are flawed.Many students miss the same critical elements of the methodthat are missing in your descriptions. Let's say we want to prove that a certain proposition is true for all integer values of n greater than zero.I'll call the proposition P(n): it says that something is truefor a *particular* value of n. We must first prove that the proposition to be proved is true for the least value of n, which is 1 in my example (as is commonly the case). That is, weprove that P(1) is true. That's the "base case." Now, weassume that P(n) is true--for some *particular* value of n. On this assumption, we prove that a *different* proposition is true -- namely, P(n + 1). We aren't assuming what we want to prove; we'reproving that one proposition, P(n + 1), follows from the truth of another, P(n).

What we have is a sequence of propositions, and our goal is to prove that they are all true. Since each is a separate proposition, we may assume one of them true in proving the next.

This second part of the proof provides sort ofa template for an infinite set of proofs, an infinite line of reasoning, as follows: First, we have proved that P(1) is true. Then, using the template, we can prove that, given that P(1) is true (and we know it is), therefore P(2) must be true. The logic holds, so now we know that P(2) is true. Next, again using the template, we can prove that, given that P(2) is true (and we know it is), therefore P(3) must be true. Thus P(3) has been established as true. The path has been laid for proving in the same way that P(4) is true, and P(5) is true, and so on, up to any finite integer k for which you wish to prove P(k). Knowing HOW to prove that P(k) is true, we don't have to do it all the way out to P(1000000) or whatever; we know it's true.

You might think of our proof as writing a computer program that will carry out the 1,000,000 little proofs needed to get to P(1,000,000). Once we write the program, we know it can, in principle, keep proving as far as we choose.

We aren'tassuming what we want to prove, orproving what we already know; you're right, the first would be a logical fallacy, and the second would be unnecessary. Rather, we're proving that IF one thing is true, THEN the NEXT thing must be true.We set one step of an infinite staircase in place, and provide a blueprint showinghow to build any step while standing on the step below it. Following the blueprint repeatedly, we can build the infinite staircase one step at a time. That's what mathematical induction is. Does that make sense?

We might think of the base case as the floor, which we establish on its own; then each step is supported by the step below it. The blueprint here corresponds to the program I suggested earlier. Here is a new image for an inductive proof:

Brandon replied:

Thanks so much for the response! That was certainly not something I could have chewed through by myself! You have saved me a great deal of grief. I'm very bad at abandoning problems when I don't have any leads, and this one ended up being a little time sucker.

Next week, we’ll look at some deeper examples of inductive proof.

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We got this question in 1999:

A Fibonacci Jigsaw Puzzle For a homework problem in math, our geometry teacher showed us a problem dealing with the area of a square and making it equal to a rectangle. Thearea of the square was 64but by the time we got to the rectangle (using Area formula A = b(h)) it equaled65. The question iswhere did the mistake occur? He had an 8*8 square, which he cut into 2 pieces; the areas of these were 24 and 40. Then he cut the rectangular 24 piece into a right triangle, and the 40 piece into a trapezoid. Each time both pieces were equal. (There were2 equal right trianglesand2 equal trapezoids.) Each trapezoid and triangle was then matched, making a proportionally bigger right triangle. These two triangles were then put together to make a rectangle with a width of 13 and height of 5. This area then equaled 65, but it is physically impossible for this to happen. What am I missing? The figure looked like this: _ _ _ _ _ a/e _ line ac | | | entire area =65cm^2 | 40 | 24 | A_ _ _ _ _ _ _ _ _ _ _ _ _ D |f | | | | | | | | | | | | c | | e| f| | | | | | | | | | | | | | | | | - - - - - - - - - - - - - - - - - - - - - c/f B C Does this make sense to you?

The picture is hard to decipher if you are not already familiar with the puzzle. I’ll redraw it to look like our puzzles from last week, starting with the goal, which looks like this:

Can we dissect the square and cover the rectangle with the pieces? Of course not; The areas are 64 and 65 respectively. But we’ve been told we can do it this way:

It looks good! But is it? (Hint: I cheated a little in my drawing.)

I answered:

Hi, Laik. My father-in-law made a copy of this puzzle out of wood, so I'm very familiar with it. Here's a site by Ron Knott that explains the puzzle, but doesn't quite give the actual answer: Harder Fibonacci Puzzles Look for "A Fibonacci Jigsaw puzzle or How to Prove 64=65!"

(I’ve updated the link, which has been moved slightly in the intervening 20 years. And the wooden puzzle has moved from my in-laws’ home to mine.)

The trick is that in the rectangle,the diagonal line isn't quite exact: calculate the slopes of the segments Ae and eC, and you'll find that they are not the same; AeCf is really avery narrow parallelogram with area 1. See if you can calculate the area and prove me right. This isrelated to the Fibonacci sequence, as explained by Ron Knott, because the ratios of successive Fibonacci numbers, in this case 5:8 and 8:13, are close, approaching thegolden ratioas you use larger numbers; that allows the slopes to fool you.

We’ll be looking at both slope and area calculations below.

The Fibonacci sequence is 1, 1, 2, 3, 5, 8, 13, … , where each number is the sum of the two before it. These numbers have many interesting properties. At the bottom of this post, I’ll include proofs of a more specific Fibonacci fact that applies to this puzzle. Observe, though, that \(\frac{5}{8} = 0.625\), while \(\frac{8}{13} = 0.61538…\). Close, but not quite. And the golden ratio is \(\phi=\frac{\sqrt{5}-1}{2} = 0.61803…\). (If you are familiar with \(\Phi = 1.618\dots\), that is the reciprocal of this lower case phi!)

We got a similar question in 2001, but without the actual pieces:

Fibonacci Riddle An 8x8 square has an area of 64. We can cut the square into four pieces and reassemble to get a 5x13 rectangle with an area of 65.Where does the extra 1x1 square come from?

Doctor Tom answered first:

The pieces don't fit EXACTLY. I don't know what version of this problem you're working on, but you'll probably find thatthe triangles are not perfectly aligned, but it's hard to see the error, since it's spread out over a long line.

The misalignment would involve the slopes we discussed.

Doctor Rob joined in 20 minutes later:

Thanks for writing to Ask Dr. Math, Daniel. If you check out the 5 x 13 rectangle, you will find that the pieces do not exactly cover the rectangle. There is avery narrow parallelogramwith two opposite corners at opposite corners of that rectangle whose area is 1 square unit. That is because theslopes of the diagonal edgesof the two kinds of pieces are not exactly the same, although they are close. o---+---+---+---+---o---+---+---o | | /| + + / + | | / | + + + | | / | o.__ + / + | `-.__ CD | / | + `-._ + AC + | AB `--._ | / | + `-o BD + | | / | + + / + | | / | + + + | |/ | o---+---+---+---+---o---+---+---o A o---+---+---+---+---+---+---+---o---+---+---+---+---o | `--.__ | | + `--.__ + + | `--.__ | | + o--._ + + | B | `--._ | C | + + `-o._ + | | `--._ | + + `--.__ + | | `--._ | o---+---+---+---+---o---+---+---+---+---+---+---+---o D

Like my drawing, this picture cheats a little, making ABCD look collinear:

The slope of AC is -3/8 = -0.375, and the slope of AB is -2/5 = -0.400. Thus <BAC is not zero, andB is slightly below line segment AC. Similarly, <BDC is not zero, andC is slightly above line segment BD. The narrow parallelogram is ABDC, whose area is 1 square unit.

In particular, the height of point B is 3, but the corresponding point on AC is \(5-\frac{3}{8}(5)=\frac{25}{8}=3\frac{1}{8}\), so the vertical error is \(\frac{1}{8}\).

Here are more accurate pictures, close up, to show that parallelogram:

The yellow 1×1 square has the same area as the yellow parallelogram in the middle of the rectangle, as hard as that is to believe! Here is an exaggerated view of it:

Here \(b = \frac{1}{8}\) and \(h=8\); moving ABE to CDF we keep the same area, which is equal to \(bh=1\).

Rather than slopes, we can also use lengths (via the Pythagorean theorem):

Another way to see this is that AB = CD = sqrt(5^2+2^2) = sqrt(29) = 5.385165..., BD = AC = sqrt(8^2+3^2) = sqrt(73) = 8.544004..., AD = sqrt(13^2+5^2) = sqrt(194) = 13.928388..., AB + BD = AC + CD, = 5.385165... + 8.544004..., = 13.929169..., > 13.928388..., = AD. This proves that neither B nor C lies on line segment AD.

If C were on segment AD, then we would expect that AC + CD = AD. That fact that it is greater implies, by the triangle inequality, that ACD is a triangle.

A similar but different puzzle was sent to us in 2002:

Magic Triangle Puzzle Hello, While surfing on the Internet I encountered a site with a triangle puzzle on it. The guy that put it on his siteclaims he has found a flaw in Euclidean geometry. You can view the problem here: Simeon's Triangle Puzzle It looks simple, but I just can't explain it. Can you tell me where that white square came from?

The site no longer exists, so I have removed the link. Several others have asked about the same puzzle by sending us a link to an image file called “bizare.gif” [sic] at various sites, and none of those still exist either. But the problem is mentioned on the R. Knott site referred to above, as Yet Another Fibonacci Jigsaw Puzzle. Here is the image from that page:

It appears that by rearranging the same four pieces they get an identical triangle, but with room for an extra square! Observing that the triangle has legs 8 and 13, which are divided up into parts of 2, 3, and 5, you might wonder if this, too, involves Fibonacci numbers. You’d be right! And the basic trick is similar. Can you spot it?

Doctor Nitrogen answered:

Hi, Pieter: I suspect the reason a blank white square appears in the bottom diagram is that in the top diagram, the green right triangle at the top and the red right triangle at the bottom arenot similar triangles. As a result, the area of the entire figure does not remain the same when you move the two triangles.

If the triangles are not similar, then their slopes will be different, and the top edge of each figure will not be a straight line. The thickness of the black line contributes to covering this up; it’s drawn as a straight line but doesn’t actually pass through the vertices of the triangles.

Go back to the image and examine for yourself. This refers to the top diagram: The acute angle at the bottom left of the topgreen triangleis tan(beta) = 2/5, and the acute angle at the bottom left of the largerred triangleat the bottom is tan(alpha) = 3/8.

These tangents of angles could just as well be described as slopes.

Put another way, the two smaller sides of the smaller green triangle are 2:5, but the ratio of the sides of the larger red one is 3:8. Clearly, 2:5 =/= 3:8, andthe red and green triangles are not similar. Why should this matter? Because if the red and green right triangles were similar, you could place the green one at the bottom left and the red one at the top right, and leave no blank white square when you moved the orange figure. But the triangles are not similar, so when you place the green one in the lower position and the red one in the higher position, you increase the total area of the larger triangle that contains the red and green triangle and the other orange figure.

Because the slope of the green triangle, \(\frac{2}{5} = 0.4\), is greater than that of the red triangle, \(\frac{3}{8} = 0.375\), the lower image has a “hump”, raising the “hypotenuse” enough to make room for the extra square, even though it doesn’t look like enough.

In fact, Doctor Ian suggests that for the two triangles to actually act as you see on that site, the big triangle that contains both of them (as well as the orange figure) would have to have a curved,concave-shaped hypotenusein one case and a curved,convex-shaped hypotenusein the other. I did not work out a formal proof for this; nevertheless, that's why I suspect the blank white square mysteriously appears in the bottom diagram. I hope this got you thinking on some intriguing math.

Here are accurate versions of the two pictures; the dotted lines are the true hypotenuse:

Another Math Doctor has suggested cutting the two figures, carefully drawn, from graph paper, and putting one on top of the other to compare more easily.

Doctor Ian simultaneously added his own ideas, filling in the proof:

Hi Pieter, Dr. Nitrogen's analysis is correct. The key is that the two triangles aren't similar. Here's how it looks to me. In the first triangle, moving from left to right, you move at one slope (3/8), and then switch to asteeperslope (2/5). This means that the 'hypotenuse' is not actually a straight line, but in fact is slightlyconcave. In the second triangle, the situation is reversed: you switch from a steeper slope to agentlerone, which makes the 'hypotenuse' slightlyconvex. In fact, then, neither of the 'triangles' is really a triangle at all, but a quadrilateral in which one of the angles is nearly 180 degrees. The thickness of the line is used to mask the change in slope; but the difference between the convex and concave 'hypotenuses' is the area of the white square in the bottom triangle. Does that make sense?

To quantify the error, in the bottom picture the top of the green triangle is at height 2; but the corresponding point on a straight hypotenuse, 5 units along at slope \(\frac{5}{13}\), would be \(5\cdot\frac{5}{13}=\frac{25}{13}=1\frac{12}{13}\), just \(\frac{1}{13}\) lower. So the “hypotenuse” bulges up by \(\frac{1}{13}\) over a width of 13 units, making a triangle with area \(\frac{1}{2}\) square unit.

Later that year, we got a comment from a reader, Michael:

Here is a numerical answer, easier to see for some kids: 13 * 5 / 2 =32.5 square units. That's the apparent size of both combined (whole) triangles. But if you calculate the areas of each of the sections, again as they appear on the puzzle, they add up to 32 square units on the top one, and 33 square units on the bottom (the one with the white square in it). This confirms that the 'slightly concave, slightly convex' analysis is right, and that the hypotenuses of both triangles are not straight lines; both are arcs, with a net area of one square unit.

So the area of the triangle we think we see is the average of the two non-triangles that are actually there; and the difference between the two is that one extra square.

Doctor Schwa replied:

Thanks! Several of the answers we link to are somewhat similar to yours, but surprisingly none of them explicitly does the area calculation. I'd add some of supporting details to your calculation: Top "triangle" has: one 8x3 triangle, area 12 two "P" shaped pieces, total area 15 one 5x2 triangle, area 5 for a total of 32. Bottom picture has those same four pieces, plus the one square unit hole, for a total of 33. Thus neither of them can actually fill the 5x13 triangle exactly.

Let’s take a little look at why Fibonacci numbers work so well in these puzzles, in addition to the fact mentioned earlier that those dimensions result in similar slopes.

In 2002 we got this variation on the question, which wasn’t archived:

This is a question from my assignment which I don't understand why? Takeany three consecutive Fibonacci numbers. Use the middle number as the sides of thesquareand the outside numbers as the sides of therectangle. Why is there adifference of onein the calculated values of the area of the square and rectangle, but when divided up into pieces right, the pieces from the square fit exactly into the rectangle? That is the question I don't understand some help would be great.

Sebastian has not just been shown one example of our first puzzle (using 5, 8, 13), but was told that the same thing can be done with *any* set of three sequential Fibonacci numbers (like 2, 3, 5, or 3, 5, 8, or 8, 13, 21) and do the same thing – and the areas of the rectangle and square will always differ by exactly 1. But why? Can it be proved?

Doctor Rick answered:

Do you need help with both parts of the question, or just one?You can understand why there is a difference of 1in the areas. First examine how it works in a few small cases: F[1] = 1, F[2] = 1, F[3] = 2 ==> 1^2 - 1*2 = -1 F[2] = 1, F[3] = 2, F[4] = 3 ==> 2^2 - 1*3 = 1 Notice that the difference is 1 in each case, but it isn't always the same area that is bigger! Once you see what is happening, you can write an inductive proof. For any k, assuming something about F[k]^2 - F[k-1]*F[k+1], you can prove something about F[k+1]^2 - F[k]*F[k+2]. Since the something is true for k=2, it is true for all kby induction.Why do the pieces from the square fit exactly into the rectangle?The answer is that they don't! If the pieces are cut the way I am picturing it, there is a diagonal cut. You'll find that the diagonal cuts are not quite parallel (how can you tell?), so thatthe pieces either overlap slightly, or they leave a narrow gap.

What we want to prove, in general, is $$F_{n-1}\cdot F_{n+1}-{F_n}^2=(-1)^n$$ That is, the difference alternates between \(+1\) and \(-1\). In our example, \(8^2=64\) is 1 less than \(5\times 13 = 65\); in the next larger size, \(13^2=169\) is 1 *more* than \(8\times 21 = 168\). We’ve had questions about the latter puzzle, too.

This fact was the subject of a question in 1999:

Fibonacci Trick I am interested inwhy the Fibonacci trick works. How do f(n-1), f(n+1), and f(n)^2 relate to each other? I am not a good math student at all but the Fibonacci sequence has really interested me. I have been working for a while on a way to solve this but I haven't gotten anywhere. Please help me! I would like to look at why this works and see if it helps to understand the ways the Fibonacci sequence shows up in nature, etc.

Doctor Anthony answered, using a known formula for the *n*th Fibonacci number in terms of the Golden Ratio φ:

For the Fibonacci sequence show: [F(n+1) * F(n-1)] - [F(n)]^2 = (-1)^n The expression for the nth term of the Fibonacci sequence is: 1 [1+sqrt(5)]^n 1 [1-sqrt(5)]^n F(n) = ------- * ------------- - ------- * ------------- sqrt(5) 2^n sqrt(5) 2^n I will write this in the form 1 F(n) = ------- * [P^n - Q^n] sqrt(5) 1+sqrt(5) 1-sqrt(5) where P = --------- and Q = --------- 2 2 Note that P+Q = 1 P-Q = sqrt(5) PQ = 1/4 - 5/4 = -1

The numbers \(P\) and \(-Q\) are, respectively, the numbers \(\Phi\) and \(\phi\) we mentioned above. We’ll be looking at some aspects of the Fibonacci sequence soon, if this mystifies you!

We are asked to show that [F(n+1) * F(n-1)] - [F(n)]^2 = (-1)^n [F(n)]^2 = (1/5)*[P^n - Q^n]^2 = (1/5)*[P^(2n) + Q^(2n) - 2*P^n*Q^n] ............(1) F(n+1) * F(n-1) = (1/5)*[P^(n+1) - Q^(n+1)]*[P^(n-1) - Q^(n-1)] = (1/5)*[P^(2n) + Q^(2n) - P^(n+1)*Q^(n-1) - P(n-1)*Q^(n+1)] ............(2)

We need to show that the difference between these is always \(\pm 1\).

subtracting (1) from (2) we get: (1/5)*[-P^(n+1)*Q^(n-1) - P^(n-1)*Q^(n+1) + 2*P^n*Q^n] = (1/5)*[-P^(n-1)*Q^(n-1)*(P^2 + Q^2) + 2*P^n*Q^n] = (1/5)*P^(n-1)*Q^(n-1)*[-(P^2 + Q^2) + 2*P*Q] = -(1/5)*P^(n-1)*Q^(n-1)*[P^2 - 2*P*Q + Q^2] = -(1/5)*P^(n-1)*Q^(n-1)*(P-Q)^2 but (P-Q)^2 = 5, so we obtain = -P^(n-1)*Q^(n-1) = -(PQ)^(n-1) but PQ = -1 and so the difference between (1) and (2) reduces to: = -(-1)^(n-1) = (-1)^n and this completes the proof.

The same fact was the subject of a question in 2001, without reference to the puzzle:

Fibonacci Identity I am trying to create aninductive prooffor the particular identity of Fibonacci numbers that: F(n-1) * F(n+1) = (-1)^n + (Fn)^2 I know I do abase caseof n being 0, and then aninductive stepof n being n+1, but beyond that I just can't get the math to work out. Is this a valid identity, and if so, how would I work out the math to inductively show this?

Doctor Anthony gave the requested inductive proof :

Well, I suppose you have been able to show thatthis works for n = 0. Now we suppose that F(n-1)*F(n+1) = (-1)^n + (F(n))^2 holds for n. We can rewrite this to F(n-1) * F(n+1) - (F(n))^2 = (-1)^n [*] Now we will replace all n by n+1 at the left-hand side of the equation, and show that it yields (-1)^(n+1). For that we use the truth we suppose of [*] as well as application of theFibonacci rule F(m+2) = F(m)+F(m+1)for all m: F(n)*F(n+2) - (F(n+1))^2 = F(n)*[F(n+1) + F(n)]- F(n+1)*[F(n) + F(n-1)]= F(n)*F(n+1) + (F(n))^2 - F(n)*F(n+1) - F(n+1)*F(n-1) = (F(n))^2 - F(n+1)*F(n-1) = -(-1)^n = (-1)^(n+1) And we get where we want to be.

We’ll be looking at inductive proofs soon. Here he has shown that if the claim is true for some given value of *n*, as it is for *n* = 0, then it is true for the next, and therefore for all subsequent values.

Our first question will provide an overview of possibilities:

World War II Window Blackout During the Second World War, families had to black out their windows. Mr. Brown had asquare window 120cm x 120cm, but the only material he could find was asheet of plywood 160cm x 90cm; same area, different shape. He drew some lines and cut out justtwo congruent shapes, which he joined to make a square of the correct size. How did he do it? I have tried drawing sketches, scaled drawings, and cardboard cutouts, but I can't solve it. I see no way of doing it. Please help!

We are to cut and rearrange the \(160\times 90 = 14,400 \text{ cm}^2\) red rectangle to exactly cover the \(120\times 120 = 14,400 \text{ cm}^2\) yellow rectangle:

Doctor Ian answered, first showing how one might experiment with various possibilities to see what happens:

My first thought was that this would end up being two L shapes, e.g., +-----------+ | | | B | +-----+ | | | | | | | | | | | | | | | | | +-----+ | A | | | +-----------+ but that doesn't quite work. You always end up leaving a hole in the middle.

This sort of dissection can be useful in some problem; in fact, we’ll see something of this sort later. But it fails, no matter how you choose dimensions:

Then I tried trapezoids, +-----------+ | | | B | + | | . | | . | | . | | . | | . | | + | A | | | +-----------+ but that doesn't quite work, either. You always end up leaving a hole in the corner.

For example,

But these ideas suggest a generalization, with more (and equal) steps. We’ll see a way to decide how many steps we need later, but trial and error can lead to this:

Here is the general idea: +-----------+ | | | B | +---+ | | | | | | | | +---+ | | | | | | | | +---+ | A | | | +-----------+ By moving part B to the right and then down, you can make a square: +---+-----------+ | | | | | B | | +---+ | | | | | | | | +---+ | | A | | | | | +-----------+---+ I'll leave it to you to work out the dimensions involved.

Here it is, with appropriate dimensions, dividing the height into 4 equal parts, and the width into 3:

I finally came up with it by forgetting about the initial rectangle, andfocusing on the final square. I knew I'd have to shift one of the pieces up and over to the left (do you see why?). I knew the base of A would be 90cm wide, which meant that the tip of B would have to be 30cm wide. The rest follows from the fact that the two pieces have to be congruent. This was a fun question to think about. Thanks for asking it!

It is very common to solve a problem by focusing on the goal, rather than what we have; or by working from both ends simultaneously. No piece can be wider than 90 cm to fit on the left, or higher than 120 cm to fit on the right; so we cut steps according to the difference between the piece dimensions and either whole.

Here is a similar question from 2004:

Using Dissections to Cut and Reshape Rectangles Cut a20x15 piece of paperinto two pieces to form a25x12 piece. I have tried and tried and tried and my parents have tried also but we can't figure it out. This may be a trick question but I can't decide what the trick is. Can you help?

Again, we want to cut the red rectangle in two pieces and rearrange them to cover the yellow one:

Doctor Douglas answered, not with a solution to the given problem, but with a similar example, leaving room for Marcia to do some problem solving of her own:

Hi Marcia. The trick is to use a stairstep-shape. Here's an example of a20 x 12 rectanglerearranged into a15 x 16 rectangle. 20 15 +----+----+----+----+ +----+----+----+ | | | | | + +----+ + + +----+ | | | 12 | | | + +----+ + + +----+ + 16 | | | | | | +----+----+----+----+ +----+ + | | +----+----+----+ Can you figure out how to adapt this for your problem? These types of problems are known as "dissections".

Here is that example:

Now, how can we decide how many steps to make, and also whether to change the orientation before reassembling? Under what conditions can this dissection be done successfully? Let’s examine this example closely before trying to solve our problem.

We see that we are multiplying the width by \(\frac{15}{20}=\frac{3\cdot 5}{4\cdot 5}=\frac{3}{4}\), and the height by \(\frac{16}{12}=\frac{4\cdot 4}{3\cdot 4}=\frac{4}{3}\). These are, of course, reciprocals or the areas would be different. We decrease the width by subtracting one step width of 5 (the difference of the two widths, which is also a common factor), and increase the height by adding one step height of 4 (the difference of the two heights, and again a common factor). In general, it would look like this:

In the problem we are trying to solve, the widths are 20 and 25, which differ by 5, which is a factor of both; so we’d like our step width to be 4 (so *w* = 5, *m* = 4). The heights are 15 and 12, which differ by 3, which is a factor of both; so we’d like our step height to be 3 (so *h* = 3, *n* = 5). Let’s do it:

What if we’d been asked to turn a \(20\times 15\) into a \(12\times 25\)? Then we’d be decreasing 20 to 12, a difference of 8 which is not a factor of either; so we’d have to turn the target into the orientation we actually used.

The next question is more complicated; we’ll see three different descriptions of how to discover it. This is from 2001 again:

Carpet Problem You have to carpet a 9x12 room, but when you go the store they only have a10x10 carpetand a1x8 piece of carpet. If you add up the number of square units, then 10x10 and 1x8 = 108 and 9x12 = 108, so you know that it can fit. However,you can only make one cutin the carpet (meaning either entering one side of the carpet and exiting any side, or entering one side of the carpet and cutting over yourself). Where would this cut be to make the carpet fit in a 9x12 room.?

Here is the problem:

This received two answers. We’ll start with Doctor Greenie’s, which is only a set of hints:

Hi, Nathan - I have seen this problem posed many times, and the instructions are virtually never clear. For example, in your presentation of the problem, I don't know what "cutting over yourself" means. Here are a couple of things I think you need in order to have a chance of solving this problem: (1)You can only cut one of the pieces of carpet, giving you a total of three pieces. Putting the two original pieces on top of each other to make the cut is not allowed, nor is folding the piece you cut so that the single cut gives you more than three pieces. (2) You can't solve the problem by making a single straight cut.It has to be some sort of fancy zigzag cut. Note that for the resulting pieces to exactly cover a rectangular floor, the cut will have to consist of segments all at right angles to each other. Those are a couple of hints that might help you find the solution; but I suspect that even with those hints only a tiny fraction of students would be able to do so. (I never figured out the problem for myself -I had to be shown how it is done...)

Although we generally like to give only hints, it’s hard to give a good enough hint that isn’t too good!

Doctor Rob had beat that answer by 3 minutes; he gave the actual solution:

Try this: o---+---+---+---+---+---+---+---+---+---o | | o---+---o + | | | + o---+---o + | | | + o---+---o + | | | + o---+---o + | | | + o---+---+---+---+---+---o + | | | + o---+---o + | | | + o---+---o + | | | + o---+---o + | | | + o---+---o | | o---+---+---+---+---+---+---+---+---+---o Slide the upper right piece two feet right and then one foot down, and see what happens.

Here is what happens (slipped side to side following the next answer we’ll see):

But back in 1996, we had had the same question (which Doctor Greenie referred to as a last resort):

Cutting Carpet Two pieces of carpet are to be used to cover a floor. The two rectangular pieces are 1 ft by 8 ft, and 10 ft by 10 ft. You are allowed tomake just one cut in one of the two pieces. The three pieces of carpet must then be arranged to fit a 9 ft by 12 ft floor exactly. Make a sketch that shows how this can be done.

Doctor Ceeks answered, describing the solution in words rather than try to draw it with text:

Because there is no way to fit the 10 by 10 piece into a 9 by 12 rectangle, it is the 10 by 10 piece which must be cut somehow. Here's how: I'll start by describing a walk starting at the lower left corner of the 10 by 10 foot square piece. up 1 foot. (no cut needed for this part) right 2 feet. up 1 foot. right 2 feet. up 1 foot. right 2 feet. up 1 foot. right 2 feet. up 1 foot. left 6 feet. up 1 foot right 2 feet. up 1 foot. right 2 feet. up 1 foot. right 2 feet. up 1 foot. right 2 feet. Here, you should have cut the square into two pieces, a left piece and a right piece. (you come out 1 foot below the upper right corner.) Now, slide the right piece over 2 feet and slide it up 1 foot. You will see a 9 by 12 foot rectangle with a 1 by 8 hole in the exact center!

What’s most important for us is his description of how he found the solution:

(I arrived at this solution by hoping for a symmetric solution...so I put the 1 by 8 strip in the exact middle of the 9 X 12 rectangle. Since I was hoping to deal only with integral length cuts, putting the 1 x 8 strip in as in the solution was the way to go.)

So this was the starting point, around which the solution crystalized:

The rest would be the same zigzag concept we’ve seen before. Symmetry is often a good start!

We’ll close with one final question, from 2001 again:

Carpet and Room Areas A man buys a roll ofcarpet 9 ft. wide and 12 ft. longto fit a room10 ft. x 10 ft.When the carpet is unrolled, arectangular hole 1 ft. wide and 8 ft. longis discovered in the exact middle of the carpet. The man decides that since the area of the roll is 108 sq.ft. and the area of the hole is 8 sq. ft., the carpet will still fit in the 100 sq. ft. room. How can he cut the roll of carpet with the hole in the middle of it into to two pieces that will fit into the 100 sq. ft. room when the two pieces are put together? I don't know where to start. Please help me.

You may see how this is just a reworking of that last problem! It’s just reversed. In fact, what we start with this time amounts to the seed of an idea that Doctor Ceeks used:

Doctor Rob answered, using a series of simpler problems to work up to the answer we’ve already seen:

Thanks for writing to Ask Dr. Math, Matt. Let's start with a smaller case: a 4-by-4 room and a 3-by-6 carpet, with a 1-by-2 hole in the middle: o---+---+---+---+---+---o | | + o---+---o + | | | | + o---+---o + | | o---+---+---+---+---+---o One should cut the carpet thus: o---+---o---+---+---+---o | | | + o---+---o + | | | | + o---+---o + | | | o---+---+---+---o---+---o Then move the right-hand piece 1 foot up and 2 feet to the left (to make the result 1 foot taller and 2 feet narrower), and you'll have o---+---+---+---o | | o---+---o + | | | + o + | | | + o---+---o | | o---+---+---+---o Ta-daaaa!

Now the next larger case, a 6-by-6 room and a 5-by-8 carpet with a 1-by-4 hole in the middle: o---+---+---+---+---+---+---+---o | | + + | | + o---+---+---+---o + | | | | + o---+---+---+---o + | | + + | | o---+---+---+---+---+---+---+---o The cuts should look like this: o---+---o---+---+---+---+---+---o | | | + o---+---o + | | | + o---+---o---+---o + | | | | + o---+---o---+---o + | | | + o---+---o + | | | o---+---+---+---+---+---o---+---o Then move the right-hand piece 1 foot up and 2 feet to the left (to make the result 1 foot taller and 2 feet narrower), and you'll have o---+---+---+---+---+---o | | o---+---+ + | | | + o---+---o + | | | + o---+---o + | | | + o---+---o + | | | + o---+---o | | o---+---+---+---+---+---o

Now you try the same ideas on the 10-by-10 room and 9-by-12 carpet with a 1-by-8 hole in the middle.

Of course, we’ve seen the solution already!

]]>We usually see limits applied to functions in a calculus class. An interesting question from late October deals with a limit in a geometrical construction based on a function. We’ll be seeing how to discover a proof, then several alternative proofs, and finally what the answer means.

Here is the initial question:

Hello, I am currently in my college holidays and I have decided to do some maths to improve. My weakness is graphing and I am hoping to get some help or the solution on this question.

Question:

Let P(k, k^2) be a point on the parabola y = x^2 with k > 0.

Let O denote the origin.

Let A(0, a) denote the y-intercept of the perpendicular bisector of the segment OP.What is the limit of

aas P approaches O?Thank you!!

The graph provided isn’t quite to scale (the two lines don’t look quite perpendicular), but you can try to imagine, as P moves toward O, what might happen to A. Would it just keep moving up toward infinity as AM gets steeper, or would it approach some particular point on the *y*-axis? Not obvious, is it?

Jalal had submitted the question under the category Geometry, though it involves an algebraic equation, and he specifically mentioned graphing; the limit concept is from calculus, and Jalal’s reported age range suggested the beginning of college. So I couldn’t be sure what approach to suggest.

Having explored the problem enough to find a solution, I answered:

Hi, Jalal.

This is an interesting problem, and is about a lot more than mere graphing!

Here’s how I’d start:

On the picture, label the

coordinatesof point P, and of M, the midpoint of OP. Then you might think about the slope of AM and how it relates to that of OP. There are several ways you can use this to find the value ofaas a function ofk; you couldwrite the equation of line AM, or you could usesimilar triangles, or theslope formula. How you do that is up to you; so I’d like to see what you can do with the problem in order to see how best to help.I’ll also want to know

what you have learned about limits, since that is a topic beyond geometry. (But the limit involved here is an easy one.)I look forward to hearing more from you!

We’ll eventually be looking at all three of the methods I suggested.

Jalal wrote back:

I haven’t learned much limits but I have learned about gradients, differentiating normal line and tangents.

Thank you

So he knew a little calculus; all we need of it here will be the mere concept of a limit.

I replied:

This doesn’t require any calculus, just analytic geometry, and in particular the relationship between slopes (gradients) of perpendicular lines. If you use the similar triangle approach I suggested as an alternative, you don’t even need that. (The limit is very simple.)

Here is the first step I suggested, with coordinates of P and M labeled:

Please show what work you have done, using any method, so I can see where you need help.

The coordinates of P come from the equation of the parabola; those of M come from the midpoint formula, averaging the coordinates of the endpoints.

Jalal now showed work:

I will show you what I have done before you sent me the last message.

Jalal has correctly found the slopes, but has not dealt with A yet. The last half finds the lengths of OP and then OM, which he won’t actually be using – but we’ll see it at the end when I show another method.

I answered:

I don’t think you need to find the distance OP. But you have the slope of AM and the coordinates of M; can you write an equation for the line AM?

Jalal took a slightly different approach than I had in mind, writing equations for *two* lines and intersecting them.

OP = kx

AM = (-1/k)x + a

kx = (-1/x)x + a

k

^{2}x – -x + ka

x(k^{2}+1) = kaSub k/2 into x and @ M, x = k/2k/2(k

^{2 }+ 1) – ka

k(k^{2 }+ 1) = 2ka

k^{2 }+ 1 = 2aTherefore, a = k

^{2 }+ 1/2

a = k^{2}/2 + 1/2k –> 0, OP–>0, AM –> infinity

Lim a = lim (1/2(1 + k

^{2})) = 1/2

k–>0 p–>0

The approach Jalal chose was to find the equations of line OP and of line AM (given its *y*-intercept A), and using the fact that that intersection is M. The answer was right, but the work was hard to follow and had a few errors in detail.

I approved, and fixed up minor details in the work:

You got it!

I’ll make a couple comments on your work, and then show my approach, which was a little different.

Your work:

OP = kx <— I think you meant that line OP has equation y = kx.

AM = (-1/k)x + a <— Similarly, line AM has equation y = (-1/k)x + a.

kx = (-1/k)x + a <— Finding the intersection, which should be point M; I’ve corrected a typo.

k

^{2}x = -x + ka <— You multiplied by k; I’ve corrected a typo.

x(k^{2}+1) = kaSub k/2 into x and @ M, x = k/2<— Now you use the fact that this point is M.k/2(k

^{2 }+ 1) = ka <— I’ve corrected a typo.

k(k^{2 }+ 1) = 2ka

k^{2 }+ 1 = 2aTherefore, a = (k

^{2 }+ 1)/2 <— You omitted these parentheses.

a = k^{2}/2 + 1/2 <— Good.k –> 0, OP –> 0, AM –> infinity <— Line OP approaches line y=0, and AM becomes vertical ??

Lim a = lim (1/2(1 + k

^{2})) = 1/2 <— Good.

k–>0 p–>0The part about OP–>0, AM –> infinity isn’t needed, fortunately. All you need is what you did at the end, that a approaches 1/2.

The only actual errors were mere typos.

I then showed the method I’d referred to as “using the slope formula”:

Here’s what I did, which is essentially the same thing turned inside-out:

Slope of OP is k, so slope of AM has to be the negative reciprocal, -1/k.

Slope of AM is Δy/Δx = (k

^{2}/2 – a)/(k/2) = k – 2a/k.Setting these equal, k – 2a/k = -1/k.

Multiplying by k, k

^{2}– 2a = -1.Solving for a, a = (k

^{2}+ 1)/2.As P approaches O, k approaches 0, a approaches 1/2.

Your work, with my little corrections, is as good as mine.

I hope you enjoyed that!

We both obtained the equation \(a=\frac{k^2+1}{2}\) for the *y*-intercept as a function of *k*. [Jalal, if you are reading this, I discovered a sign error in what I actually wrote to you, which I have fixed here! I hope that didn’t confuse you.]

In the course of the work, the derivation was invalid when \(k=0\), but valid otherwise; the equation we ended up (multiplying by *k*) with removes that discontinuity, so we just have to set \(a=0\) to get the answer. As a result, the concept of limit only shows up if you are reading carefully!

Let’s try a couple more ways to get the solution.

Probably when I suggested writing the equation of line AM, I had in mind using the point-slope method with point M. We know the slope \(-\frac{1}{k}\) and the point M \(\left(\frac{k}{2},\frac{k^2}{2}\right)\), so the point-slope form $$y-y_0=m(x-x_0)$$ becomes $$y-\frac{k^2}{2}=-\frac{1}{k}\left(x-\frac{k}{2}\right)$$ Putting this into slope-intercept form by expanding and solving for *y*, we get $$y=-\frac{1}{k}x+\frac{1}{2}+\frac{k^2}{2}$$ Therefore the *y*-intercept is $$a=\frac{1}{2}+\frac{k^2}{2}$$

This is the same formula we got above, and the limit is again \(\frac{1}{2}\).

The yellow (AMO) and green (OXP) triangles here are similar, because each side is perpendicular to the corresponding side of the other.

This gives us the proportion $$\frac{\text{OA}}{\text{OM}}=\frac{\text{PO}}{\text{PX}}$$ so that $$\frac{a}{\frac{1}{2}\sqrt{k^2+k^4}}=\frac{\sqrt{k^2+k^4}}{k^2}$$ Solving for *a*, $$a=\frac{k^2+k^4}{2k^2}=\frac{1+k^2}{2}$$ which again is the same formula.

Let’s take a look at what’s happening. Here is a graph of the problem showing P closer to A than in my image above; I have superimposed on the figure the graph (in green) of our function \(a=\frac{k^2+1)}{2}\), so you can see how it approaches \(\frac{1}{2}\):

For instance, the point B on the green curve directly above P is directly to the right of A. As P moves toward O, A moves down to \(\frac{1}{2}\), the vertex of the parabola \(y=\frac{x^2+1)}{2}\).

In the next picture,I have moved P even closer to O, and added a circle centered at the limit point \(\left(0,\frac{1}{2}\right)\). This is called the **osculating circle** to the parabola at the vertex:

The osculating circle touches the parabola as closely as possible for as long as possible (“osculate” is Latin for “kiss”); its radius is called the **radius of curvature** of the parabola, and its center, our limit point, is called the **center of curvature**. You can see why, with P nearly on this circle, the perpendicular would nearly pass through that center.

We’ll start with this question from 2001:

Five Equal Pieces of a Square Cake Ravina has a square birthday cake. Its side length is 20cm. She wishes to cut the cake intofive pieces, one for each of her four friends and one for herself. She wants to usestraight vertical cutsto make five pieces ofequal volume. Suppose that Ravina makes the first cut from the cake's center to a point from the top left corner. Where must she make the other four cuts ifthey all start for the cake's center? I haven't even started... I would not even know where to start. Please help me!

We’ll be seeing several variants of the problem, of which this is relatively simple: We are told how big the square is (but not the cake’s height), and want to cut it into 5 equal parts. But rather than make 4 cuts parallel to the side, as we would expect (especially if two people want most of the icing!), we are required to make wedges that meet at the center:

Because the cuts are made vertically, it doesn’t matter how tall the cake is; the volume of each piece is proportional to the area on top. The only question is, how do we find the right locations?

I answered briefly, just offering a hint:

Hi, Katie. A good place to start is to think about what theareawill be for anywedgecut from the cake, like this: +-----------+ | | | | | + | | / \ | | / \ | +--------+--+ |<---d-->| Once you've done that, think about the area of a wedge that goesaround a corner, like this: +-----------+ | | | | | + | | / \ | | / + -- +-----------+ ^ |<-------d-----+ You will find thatthe distance d along the outside is all you need to know, and using that you can find the correct value of d for which each piece is 1/5 of the cake. The answer is surprisingly simple!

The area of the triangular piece can be found by the formula \(A = \frac{1}{2}bh = \frac{1}{2}d\cdot 10 = 5d\). The area of a wrap-around piece can be thought of as the sum of two triangles. I’ll let you think about the rest for now.

Two months later, we got a similar question with an additional requirement:

Dividing a Square Cake into Five Equal Pieces How can you divide a square-topped cake that is a rectangular solid and isfrosted on all facesinto five pieces so that everyone receives thesame amount of cake and icing? All cuts must beperpendicular to the surface. I can not figure this out. All I can do is divide the cake four ways and eight ways. Please help me or tell me a way that I can try to solve this problem.

We still need vertical cuts, which prevents some trickery; but now we need to make both the volume (cake) and the amount of surface area (icing) the same for each piece. That seems like a much bigger challenge. Pam has presumably used symmetry to make 4 or 8 pieces that are all congruent, and therefore have the same volume and area:

But you can’t do that with five pieces, can you?

I answered again, with more or less the same hint:

Hi, Pam. This problem is easier to solve than you would think. I suggest you start by thinking about wedge-shape pieces like this: s +---------------+ | | | | | | | + | | / \ | | / \ | | / \ | +---+-------+---+ a Then you can extend it to wedges that wrap around a corner like this: s +---------------+ | | | | | | | + | | / \ | | / + | / |b +---+-----------+ a You can find this area by drawing the line from the center to the corner to divide it into two triangles, and finding the area of each triangle. Once you've done this, I think you will have a good idea what to do.

Let’s take it a little further this time. Here we weren’t given specific lengths, so a general description is all that is expected; but let’s keep the side lengths at 20 cm from the first problem. We saw that the area of a triangular piece is \(5a\), using the variable we have here; the area of the quadrilateral piece is \(5a + 5b = 5(a+b)\), still 5 times the amount of perimeter used by the piece (the *d* in my first answer). This is the key idea.

Now we want to divide the area of the top of the cake, \(20\times 20 = 400\text{ cm}^2\), into 5 equal parts, so each has to be \(80\text{ cm}^2\). Setting \(5d=80\), we find that the distance along the side has to be 16 cm. And since each piece will have the same distance along the perimeter, they will not only have the same area of icing on top, but also around the side.

At the end of that answer, I referred to the following answer, from 1999, which we can look at now:

Cutting a Square into Five Equal Pieces Hi Dr. Math, I was wondering how it is possible to divide a square cake intofive equal parts. There are some restrictions: you cannot divide the cake into ten parts and give two to each person, andit must be cut through the center point. ___________ | | | | | . | | | | | ----------- What I have done is these but they were rejected. ___________ ___________ | | | | | | | _|__ | | \ | / | |---| |--- & | \|/ | | |__| | | / \ | | | | | / \ | --------- --------- Can you help me please? Thank you, Doctor.

The first attempt doesn’t use the center; but the second is just what was described – if the measurements are right.

Note that we are back to a question that doesn’t worry about the icing, though the answer will.

Doctor Rob answered:

Thanks for writing to Ask Dr. Math. Your last drawing above contains the right idea. Measure the length of the sides of the square, which is thebase of the cake. Call that s. The perimeter of the square is then 4s. Divide that by 5. Now from any starting point (such as the center of the top edge in your picture),measure around the edge a total distance of 4s/5. Mark this point, and measure around the edge again a total distance of 4s/5. Repeat this until you return to the starting point. Now from the center, make cuts to all five points marked. That will divide the cake into five equal volume pieces, and furthermore,each piece will have the same amount of icing, too.

Here’s what this looks like when \(s=20\) as in the first question:

Here the perimeter is 80, so we moved \(\frac{1}{5}\) of that distance from one point to the next, that is, \(\frac{4}{5}\) of 20 = 16. I marked each side in tenths (2 cm).

So that’s the answer. But is it true?

To prove that thevolumes of the pieces are all the same, it is enough to prove that the areas of the four quadrilaterals and one triangle are all the same. Draw the diagonals of the square. That will cut the quadrilaterals into two triangles each. All the nine triangles will have height s/2, and their bases will be s/2, 3s/10, 7s/10, s/10, or 4s/5. Starting at the starting point and moving around the perimeter of the square, you'll see that s/2 + 3s/10 = 4s/5 7s/10 + s/10 = 4s/5 4s/5 = 4s/5 s/10 + 7s/10 = 4s/5 3s/10 + s/2 = 4s/5

In our example, the individual bases are, respectively, \(AB = 10 + 6\), \(BC = 14 + 2\), \(CD = 16\), \(DE = 2 + 14\), and \(EA = 6 + 10\), all equal to 16 cm.

Starting at one corner and moving around the perimeter of the square, you'll see that s/2 + s/2 = s 3s/10 + 7s/10 = s s/10 + 4s/5 + s/10 = s 7s/10 + 3s/10 = s

Each side adds up to \(s = 20\). Here we see the 5 pieces split into 9 triangles, along with the fact that they all have the same altitude.

For each triangle, the area is \(A = \frac{1}{2}bh = \frac{1}{2}b\cdot \frac{1}{2}s = \frac{1}{4}bs\).

Thus the triangle areas will be s^2/8, 3s^2/40, 7s^2/40, s^2/40, or s^2/5. When you add up the areas of the triangles to get the areas of the quadrilaterals, all quadrilaterals will have area s^2/5, or 1/5 of the total area of the square: s^2/8 + 3s^2/40 = s^2/5 7s^2/40 + s^2/40 = s^2/5 s^2/5 = s^2/5 s^2/40 + 7s^2/40 = s^2/5 3s^2/40 + s^2/8 = s^2/5 Now the volume is the area of any of these polygons times the height of the cake, soall volumesare equal to hs^2/5, andall icing areasare equal to (4h+s)s/5.

Each piece has a volume equal to its area times the height *h* of the cake, so they are all equal, \(\frac{1}{5}s^2\); and the icing on each piece is its top area plus the area of a rectangle with width \(\frac{4}{5}s\) and height *h*.

Notice that if you had started at, say, a corner point, the quadrilaterals would be different shapes, but the area of each would still be s^2/5, because the triangles forming them havecommon altitudes/2 and thesum of their basesis 4s/5, so the total area is A1 + A2 = ab1/2 + ab2/2 = a(b1+b2)/2 = (s/2)(4s/5)/2 = s^2/5 For the same reason,anystarting point could have been used.

Here is a cake with the first cut at a corner:

Now, it’s curious that all these questions were about five pieces; we’ve had a couple with different numbers, too. What if there were 13? Consider this question from Jim, in 1998:

Dividing A Cake - A Math Puzzle You have a nine-inch-square cake. It is a two-layer cake. Each layer is one and one-half inches high. Each layer is covered with 1/4-inch-thick frosting and so are the sides. There is no frosting on the bottom. Using straight knife cuts, how can you divide the cake into13 piecesso thateach piece has exactly the same amount of cake and frosting? After pieces are cut they may be put back together, but you may not remove any cake or frosting.

This is the most detailed version of the problem we have, with complete dimensions, and the requirement that both cake and frosting (also called icing) be divided equally. We don’t really need any of the dimensions, though we’ll see a little issue with the thickness of the frosting.

Doctor Wilkinson answered:

This is a very nice puzzle. In case you want to work on it a little longer on your own,I'll give you a couple of hints. If you get stuck or you just want the answer, let me know. 13 is too big; it's hard to visualize the problem with such a large number of pieces.Try some smaller number like 5(4 or 2 would be too easy).If the cake were circular instead of square, it would certainly be easy. You would just mark off points on the circumference that would divide it into 5 equal parts, and cut from the center of the circle to these points, right? Now see if you can do something similar with a square cake.

“Try a smaller number” is a standard piece of advice when you face a big problem. Can you see why 5 is the best choice?

“Try a simpler but similar problem” is also standard; thinking “what if it were round?” can lead to the idea of dividing the perimeter equally, though one might not at first think it could work.

Jim replied, wanting more than just a hint (without directly saying that he really needs more than the hint!):

Thank you very much for your response. These hints are very helpful. I think that I know how to go about solving the problem, but could you please send me an explained solution?

Doctor WIlkinson responded:

The hint I gave you was really a very good one. You just need tobe bold and follow it exactly. If youdivide the circumference of the cake into 13 equal partsand then make your cuts from the center of the cake to each of the 13 points on the circumference, all the pieces will have equal amounts of both cake and frosting! To see this, notice first that theamount of cakein a piece is just thearea of the topof the piecetimes the thickness of the cake. Theamount of frostingfrom the top is just thearea of the toptimes the thickness of the frosting, and the amount of frosting from the side is just thelength of the part of the piece on the edgeof the cake times the thickness of the cake times the thickness of the frosting. Now if you divide the circumference into 13 equal parts, you've taken care of the frosting on the side, and you just need to be sure that the pieces have equal area to take care of the cake and the frosting on the top.

There’s one little issue with the thickness of the frosting: we are cutting through the frosting at an angle, so the parts on the sides are not exactly rectangular, and therefore *not quite* proportional to their width. You can see the issue here:

Pieces closer to the center have a wider angle, making trapezoids with a little more area. But the standard puzzle assumes the icing is very thin (perhaps just a glaze), minimizing this problem!

There are two kinds of pieces of cake if you divide it as I suggested: triangular wedges, and pieces that go around a corner of the cake. For thetriangular pieces, the area is one half the base times the height. The height is just the distance from the center of the cake to the edge, which is the same for all the pieces, andthe base is one of those equal divisions of the circumference, so that's the same for all the triangular pieces also. I'll let you think about the case of the around-the-corner pieces, but they work too. Here's a picture:

We’ll close with a question about only 3 pieces; by now you know what to do, but what if you had to identify the angles for the cuts? This question is from 2001 again:

Dividing a Square in Thirds I want to take a square and divide it into three equal pieces using three lines radiating from the center of the square. I determined that each line has to have an angle of 120 degrees to complete a circle. The difficulty is in getting the areas of the square to be equal.

This one is not posed in terms of a cake, and doesn’t mention the equivalent of icing; Frank is apparently thinking of the fact that three sectors of a circle with central angles of 120° will be equal. What would the angles be for a square?

Doctor Rob answered, starting with the same idea we’ve already seen several times, and then calculating the angles. For simplicity, he uses a coordinate system and makes each side 2 units long:

Thanks for writing to Ask Dr. Math, Frank. There is more than one way to do this. Here are two: A(-1,1) G B(1,1) o-----------+-----------o | -_ . | | -_ . | | -_ . _,o E | -_ . _,-' | | -_. _,-' | H + - - - - - o - - - - - + I | /.O | | / . | | / . | | / . | | / . | o-----o-----+-----------o D(-1,-1) F J C(1,-1) Thefirst wayis to make one line from A to O. Then the area of AGO and AHO are both 1/2 square unit. That means the area of GBEO and HDFO both have to be 5/6 square unit to make a total of 4/3 square unit for ABEO and ADFO. That leaves OEI and OFJ with an area of 1/6 square unit, so E must have coordinates (1,1/3) and F(-1/3,-1). Then: <AOF = <EOA = arccos(-1/sqrt[5]) = 116.565 degrees <FOE = arccos(-3/5) = 126.870 degrees These angles can be found by using the Law of Cosines and the lengths of the sides of the triangles AOE and EOF.

Using the technique we’ve seen already, with a total perimeter of 8, we need \(AB + BE = \frac{8}{3}\), so \(BE = \frac{8}{3} – 2 = \frac{2}{3}\), which confirms that \(E = (1, \frac{1}{3}\).

Let’s find one of those angles using the Law of Cosines: The sides in triangle AOE are $$|OA|=\sqrt{1^1+1^2}=\sqrt{2}\\ |OE|=\sqrt{1^2+\left(\frac{1}{3}\right)^2}=\sqrt{\frac{10}{9}}=\frac{\sqrt{10}}{3}\\|AE|=\sqrt{2^2+\left(\frac{2}{3}\right)^2}=\sqrt{\frac{40}{9}}=\frac{2\sqrt{10}}{3}$$ By the Law of Cosines, $$|AE|^2=|OA|^2+|OE|^2-2|OA||OE|\cos{\alpha}\\\cos{\alpha}=\frac{|OA|^2+|OE|^2-|AE|^2}{2|OA||OE|}=\frac{\left(\sqrt{2}\right)^2+\left(\frac{\sqrt{10}}{3}\right)^2-\left(\frac{2\sqrt{10}}{3}\right)^2}{2\sqrt{2}\frac{\sqrt{10}}{3}}\\=\frac{2+\frac{10}{9}-\frac{40}{9}}{\frac{4\sqrt{5}}{3}}=\frac{-12}{12\sqrt{5}}=-\frac{1}{\sqrt{5}}$$

The rest, of course, are similar.

Here is an accurate picture (sides marked in thirds):

Starting vertically, we get different angles:

A(-1,1) G B(1,1) o-----------+-----------o | . | | . | F o._ . _,o E | `-._ . _,-' | | `-._ . _,-' | H + - - - - - o - - - - - + I | |O | | | | | | | | | | | | | o-----------+-----------o D(-1,-1) J C(1,-1) Thesecond wayis to put one line from J to O. Then the areas of OIE and OHF must be 1/3 unit, so the coordinates of E and F are E(1,2/3) and F(-1,2/3). Then the angles are: <JOE = <FOJ = arccos(-2/sqrt[13]) = 123.690 degrees <EOF = arccos(-5/13) = 112.620 degrees These angles can be found by using the Law of Cosines and the lengths of the sides of the triangles JOE and EOF.

Here is the picture:

It’s easier, of course, just to measure around the perimeter; but it’s worth seeing how the angles differ from the 120° of a circle.

]]>A question from last month provides an opportunity to show how to develop an algebraic proof of a combinatorial identity involving factorials. We’ll be looking over Doctor Rick’s shoulder as he guides a student through the maze. I’ll also add in a previously published version of the same proof for comparison, and a combinatorial proof of the same fact.

Here is the question, from Frankie:

Hi. I’m having some problems answering this question.

I’ve included a picture of my workings so far but I’ve hit a wall and can’t think how to carry my work on to finish the proof. Any hints or tricks for this question would be appreciated.

Also I will include a picture of a possible way to proceed, however I’m unsure if it is numerically sound.

The top third of the first page restates the problem in terms of factorials, using the fact that $${n\choose k}=\frac{n!}{k!(n-k)!}$$ Then each term has been rewritten, with some invalid moves.

Let’s first think about what the identity says: $${n\choose k} + {n\choose {k+1}} = {{n+1}\choose{k+1}}$$

This says that the number of ways to choose *k* + 1 out of *n* + 1 items is the same as the number of ways to choose *k* out of *n* items, plus the number of ways to choose *k* + 1 out of *n* items. We can see examples of this in Pascal’s Triangle:

As illustrated in red and in green, the identity says that the sum of two adjacent numbers in the *n*th row is the number between them in the (*n*+1)st row. This is the quick way to create the triangle.

The green case, for example, which I’ll be demonstrating more fully later, shows that $${4\choose 2} + {4\choose 3} = {5\choose 3}$$

Doctor Rick answered:

Hi, Frankie.

You started well, expanding each combination in terms of factorials, but after that I can see that you are floundering, not finding a way to make any progress toward connecting the two sides.

The secret ingredient you’re searching for is this:

(n+1)! = (n+1) n!

which could be taken as a recursive definition of the factorial, along with a starting point, 1! = 1 (or 0! = 1).

This may be all the help you need. Give it another try, and let me know if you get stuck again. (I’d like to know if you succeed, also!)

This identity simply reflects the fact that $$(n+1)! = (n+1)(n)(n-1)…(3)(2)(1) = (n+1)\left((n)(n-1)…(3)(2)(1)\right) = (n+1)n!$$ For example, \(5! = 5\cdot 4\cdot 3\cdot 2\cdot 1 = 5\cdot (4\cdot 3\cdot 2\cdot 1) = 5\cdot 4!\). It is very useful in simplifying expressions using factorials.

Frankie wanted to confirm a very relevant example:

Could you write (n-k)! = (n-k) (n-k-1)! ?

Doctor Rick replied,

Yes … now, do you see how to use this idea to pull out common factors in the denominators, etc.?

Frankie wanted a little more confirmation:

I think I have the main idea; before I continue, though, could you see if the picture I’ll include is correct?

Doctor Rick confirmed its truth while recommending against bringing division into the work, if only because division makes things a little more complicated:

Yes, that’s correct, although I am more inclined to work with multiplication and factoring rather than division, as in

(n–k)! = (n–k) (n–k–1)!

Frankie got back to work, and asked for a quick check:

I’m a bit stuck again as I feel like whatever I do I’m ending up with more terms on the fraction. I’ve simplified the fraction for the picture; if it is right could you perhaps give me a hint on where to go from here?

Each term has an unclear provenance. It’s time to dig into details. Doctor Rick answered,

I don’t see where this could have come from – how you could get a product of two factorials in the numerator, or only one in the denominator.

We started with this as what we want to prove (it’s what I got, and I believe you got it too):

n! n! (n+1)! --------- + --------------- =? ------------ k! (n-k)! (k+1)! (n-k-1)! (k+1)! (n-k)!Now, on the left side, we want to find the

greatest common factor, as an aid in choosing theleast common denominator. For this purpose, I would write (k+1)! as (k+1) k!, and (n-k)! as (n-k) (n-k-1)!, so that we can see the common factor k! (n-k-1)!.

This gives us a specific goal. We have the tool that lets us express a factorial in terms of a smaller one; now we have a reason to use it, namely to break down each factorial into smaller ones that will be shared by the terms.

Frankie took that starting point and gave it a new try:

Okay, I restarted and here is what I’ve got so far:

Frankie is working on both sides of the identity at once (which we’ll be commenting on later), applying our key fact and then forming a common denominator. All the steps are valid, so we’re making progress. Then, ten minutes later:

Here is what I’ve got to now. I think I’m pretty close to the answer now?

Frankie has distributed and combined terms, and does indeed seem close. Doctor Rick replied with a suggestion for making the work clearer (and therefore safer):

Your work is somewhat hard to follow because (I think) you keep

writing both the left and right sidesof what we are trying to prove, thoughyou haven’t yet shown that they were equal. When you do that, it’s hard to be sure which expressions you’re saying youknowto be equal.The usual practice, in presenting a proof like this, is to

start with the expression on one side, and rewrite it step by step until what you’ve got is the right side of what you were to prove. I understand that you aren’t at the stage or writing up your proof yet, and while you’re exploring, it is helpful to keep in view the target you’re aiming at — the right side. There aretwo things I might doto help me keep things straight.

- I might write those equal signs as you did, but put a
question markabove the equal sign if it’s one I want to prove rather than one I know to be valid. (In my previous message I wrote =? with that idea.)- I might write the identity to be proved at the top of the page, then
draw a vertical linedown the page where the equal sign is, to keep the two sides separate until they turn out to be the same.Anyway, back to the work you show. I can’t make sense of the work that you circled, but ignoring that,

your work seems to be correct. Having gotten a common denominator, you needed to add the numerators, which you did, but it could have been done a bit more simply like this:n!(k+1) + n!(n-k) = n!(k+1+n-k)

= n!(n+1)

= (n+1)!

We’ve previously discussed proofs of a trigonometric identity, and alternate ways to present them; some of the ideas here are similar. The work at the end replaces Frankie’s distribution with it opposite, factoring. Both are valid.

Frankie took the line-down-the-middle approach, but still worked on both sides:

Hi I rewrote it out neatly here with anything to the left of the red line being my work so to speak and to the right is the RHS (i.e. what the LHS needs to look like). If my methodology is correct I will rewrite the proof to be more presentable; however, for now I just want to see if I’ve got it right.

Doctor Rick acknowledged that the work is (mostly) correct, with a couple comments:

You miswrote the fourth line, I now notice: the denominator of the second fraction has two factors of (k+1), one of them was supposed to be (n-k). But I know you were thinking correctly, and just wrote something different, because the next line is correct.

On the second page you correctly wrap up the proof, essentially

reading up the right column on the first page. That’s how I think of this sort of proof: as a U in which the final proof starts at the upper left, goes down around the bottom of the U and back up to the top right.Looks good!

The U idea is also discussed in the trig identity post. Below, we’ll be seeing what amounts to a final cleaned-up version of this proof.

In working this into a post, I was going to just type up the final version for clarity, but then discovered that we have a similar algebraic proof of the same identity in *Ask Dr. Math*. It may be enlightening to compare it.

Here is the question, from 1998:

Pascal's Triangle and Combinations InPascal's Triangle, there is a pattern where any given number from the triangle is the sum of two numbers in the row before it that are closest to the number. So in the fourth row down from the top, in the third column, the number is 6. The two numbers closest to it in the previous row are 3 and 3. So 3 + 3 = 6. Each number from the triangle can represent a combinations problem. In this case, the 6 is equal to 4 C 2. The threes are equal to 3 C 1 and 3 C 2.Whyis 4 C 2 equal to the sum of 3 C 1 and 3 C 2? I realize that 3 plus 3 is 6, but I need a detailed explanation that shows how 3 C 1 + 3 C 2 equals 4 C 2. I have come up with a formula: x combinations taken y at a time = (x-1) C (y-1) + (x-1) C y However, this hasn't helped much in my understanding of the relation. Please give me any information you have on this relation of a number in Pascal's Triangle and the two numbers closest to it in the previous row, in terms of how they are represented in combinations.

Jesse had been shown this pattern in Pascal’s Triangle without proof, and has been able to express it as a formula, using the notation \(_n\text{C}_k = {n\choose k}\), but now wants to understand it. Bravo!

Doctor Pat answered, giving a proof in terms of factorials as above:

Jesse, I'll try to do all this in a way that you can see it develop. Here goes: I hope the value of n C r = n! / ( r! (n-r)!) is already known to you. I will use it to show why the relation you observe is true. First, though,we'll do an exampleto make a believer of you. Let's look at 8 C 3, 8 C 4 and note that the number below and between them is 9 C 4. Can you see that it will always be true that if the first two numbers are n C r and n C r+1, the number between and below will be n+1 C r+1?

Jesse has already seen numerical examples, but here the goal will be to prove it for this numerical case, as a template for the proof. I like this approach to formulating a proof in an unfamiliar area, by doing a “dry run” before diving in to formal details. In particular, by using specific numbers, we can more easily see what is happening with the factorials.

Back to our example: 8!/(8-3)!*3! ends up looking like this: 8*7*6 ------- 3*2*1 Note that there will be thesame number of values on the top and bottom, and with n C t, there are t numbers on top and bottom. So 8 C 3 + 8 C 4 looks like: 8*7*6 8*7*6*5 ------- + ------- 3*2*1 4*3*2*1 From here we have to do some simple fraction work.The two denominators are alikeexcept for the first term of the second (4). If we multiply the left fraction top and bottom by four (multiply by 4/4 = 1) we get: 4*8*7*6 8*7*6*5 4*(8*7*6) + (8*7*6)*5 ------- + ------- = --------------------- 4*3*2*1 4*3*2*1 4*3*2*1 Notice that on the right side I used parentheses to make a grouping show up. If we factor the (8*7*6) we end up with 9 (8*7*6) which is the numerator of 9 C 4. Thus the whole thing is equal to 9 C 4.

So the work looks like this, finishing it up: $${8\choose 3}+{8\choose 4}=\frac{8\cdot 7\cdot 6}{3\cdot 2\cdot 1}+\frac{8\cdot 7\cdot 6\cdot 5}{4\cdot 3\cdot 2\cdot 1}=\frac{4\cdot (8\cdot 7\cdot 6)}{4\cdot 3\cdot 2\cdot 1}+\frac{(8\cdot 7\cdot 6)\cdot 5}{4\cdot 3\cdot 2\cdot 1}=\frac{9\cdot (8\cdot 7\cdot 6)}{4\cdot 3\cdot 2\cdot 1}={9\choose 4}$$

Now rewriting this with (n C r) + (n C r+1) to show it is equal to n+1 C r+1 looks full of variables, butremember the variables just stand for the numbers we used above: (n C r) + (n C r+1) n*(n-1)*...*(n-r+1) n*(n-1)*...*(n-r+1)*(n-r) = ------------------- + -------------------------- r*(r-1)*...*1 (r+1)*r*(r-1)*...*1 Now we multiply the left term by (r+1) and use square brackets for the special grouping to factor: (r+1)*[n*(n-1)*...*(n-r+1)] [n*(n-1)*...*(n-r+1)](n-r) = --------------------------- + --------------------------- (r+1)*r*(r-1)*...*1 (r+1)*r*(r-1)*...*1 With the square brackets, we see that the numerator is: [(r+1)+(n-r)]*[n*(n-1)*...(n-r+1)] = [n+1]*[n*(n-1)*...(n-r+1)] which is (n+1)! divided by (n-(r+1))! The denominator is (r+1)!, giving us: (n+1)! ---------------- = n+1 C r+1 (r+1)!(n-(r+1))!

This is the same proof we saw above, written out in one chain from the left-hand-side to the right-hand-side as we prefer. And as before, the key step is the addition \((r+1)+(n-r) = n+1\), which in the example was \((3+1)+(8-3) = 4+5=9\).

In typeset form, we have this: $${n\choose r} + {n\choose {r+1}} = \\ \frac{n(n-1)\cdots(n-r+1)}{r(r-1)\cdots 1} + \frac{n(n-1)\cdots(n-r)}{(r+1)r(r-1)\cdots 1} = \\ \frac{(r+1)[n(n-1)\cdots(n-r+1)]}{(r+1)r(r-1)\cdots 1} + \frac{[n(n-1)\cdots(n-r+1)](n-r)}{(r+1)r(r-1)\cdots 1} = \\ \frac{[(r+1)+(n-r)]n(n-1)\cdots(n-r+1)}{(r+1)r(r-1)\cdots 1} = \\ \frac{(n+1)n(n-1)\cdots(n-r+1)}{(r+1)r(r-1)\cdots 1} = \\ {{n+1}\choose{r+1}}$$

You will observe that factorial notation was not used for most of the work here, as we did in our work above. Putting it in the terms we used, the same proof looks like this: $${n\choose r} + {n\choose {r+1}} = \\ \frac{n!}{r!(n-r)!} + \frac{n!}{(r+1)!(n-r-1)!} = \\ \frac{(r+1)n!}{(r+1)r!(n-r)!} + \frac{(n-r)n!}{(r+1)!(n-r)(n-r-1)!} = \\ \frac{(r+1)n!}{(r+1)!(n-r)!} + \frac{(n-r)n!}{(r+1)!(n-r)!} = \\ \frac{[(r+1)+(n-r)]n!}{(r+1)!(n-r)!} = \\ \frac{(n+1)!}{(r+1)!(n-r)!} = \\ {{n+1}\choose{r+1}}$$

This amounts to Frankie’s proof, streamlined and prettied up for presentation!

The algebraic complexities obscure what it is that we proved. Can we prove it in terms of the definition of the binomial coefficient (combinations) as the number of ways to choose *k* out of *n* items? Yes.

Suppose we have a set of *n* items (5 in the example below), and we extend the set by adding one more, shown here in red:

Now we want to choose *k* + 1 of them (here, *k* = 2). How many of these subsets will contain only the original (yellow) items? We are choosing all 3 from among the 4:

How many subsets will contain the red item, along with *k* from the yellow ones? We are choosing only 2 from the 4, plus another:

How many are there in all? $${n\choose k} + {n\choose {k+1}} = {{n+1}\choose{k+1}}$$

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