While I was researching for the post on uncountable sets, I ran across a discussion that didn’t quite fit, but raises interesting questions about how countable and uncountable sets can fit together. How can the rational numbers be countable, but the irrational numbers, which are closely intertwined with them, are uncountable?

Here is the question, from 2010:

Uncountable Infinitude, Illogically Concluded Regarding the question that I have seen here: Which set is bigger, the set of rational or irrational numbers? I understand both proofs: the countability of rationals and the uncountability of the irrationals. However,between each two irrational numbers we can always find a rational number. So -- to follow the same kind of reasoning that says there are the same infinite number of odd and even integers becausebetween each two odd numbers there is an even number-- I would expect that there will be the same number (infinite of course) of rational and irrational. What is the solution to this contradiction? Where is my mistake?

To illustrate the ideas here, first think about the familiar irrational number \(\sqrt{2} = 1.4142135623730950488016887242097\dots\). We can see that this is between the rational numbers 1.4 and 1.5; or between 1.414 and 1.415; and so on. And between this and some other irrational number, there will be rational numbers like 1.414213562373. So there are irrationals between rationals, and rationals between irrationals.

Below I will be including proofs of the facts alluded to here, but it should make some sort of sense that if you choose *any* two irrational numbers there is *at least one* rational number (and in fact infinitely many of them) between, and likewise if you choose *any* two rational numbers there is *at least one* irrational number (in fact, *uncountably* infinitely many of them) between. Both rational and irrational numbers are “dense” among the real numbers, meaning that both types are within any interval you name. They are all packed tightly together – so much so that no picture I could make would give the idea accurately.

Boaz is picturing rational and irrational numbers sort of alternating the way even and odd numbers do:

But it is not, as he suggests, the mere presence of an even number between two odd numbers that shows that the even and odd numbers have the same cardinality; rather, we use a *one-to-one correspondence* that is implied by the very rigid *alternation* of the two (that is, that because there is *exactly one* odd number between any two consecutive even number, each odd number can be paired up with the immediately following even number). Boaz has not shown such a correspondence between rationals and irrationals! Furthermore, the real picture doesn’t look remotely like that.

I replied briefly, just noting the lack of a proper reason:

You seem to be assuming that because there is AT LEAST ONE rational between any two irrationals, and vice versa, that there are THE SAME NUMBER of each between the other. There are in fact countably (infinitely) many rationals between any two irrationals, but UNcountably many irrationals between any two rationals!

Assumptions in math can be dangerous; you need a justification for any conclusion you make! This is all the more true when dealing with infinities, which we can’t visualize. The comparison to even and odd doesn’t hold up, because those are discrete, with a definite gap between them, and only a finite number of odd numbers between any two even numbers. Infinity is much more slippery than that.

Boaz wasn’t satisfied, because there still seemed to be a reason for concern. He wrote back:

Many thanks for your reply.Between any two of your UNcountable irrationals, I can find a rational number. Doesn't that mean that the rationals are also UNcountable? I have seen some proofs regarding the countability of rationals, but they were not convincing. Maybe I need a more accurate proof.... (actually, I think I have two proofs, but they are probably not valid.) I hope I managed to explain my difficulty correct. Again, thanks for the prompt response!!!

This is much more convincing, on the surface: It would seem that there are **uncountably many spaces** between irrationals, so there must be **uncountably many rationals in those spaces**! Hmmm …

I replied, still pointing out that something that seems obvious from our finite experience may not be true of infinite sets, and would need proof:

You're assuming that if every member of a set A lies between two members of an uncountable set B, then set A is also uncountable. Do you have a proof of that? An uncountable set is one thatCAN'T be counted. All you've done is essentially to show ONE way you could TRY to count the rationals and fail -- namely, by putting them between irrationals and realizing that you can't count those. That doesn't mean that another way does not exist. And in fact,proofsdo provide a way to count the rationals. Infinities can't be trusted not to surprise you. You have to watch out for hidden assumptions such as the kind you've made.

The subtlety here lies in the negative nature of “**un**countable”. To show a set is uncountable, you have to actually show that there is **no** way to count it, not just that one way doesn’t work, or even that no way *you have found* can do so.

Boaz answered the way we like to hear:

Many thanks. I think I can sleep now :-)

I know it still feels like we’ve failed to fully answer the question. We’d like a way to properly visualize what the relationship *is* between the rationals and the irrationals! Try this:

One way to count the rationals (within an interval, such as from 0 to 1) is to take them in order of their denominators, so we first list the halves (1/2), then the thirds (1/3, 2/3) then the quarters that are not already listed (1/4, 3/4), then the fifths (1/5, 2/5, 3/5, 4/5), then the new sixths (1/6, 5/6), and so on. As we do this, one layer at a time, each new interval we form will contain infinitely many rationals, and infinitely many irrationals. We keep descending deeper into the real number world with each increase in the denominator, picking up new rationals one by one (countably), but always leaving uncountably many irrationals still hidden between them. Here the solid dots at the rationals in the order we count them, left to right and bottom to top; the open dots are duplicates we don’t recount.

But there’s no way to put in red dots for the irrationals! At every step, there are infinitely many of both rational and irrational in every space between. All we can say is that because the rationals are countable but the reals are not, there are a lot of points in between that will not be touched, and they are “everywhere”.

Now, Boaz imagines looking at *all* the irrationals and putting a rational in each space between. But the uncountability of the irrationals means that we *can’t* carry out the process I just did with the rationals, looking at the spaces between at each step. All I can say to Boaz’s question is that it is a paradox: We can’t imagine uncountable infinity, but know that the conclusion must be true.

For more about counting the rationals, see

Set Theory and Orders of Infinity Infinite Sets Counting Rationals and Integers

Each of these uses a slightly different correspondence; the first, like mine above, deals only with numbers between 0 and 1, while the others deal with *all* rational numbers, positive and negative.

A similar question was asked in 2000:

Do Rational and Irrational Numbers Alternate? It is known that: 1. Any two non-equal real numbers "contain" an irrational. 2. Any two non-equal real numbers "contain" a rational. These real numbers can of course be rational or irrational.Does this mean rational and irrational numbers alternate?

That is, between any two reals (including rationals) there is an irrational, and between any two reals (including irrationals) there is a rational.

That time, Doctor Rob replied, with more formal details than I gave above, spending most of his time actually proving the two statements, and only briefly dealing with the actual question:

Yes, both statement 1 and statement 2 are true.The answer to your question, however, is "No."If that were so, the two sets would have the same size, and they don't (the irrationals are uncountable, the rationals are countable).

That was quick!

The proofs of these depend on theArchimedean Ordering Principle: if r > 0 and s > 0 are real numbers, then no matter how small r is and how large s is, there is a positive integer n such that n*r > s.

Let’s think about this, also called the Archimedean Property of the real numbers. What it says is that if I give you two numbers, say *r* = 0.00007 and *s* = 5,000,000, you can find an integer *n* such that 0.00007*n* > 5,000,000. How would you find it? Just divide 5,000,000 by 0.00007 to get 71,428,571,428.57…, and round up to 71,428,571,429. (Or anything bigger than that, like 75,000,000,000!) This says that no number is so large, and no number is so small, that you can’t exceed the former by taking steps of the latter size.

To prove thestatement 2, for example [there is arationalbetween any two reals], let the two unequal real numbers be x < y. Then let r = y - x > 0, and s = 1 + r. Then by the A.O.P., there is a positive n such that n*r > 1 + r, n*y - n*x > 1 + r, n*x < n*y - r - 1. Now consider integer m = [n*y-r]. (Here [x] means the greatest integer less than or equal to x, that is, the integer you get by rounding down from x. It satisfies x-1 < [x] <= x.) Then we can see that n*x < n*y - r - 1 < m <= n*y - r < n*y, x < m/n < y. Thus m/n is a rational number in the interval (x,y). To get anirrationalnumber in the interval (x,y) [statement 1], let z be any positive irrational number. Now use the above argument to find a rational number m/n in the interval (x/z,y/z). Then z*m/n will be in (x,y) and irrational.

To see what’s going on here, let’s find a **rational number** between the real numbers \(x=\sqrt{50}\) and \(y=\sqrt{51}\). (That can be done more easily using specific properties of radicals; I’ll do it here using Doctor Rob’s generally applicable method.)

Let \(r=\sqrt{51}-\sqrt{50}\approx0.07036\). The Archimedean Property implies that there is some integer *n* by which we can multiply this to get a product greater than \(s = 1 + r\) (which for our example is 1.07036); we can find it by dividing and rounding up:\(16\cdot0.07036\approx1.12577>1.07036\)). Now we let \(m=[ny-r]=[16\cdot\sqrt{51}-0.07036]=114\). Our fraction is therefore \(\frac{m}{n} = \frac{114}{16} = 7.125\). This is in fact between \(\sqrt{50} \approx 7.07107\) and \(\sqrt{51} \approx 7.14143\).

Next, let’s find an **irrational number** between, say, \(x = \frac{7}{52}\approx0.134615\) and \(y = \frac{7}{51}\approx0.137255\). Let \(z = \sqrt{2}\), arbitrarily. We want to find a rational number between \(\displaystyle\frac{x}{z} = \frac{\frac{7}{52}}{\sqrt{2}}\approx 0.095187\) and \(\displaystyle\frac{y}{z} = \frac{\frac{7}{51}}{\sqrt{2}}\approx 0.097054\).

Doing the same things we did in the last paragraph, we let \(r=0.097054-0.095187 \approx 0.001866\). Our *n* will be 1.001866/0.001866, rounded up to 537.

Now we let \(m=[ny-r]=[537\cdot0.097054-0.001866]=52\). So our m/n is 52/537; and our irrational number is \(\displaystyle z\left(\frac{m}{n}\right) = \sqrt{2}\left(\frac{52}{537}\right) \approx 0.136944\). And this is an irrational number in the right interval.

(Note that there is no actual value in these calculations themselves, because the numbers we found are just approximations to the irrational numbers we’re really talking about. What matters is that the exact values are demonstrably just what we say they are.)

]]>A recent question from a student working beyond what he has learned led to an interesting discussion of alternative methods for solving a minimization problem, both with and without calculus.

The question came from Kurisada a couple months ago:

f(x, y) = x

^{2}– 4xy + 5y^{2}– 4y + 3 has a min value. Find the value of x and y when f(x, y) is minimum.This is the first time I met this question, and here is my way:

First I made f'(x) = 2x – 4y = 0. (I’m not sure if I can make it to f(x) while it is actually f(x, y).)

Therefore x = 2y.

Then I input x = 2y to f(x, y) = y

^{2}– 4y + 3 = (y – 2)^{2}– 1Thus min value = -1

y = 2

x = 4

Apparently this student has not done multivariable calculus, but has invented some of its basic concepts, particularly partial derivatives. The method is nonstandard, but correct. Now we need to show why!

Doctor Rick replied:

From your parenthetical comment, it appears that you may not have learned about partial derivatives; but what you have done is perfectly valid.

The

partial derivativeof f(x, y) with respect to x, ∂f/∂x, is what you get when youdifferentiate while interpreting y as a constant rather than a variable. The minimum of a function of two variables must occur at a point (x, y) such that each partial derivative (with respect to x, and with respect to y) is zero. (Of course there are other possibilities akin to those in calculus of one variable — if the derivative is not defined, etc. They don’t apply here.)You found the locus of points on which ∂f/∂x = 0, then wrote a function in one variable representing the value of f(x, y) on that locus, and minimized that function. It worked — good job!

Kurisada just hoped that it would be valid to temporarily pretend that *y* was constant; in effect, this amounted to slicing the surface defined by function *f* with a plane \(y = k\), and finding the minimum point of the resulting curve. This is what a partial derivative does: it gives the slope of such a curve.

Here is a graph of the surface defined by *f* (light blue), showing the intersection with the (arbitrarily chosen) plane \(y = 3.5\) (red) and its minimum point, A. The slope of this curve at any point is \(\frac{\partial f}{\partial x}\). M is the absolute minimum we are seeking.

Kurisada had found that the ordered pairs for which \(\frac{\partial f}{\partial x} = 0\) satisfy the equation \(x = 2y\), which is the equation of a plane; so the low points on the *y*-slices lie on the intersection of this plane with the surface. This curve is the locus (set of points) Doctor Rick referred to. Replacing \(x\) with \(2y\) in the equation \(z = x^2 – 4xy + 5y^2 – 4y + 3\) yielded \(z = (2y)^2 – 4(2y)y + 5y^2 – 4y + 3 = y^2 – 4y + 3\), which is the equation of the projection (“shadow”) of the locus on the *yz*-plane.

Here I have added in the intersection of the surface with the plane \(x = 2y\) (blue), showing how our point A lies on this locus; M is the minimum point of the locus, and therefore the minimum point on the surface.

Kurisada had questions about the reference to *each* partial derivative:

Is it possible if I regard y as the variable and x as the constant?

Does it mean that actually I need to do both the partial derivatives with respect to x and with respect to y?

Or is it only done to check whether the answer is true?

Doctor Rick replied first to the suggestion to treat only *y* as the variable:

That should work also. That is, find the locus of points at which ∂

f/∂y= 0, then minimize f(x, y) constrained to this locus.

Let’s try that, using Kurisada’s method with *y* rather than *x*. We have $$f(x, y) = x^2 – 4xy + 5y^2- 4y + 3,$$ so $$\frac{\partial f}{\partial y} = -4x + 10y – 4,$$ which is zero when $$y = \frac{2x+2}{5}.$$ Putting this into \(f(x,y)\), we get $$f(x, y) = x^2 – 4x\left(\frac{2x+2}{5}\right) + 5\left(\frac{2x+2}{5}\right)y^2 – 4\left(\frac{2x+2}{5}\right) + 3.$$ This simplifies to $$\frac{1}{5}(x-4)^2 – 1,$$ whose minimum again is -1, when \(x = 4\) and \(y = \frac{2(4)+2}{5} = 2.\)

Here we have the intersection of the plane \(x = 7\) (green) with the surface, showing its minimum (B), and the locus of these minima (purple), which again passes through M:

As to whether the method using *both* partials is *needed*:

What you did is sufficient. You minimized the function in two directions: the x direction, and along the “valley” you had identified.

My description says that you can

alsochoose to minimize the function in the x and y directions. If you want to try solving the problem this way, use your equation that says ∂f/∂x= 0, and write another that says ∂f/∂y= 0, then solve these two equations simultaneously. This is no more difficult than what you did.

I like this description of the locus of what we might call “east-west” minima as a “valley”; the solution is the lowest point along this valley.

The usual method, as explained here, is to set both partial derivatives to zero, and combine the two resulting equations. The equations, as we’ve seen, are \(2x – 4y = 0\) and \(-4x + 10y – 4 = 0\). Multiplying the first by 2 and adding, we get \(2y – 4 = 0\), giving \(y = 2\); putting this into the first equation, we get \(x = 4\). So the minimum is at \((4, 2)\), and \(f(4, 2) = 4^2 – 4(4)(2) + 5(2)^2 – 4(2) + 3 = -1\) yet again. In effect, here we are finding the intersection of two “valleys” (curves along which we find minima in the east-west and north-south directions).

Here is a picture of this method, showing the two loci of minima intersecting at M:

Doctor Rick continued, referring to a previous question from Kurisada that specifically asked for multiple ways to solve a problem (an excellent idea!):

Now, in the spirit of your last question, I’ll point out that the problem can also be solved without calculus! We know that a square is minimum when the quantity being squared is zero, since a square can’t be negative. Thus, if you can rewrite the function as a sum of squared quantities and a constant, the minimum will be that constant, and will be attained when each of the squared quantities is zero. See what you can do with this idea!

Kurisada had already done this with \(y^2 – 4y + 3\), and now observed that completing the square on the first two terms “reduced the problem to one already solved”, as we say:

I changed it to (x – 2y)

^{2}+ y^{2}– 4y + 3And because (x – 2y)

^{2}to make it minimum, it becomes y^{2}– 4y + 3It is the same to my result! (This way makes me understand more about something I don’t really understand before!)

Doctor Rick finished the work, putting everything together:

Thus far you have changed the function-defining equation

f(x, y) = x

^{2}– 4xy + 5y^{2}– 4y + 3to

f(x, y) = (x – 2y)

^{2}+ y^{2}– 4y + 3Now we want to complete the square on the last three terms. But you have already done this! You said in your original posting on this thread that

y

^{2}– 4y + 3 = (y – 2)^{2}– 1That’s exactly what we need. Putting this into the function definition, we get our final result

f(x, y) = (x – 2y)

^{2}+ (y – 2)^{2}– 1Now we can see immediately that the least possible value of f(x, y) is –1, attained when both squared quantities are zero, that is, when the following system of equations is satisfied:

x – 2y = 0

y – 2 = 0

The rest is easy.

We have often recommended trying multiple methods as a way to learn math more deeply. As Kurisada mentioned, the non-calculus method helped to see the problem and its answer from a new perspective. I hope my addition of the graphs (made with GeoGebra) adds yet another dimension to your understanding.

]]>Recently we looked at the question of how likely a two-child family with a boy is to have another boy (or, to the contrary, to also have a girl). Searching for those questions turned up another one of interest involving the gender of a pair of siblings: How do things change when they are twins?

Here is the question, from 2006:

Gender Probabilities for Twins Hi Doctor Math. I am pregnant with twins--sex unknown.Since neither is older than the other, what is the probability of having any gender combination?I remember from college genetics that each birth is mutually exclusive, therefore the probability of any combination of boy or girl is 1/2, i.e. each child will either be a boy or a girl. My husband is using the bb, gg, bg, gb theory of probability, and that would increase the odds of having a boy and a girl to 1/2, while having a bb or gg would be 1/4.My point is that bg and gb are the same combination, so his probability isn't correct. What do you think? I saw your string on a similar question, but it doesn't address twin births, and I wanted to be certain that I was thinking about this correctly. Thanks!

I think that Melissa meant that each birth is *independent* (so that one being a boy doesn’t influence whether the other is a boy), not that they are *mutually exclusive* (that is, one child can’t be both a boy and a girl!).

Her husband used the same basic model we used in The Other Child: there are four equally-likely possibilities in a two-child family, based on birth order. But Melissa disagrees with this model, because twins are the same age. Is she right?

I replied, first stating my qualifications:

As a twin myself (and my identical twin brother is also a Math Doctor), I HAVE to take this question! If there were onlyfraternal twins, then your husband would be right. Even with twins,you can distinguish them(firstborn/secondborn, favorite/nonfavorite, or whatever), so BG and GB are not the same. As an example, I was expected to be a girl (because my heartbeat was weaker, and they didn't have ultrasound yet). We could be distinguished, that is, even without knowing which was a boy and which was a girl or what our names would turn out to be.

Never tell a pair of twins (or their mother) that they can’t be distinguished! In fact, never tell them that one isn’t older than the other! My brother Rick is older by only 5 minutes, but we both know he *is* older (and that much wiser).

So you could make up a table of the four equally likely cases: Dave | boy | girl| ----+-----+-----+ boy | 1/4 | 1/4 | Rick ----+-----+-----+ girl| 1/4 | 1/4 | ----+-----+-----+ That gives probabilities of 1/4 (two boys), 1/4 (two girls), 1/2 (one of each).

(Of course, I’m using our names to stand for “older” and “younger”, regardless of who we turned out to be.)

So Melissa’s husband is right … but only if it’s true that sexes of twins are truly independent. That isn’t quite true, and this is what makes the question interesting:

Now in reality, you also have to bring in the probability that the twins turn out to beidentical--in which case they MUST both be the same sex. (I understand there are other odd possibilities, but I'll neglect those.)

According to what I’ve read, in addition to fraternal (dizygotic) or identical (monozygotic) twins, there are very rare cases like “semi-identical” (sesquizygotic) twins, as well as many variations in the genetics. None of these are really relevant to our question as far as I know.

Suppose the probability of having fraternal twins is F, and of having identical twins is I. Then,given that you have twins, the probabilities of their being identical is I/(F+I); we end up with this table: BB BG GB GG I I Identical ------ ------ 2(F+I) 2(F+I) F F F F Fraternal ------ ------ ------ ------ 4(F+I) 4(F+I) 4(F+I) 4(F+I)

The identical twins comprise \(\displaystyle\frac{I}{F+I}\) of all twins, which I’ve split into two equal parts, boy and girl. The fraternal twins consist of the remaining \(\displaystyle\frac{F}{F+I}\) of all twins, split into 4 equal parts.

I didn’t assign specific numbers to these yet, in order to separate the (reliable) mathematics from the (empirical and uncertain) statistics, and can choose what numbers to use later, after considering general principles.

So the probability of two boys, or of two girls, is [summing the first or last column] 2I + F ------ 4(F+I) and the probability of one of each is [summing the two middle columns] F ------ 2(F+I)

Now we can look at the realities, which are far less certain:

A couple sites I looked at said that F = 1/125 and I = 1/300. Another site said that 1/3 of all twins are identical. Others give different numbers. If I take that first pair of numbers, we get the probability of two boys, or of two girls, is 0.3235, and the probability of one of each is 0.3529. So each probability is about 1/3.

That came from a quick search for frequencies of twins, just looking for reasonable numbers without trying for perfection. Repeating that search now, I find that Wikipedia says that, due to fertility drugs, the overall rate of twins “rose 76% from 1980 through 2009, from 18.8 to 33.3 (or 9.4 to 16.7 twin sets) per 1,000 births.” So these numbers are not very stable. Fraternal twin rates have not only increased over time, but also vary greatly among populations, between 6/1000 and 14/1000 or more. (Other sources say 20/1000 or more.) Again, from Wikipedia, identical twins are 3/1000 of deliveries worldwide, with little regional or temporal variation. (Other say 4/1000 or 1/300.) The numbers I used in my 2006 answer correspond to 8/1000 and 3.3/1000, the former apparently being low by current standards, and the latter being about right.

Note that for the overall rate, Wikipedia carefully distinguishes between the rate of *twins* and the rate of *twin sets*; one pregnancy with twins produces one twin set, but two twins, so the rate of twins is twice the rate of twin sets. The term “birth” or “delivery” seems ambiguous to me; it might mean either one mother giving birth, or one child being born. So I’m unsure how to interpret the data I’ve found.

For my calculations above, it doesn’t matter which way we take it, as long as *F* and *I* use the same definition, since we are looking only at twins. The distinction will be more important for the extra calculations I do below.

The rate given for identical twins appears to be for individuals, but it is very hard to confirm this! The fact that it is not as easy as most of us think to determine whether twins are identical further complicates this; many sources say that 1/3 of all twins are identical, but the site just referenced says instead that 1/3 are opposite-sex twins (for whom there is no question). This is consistent with the claim that 2/3 are fraternal (since half of fraternal twin sets will be opposite-sex). So *F* should be about twice *I* – presumably meaning before modern increases, and in a general American context, presumably. So we might take ** F = 6/1000** and

Using these numbers, we find that the probabilities are:

$$P(\text{same sex} | \text{twins}) = \frac{2I + F}{2I + 2F} = \frac{2(.003) + .006}{2(.003) + 2(.006)} = \frac{2}{3}$$

$$P(\text{opposite sex} | \text{twins}) = \frac{F}{2I + 2F} = \frac{.006}{2(.003) + 2(.006)} = \frac{1}{3}$$

This confirms what I said about each of the three probabilities (two boys, two girls, boy-girl) is about 1/3. In fact, this will be true regardless of the actual probabilities, as long as *F* = 2*I*:

$$P(\text{same sex} | \text{twins}) = \frac{2I + F}{2I + 2F} = \frac{2I + 2I}{2I + 2(2I)} = \frac{4I}{6I} = \frac{2}{3}$$

(Ironically, do you notice that this is what Melissa would conclude from her wrong assumption that BG = GB?)

But taking the highest reported frequency of fraternal twins, 0.020, we get

$$P(\text{opposite sex} | \text{twins}) = \frac{F}{2I + 2F} = \frac{.020}{2(.003) + 2(.020)} = \frac{10}{23}\approx 0.435,$$

bringing the probability considerably closer to 1/2. (Of course, if all twins were fraternal, this probability would be 1/2.)

Now let’s go beyond the question, and find the probability of *any* two-child family being a boy and a girl. Here, the exact meaning matters. I defined *F* and *I* as probability of *having* fraternal or identical twins, which sounds like twin *sets*, not individual twins (the probability of *being* a twin); that’s appropriate for the question we were answering. But my impression is that most of the data refer to individuals, so let’s call it that.

Here is a chart of the probability of all possibilities, broken down by kind of pair and genders in birth order:

BB BG GB GG | Total ---------------------------------------+------- 1-I-F 1-I-F 1-I-F 1-I-F | Non-twin ----- ----- ----- ----- | 1-I-F 4 4 4 4 | | I I | Identical --- 0 0 --- | I 2 2 | | F F F F | Fraternal --- --- --- --- | F 4 4 4 4 | ---------------------------------------+------- 1+I 1-I 1-I 1+I | Total --- --- --- --- | 1 4 4 4 4 |

Not too surprisingly (if you think about it!), fraternal twins have no effect of the probability that two siblings have the same gender; *F* doesn’t show up in the bottom line.

So the probability of any pair of siblings having the same sex is simply

$$P(\text{opposite sex} | \text{twins}) = \frac{1+I}{2} = \frac{1 + .003}{2} \approx 0.5015,$$

that is, 50.15%. This is barely more than the usual assumption of 50%, which makes sense because identical twins are rare. Compare this to the fact that the ratio of boys to girls at birth is not 1:1, but about 1.05:1, so that about 51.2% of babies are boys. Twins have less effect on the ratios than other causes.

]]>Take this question from 1999:

More Than 100 Percent Please help to settle an ongoing argument with a friend: is it correct to use a value of more than 100%? I say that there'sno such thing as anything larger than 100%; my friend says that it is okay to say 200% or 300% and so on.

Doctor Rick replied:

There is nothing intrinsically wrong with using percentages greater than 100%. Whether this makes sensedepends entirely on the context. A percentage is simply another way of writing a fraction with a denominator of 100. For example, 6% = 6/100. Just as you can have an improper fraction (a fraction whose numerator is greater than the denominator), such as 4/3, so you can have what we might call an "improper percentage" like 120% or 300%.

Since percentage is just a way to write a number, it can be used for *any* number. We can write 0.06 as 6% (by multiplying 0.06 by 100%), and we can write 1.2 as 120% (by multiplying 1.2 by 100%). Below, we’ll even see 1255% (that is, 12.55). In the abstract, there are no limitations.

There are situations in which a percentage greater than 100% makes no sense. For instance, "The Math Doctorsanswered 146% of the questionsreceived last month." This makes no sense because if we received 5061 questions, we couldn't possibly answer more than all of them. It's just as nonsensical as saying "I ate4/3 of the cake."

These are situations in which 100% means “all”, and there is nothing more than all.

Of course, someone could eat 4/3 of * a* cake – that is, 1 1/3 cakes (if they are not too large).

On the other hand, sometimes percentages are used like this: "The number of questions received was up 15.7%, from 5450 in February to 6305 in March." In other words, the increase from February to March was 6305-5450 = 855, and 855 is 15.7% of 5450. (These facts are true, by the way.)

Here, 100% doesn’t mean “all possible”, but just “all that there were in February”.

Note, by the way, that in March, 1999, Ask Dr. Math got an average of 203 questions a day. We weren’t able to answer all of them, but we did what we could. Today, there are many other places to ask questions, some of them willing to just give complete answers without making sure you could solve the problem yourself, so we get far fewer. We do answer every question that can be answered.

Now, what if the number of questions received went up to 14000 in April? (It didn't.) This would be anincrease of 122%from March to April. There is nothing wrong with this - no law says that the number of questions can't do more than double from one month to the next. So the answer is, there is such a thing as a percentage larger than 100%, but not everything can have a percentage larger than 100%.

This is just one example of a place where percentages can meaningfully be greater than 100%.

A similar question came from Samantha in 2003:

Percentage as Standard for Comparison How can you have more than 100% of something? 100% indicates you have a complete item, sohow can it be more than complete?

I answered this one:

Good question! In some situations, a percentage refers to apart of all there is: I can't do more than100% of a job, or eat more than100% of the one pizzaI ordered. More than100% of the skycan't be covered by clouds, and more than100% of the populationcan't be poor.

Each of these fits Samantha’s image of “a complete item”. We might picture it like this, where 100% is an entire bar; 75% is 3/4 of the bar, but 125% of it makes no sense:

But at other times, a percentage refers to something that is not an absolute limit, butjust a standard for comparison. For example, the orange juice you drink in the morning may contain100% of your daily requirementof vitamin C; if you eat any more foods containing vitamin C during the day, you will have eaten more than 100% of the RDA. For some vitamins, if you ate too much, it might be bad for you; however 130% in this case just means that you didn't need to have as much as you did, but it's okay to go beyond that. You've just taken in more than the recommended amount.

Here, 100% can be thought of as just a mark on a number line; 75% is 3/4 of the way to the mark, and 125% is 1/4 beyond the mark:

As another example, you might plan to work 10 hours to put together a project. If you keep track of the time you've used, when you have spent 6 hours on it you will have used60% of the planned time. But you might find that you aren't quite finished when the planned time is used up. If you spend 3 extra hours to make everything just right, making a total of 13 hours of work, you will have used130% of the planned time.

Sometimes, there is a fine line between the two cases:

Now, if there had been only10 hours leftbefore the project was due, you couldn't have taken 130% of that time. But you had justPLANNED to use 10 hours, and it was possible (though possibly bad for your health, or your other activities) to use extra time for the project. Just like the vitamin C requirement, 100% didn't mean "all there is," but just "all we expect." The 130% means that you are "30% over budget." Do you see the difference in these two kinds of situations? The same thing happens with fractions. I can eat1/2 a pizza, or 2/3 of a pizza; but I can also eat2 3/4 pizzas, because the one pizza isn't all there is. I can fill 1/2 of my stomach, or 2/3 of my stomach, but not 2 3/4 stomachs, because I only have one. Fractions can be greater than 1, and percentages can be greater than 100. It's only when the fraction or percentage refers to a part of a whole that we can't go beyond the whole.

(Of course, sometimes “whole” doesn’t mean all there is, but just a whole item of which there are more, as in the pizza example.)

Back to the nutrition label: a bag of potato chips can't bemore than 100% fat, because the fat is PART of the chip, and the part can't be bigger than the whole. But it might conceivably havemore than 100% of your daily requirement of fat, because the fat in the chips is only being COMPARED to the amount you're supposed to eat. Nothing prevents you from eating more. And finally, the number of calories you get from fat today can't bemore than 100% of the total caloriesyou consume, because they are part of that total. That gives us three different ways to express the fat in the chips as a percentage, and each percentage has a different meaning.

This is a key concept with both fractions and percentages: they are always relative to some “unit”. In my three examples above, the unit was (a) the total amount of fat in a chip; (b) the RDA of fat for a person; and (c) the total calories actually eaten in one day.

Here's yet another way that a percentage can be used: A copier may have a setting for enlarging or reducing what you copy. The normal setting would be 100%, meaning that what comes out is the same size as what went in. You can set it to 50%, meaning the copy will be half the size of the original. Or, you can set it to 200%, andthe copy will be twice as large. There's nothing wrong with that, because the copy is not part of the original; we are onlycomparing it to the original. And that's what makes the difference.

Here, the unit is the original size, to which we are comparing the copy size.

There are a lot of ways percentages can be more than 100%, aren’t there?

Here’s one last question from 2008, illustrating the point I made last time about being fooled by large percentages:

Using Estimation to Check Your Answer What was the percent of increase in population from 1950 to 2000. In 1950 it was 38,000,000 and in 2000 it was 515,000,000. The increase was 477,000,000 but when I try to get a percentage I keep getting like 1,255%. What I find difficult is the fact thatmy percentage is so highand I don't think that my answer is correct and I want to find out what I might have done wrong in the problem. I've worked this particular problem about 3 times and still get the same answer and I'm so frustrated.

We recommend that students check whether an answer seems reasonable; but when your reasonableness alarm is going off because it is miscalibrated, it can lead to frustration.

I replied:

Your answer sounds reasonable to me! An increase of 1000% would mean that the increase was 10 times the original value, so with 1255% the increase would be more than 12 times the original. Since 12 times 38 is about 12 times 40, or 480, and 515 is over that, it's certainly in the right ballpark, so to speak.

I’ve changed one word from the original, where I made a small misstep in the second clause. I’m assuming that non-American readers know what “in the right ballpark” means!

Percentages this high are unusual, but that doesn't make them wrong. You just need to look at the data andsee whether an unusual answer is to be expected. Since 515 million is many times 38 million, we can expect the percent increase to be huge. Knowing how to do such a "sanity check" is a good way to preserve your sanity when you're doing a problem like this!

Chanice responded:

Thank you Dr. Peterson for helping me with this problem because it was really beginning to drive me crazy. I learned new ways to look at problems and to do a "sanity check". So again I just wanted to say thanks.

If it feels wrong, it might be wrong, or you might just be misjudging it. Leaving that possibility open is helpful.

]]>**Markup** and **margin** refer to the profit (difference between cost and selling price) as a percentage of either cost or price, respectively. These are closely related to percent change. Consider this question from 2007:

Profit Margin and Percentage Markup I have a small business and need to be able to compute profit margins and percentage increases. There seems to be a difference of semantics when I try to discuss this with customers. Specifically, suppose I have an item that my customer can buy at the wholesale price of $60.00, and then they can sell it at the retail price of $125.00. What is thepercentage mark-upand is this a different number than what they would consider theirprofit margin? I believe to find the increase I would just take 60 divided by 125 which would equal .48 which would mean a 48% increase on price. Is this also considered the profit margin? I don't think so! I could say to the client that they have made a profit of what percent?

Evelyn has actually calculated what percent the cost is of the price. Doctor Ian responded, first giving a (now broken) link to a source of definitions, then continuing:

```
Following these definitions,
gross profit
gross profit margin = ------------
sales
125 - 60
= --------
125
65
= ---
125
= 0.52
or 52 percent. And
sales price - cost
markup = ------------------
cost
125 - 60
= --------
60
65
= --
60
= 1.08
or 108 percent.
The difference is what you're comparing the profit to: If you express the profit
```**as a percentage of the sale price**, you're computing the profit margin. If you express the profit **as a percentage of the cost**, you're computing the markup.

In other words, the *markup* is the percent *increase* from the cost to the selling price; the *margin* is the percent *decrease* from the selling price to the cost, because the selling price is taken as the base for the percentage.

Here are the formulas, again: $$\text{markup} = \frac{\text{selling price}- \text{cost}}{\text{cost}}\cdot100\%$$ $$\text{margin} = \frac{\text{selling price}- \text{cost}}{\text{selling price}}\cdot100\%.$$

Applying a markup, when you know the cost, is easy; you are just **applying a percent increase** to the cost. But when it is the margin you know, finding the price is the same as ** undoing a percent decrease**.

Here is a question from 2006:

The Difference between Calculating Markup and Profit Margin I have a question about how to calculate [selling price from] profit margin. We have been calculating this as fixed costs multiplied by whatever the margin percentage is. For example, with costs of $50,000 and a desired margin of 25% we do $50,000 x 1.25 = $62,500, so we would sell the product for $62,500. But I am told that the correct way to calculate this is to take the fixed costs and divide by the desired margin percentage subtracted from 100%. So that would be $50,000 / .75 = $66,667. I am having some trouble understanding how this works. Can you explain this is simple layman's terms?

I answered, emphasizing the distinction:

You have no idea how common this question is! Profit margin means "what percentage OF THE PRICE is profit?" That is different from "what percentage OF COST is the profit?"--that's the "markup". When wemark up, we add some percentage of what it costs us.Marginis seen from the customer's perspective: how much of what I'm paying is going into their pockets? So you want to increase your fixed cost of $50,000 by some amount that will be 25%of the amount you'll be charging--which you don't know yet! You can't just add on 25% of the fixed cost, because that's the right percentage of the wrong thing. (You're actually calculating a 25% markup rather than a 25% margin.)

When you have to work with a number you don’t know yet, that is where algebra is useful!

So what can we do? Well, we think backward. Suppose we knew the price we're going to charge; I'll call it X. Then the margin would be 25% of that, or 0.25 times X. The cost would be that much less than X: cost = X - 0.25X But 1 times X, minus 0.25 times X, is 0.75 times X. That is, 100% - 25% is 75%. So cost = 0.75X But now we can find X, because we know the cost is $50,000! 50,000 = 0.75X X = 50,000 / 0.75 That is, we undo the multiplication by 0.75, by dividing 50,000 by 0.75. So X = $66,666.67

This is just what we did to reverse a percent decrease: divide by the complement of the percentage.

We can check the result, of course. A 25% margin would mean that the profit is 25% of this, namely \(0.25\times \$66,666.67 = \$16,666.67\). And, in fact, the profit is the selling price minus cost, which is \(\$66,666.67 – \$50,000 = \$16,666.67\).

Note that this is more than what you were calculating, because you were adding 25% of a smaller number--here we're adding 25% of $66,667 rather than only 25% of $50,000. By charging only $62,500, you had a 25% markup but only a 20% margin (12,500/62,500).

Here are the formulas for applying a markup and a margin: $$\text{selling price} = \frac{100 +\text{markup}}{100}\cdot\text{cost}$$ $$\text{selling price} = \frac{100}{100 – \text{margin}}\cdot\text{cost}.$$ Or, using decimals rather than percent values, $$\text{selling price} = \left(1 +\text{markup}\right)\cdot\text{cost}$$ $$\text{selling price} = \frac{1}{1 – \text{margin}}\cdot\text{cost}.$$

Here is a 2004 question that initially dealt with the common confusion between markup and margin, then added a twist:

Calculating Percentage Markup Versus Profit In order to figure a profit of 10% I would normally take the amount and thenmultiply by 1.1to find the selling price. For example, 800 x 1.1 = 880 so I say we should charge $880 to make a 10% profit. However my employer tells me that in order to find a 10% profit I shoulddivide by .9to get 800/.9 = $888.89. I do not understand how this is 10%. Can you please explain? Thank you very much.

They are both calculating the selling price to yield a given percent profit, but they are talking past one another by not using clear terms. I replied (somehow neglecting to use the term “margin”, which I am adding here for clarity):

You and your employer are calculating two subtly different things: amarkupversus a profit [margin]. Your calculation is correct when you are given the cost and want to find a sale price that will give you aprofit of 10% OF THE COST TO YOU. If the cost to you is C and you sell it for 1.1C, your profit is 1.1C - C = 0.1C, or 10% of C. which is 10% of what you paid for the item, or a10% markup. But your employer is calculating the price that will give you aprofitof 10% OF THE SELLING PRICE. If you sell the item for C/0.9, then your profit is C/0.9 - C = (C - 0.9C)/0.9 = 0.1 * C/0.9, or 10% of C/0.9. which is 10% of the selling price, C/0.9. What he is doing is solving the equation P - C = 0.1P which says the profit (price minus cost) is 10% of the price, this way: P - 0.1P = C 0.9P = C P = C/0.9 You method solves the equation P - C = 0.1C Your employer's method is correct for a10% profit marginon the sale. Your method is correct for a 10% markup over the cost. It's all a matter of words.

We have found that this sort of miscommunication is common; that’s why we have different words for the two cases, and why the base must always be specified when we talk about percentages.

Lisa had another question:

Thanks! I can follow this now that I see the mathematical equations written out. This has been a great help. I have another similar question for you, if I could, again involving us buying and then selling an item, but now figuring tax in as well. For example, the price we pay is $250.00 (tax not included). We would then figure in 7% tax (250 * 1.07 = 267.50) to find our total cost. We then proceed to figure out our selling price by calculating 267.50/0.9 = 297.22. This selling price will give us 10% profit. Other people believe that we should do it the following way. Take 250.00 (our cost)/0.93 = 268.82 (price with tax) and then calculate 268.82/0.9 = 298.69 for the selling price with 10% profit. I have always figured sales tax as a multiplication factor but then again I had never thought about the difference between a markup of 10% and a selling price with a 10% profit either, so I wanted to verify that I truly am understanding everything correctly.

Lisa is calculating tax on their cost as a markup, and profit as a margin. Others are treating both like margins. I answered:

There could be tax issues that are beyond my knowledge, but the math aspect seems straightforward. You would divide by 0.93 if, as in the case of the profit, you wanted 7% of the price you charge to go for tax. But that isn't how tax works. The tax part of the transaction, as I understand it, is simply your payment when you buy the item, based on the amount charged to you, so your calculation is clearly correct, and the fact that the other version gives a different answer shows that it is wrong. You can always check something like this (assuming you have defined clearly what you want to do) by seeing how much the actual tax and profit are based on the number you calculate. If you charge $297.22; 10% of that is $29.72, which leaves $267.50. That is $250 plus the 7% tax. The other number will not check out this way. To put it simply, you multiply by 1.07 because thetax is calculated forward, from the cost to you; you divide by 0.9 because theprofit margin is calculated backward, from the price to your customer.

Now, this question was significantly edited from the original longer discussion, and looking back at that, I suspect I may have been wrong. In my answer I emphasized my understanding that the tax is on something they are buying (paying the tax themselves), and then reselling to the customer, who may then also have tax to pay. If the tax involved was paid by the customer as a percentage of *their* cost, things might be different.

Here’s a longer question about a similar issue, from 2011:

Mark-ups, Merchant Fees, and Multiplication ... or Division?How do you calculate a 30% mark-upon an item that normally retails for $300? If I take $300.00 and multiply it by 30% (= .30), this equals $90.00; and adding that mark-up to the original price makes it $390.00. On the other hand, 100% - 30% = 70%, and if I take $300.00 and divide it by 70% (= 0.70), this equals $428.57. Which is the proper 30% mark-up?

By now, you know the answer. What would you say?

But there’s more, similar to the case above including tax:

Similar question for an item that includes the3% merchant feeimposed by credit card companies for all purchases made by plastic. In this case, I think the purchase becomes 100.00 x 0.03 = 3.00 100.00 + 3.00 = 103.00 The credit card company charges the merchant 3% of $103.00: 103.00 x 0.03 = 3.09 103.00 - 3.09 = 99.91 So the store would lose nine cents on this transaction? However, you could also calculate the mark-up this way: 100.00 / 0.97 = 103.09 The credit card company charges 3%: 103.09 * 0.03 = 3.09 103.09 - 3.09 = 100.00 With this method, the store does not incur a loss on the sale. Why does one have to divide the $100.00 by .97 to make the credit card charge equal out? Does this type of calculation have a name? And why does dividing by .70 add more to the $300.00 sale than does the amount derived from multiplying by 1.30? Does that calculation have a name?

The fact that the store would not get the desired profit means that the first calculation is wrong!

Doctor Wallace answered:

What a great question! The correct wayto mark something upby a percentage of its value is the first way you mentioned. That is, an item costing $300, marked up 30%, would be $390. With the credit card example, there wouldn't be a problem if the credit card company actually charged3% of the selling price-- but they don't. They charge3% ofwhateverthe transaction amountis that was placed on the credit card. So, trying to account for the credit card company's merchant fee by offsetting it with a 3% markup is not going to work out. The reason is that the credit card company is taking 3% of the 3% the store is adding on, which results in a little extra.

The fee has to be treated as a margin, a percentage of the total after adding the fee, not a markup.

For example, on a $100 item, the store adds 3% to the purchase price, giving $100 + .03 ($100) = $103 Now when the credit card company takes 3% again, they're taking 3% of the 3%. This amounts to 9 hundredths of a percent (.03 * .03 = .0009) .03 * [$100 + .03($100)] .03*100 + .03(.03)(100) 3 + .0009(100) 3 + .09 = 3.09 In order for the store to not incur a loss of 9 cents on the transaction, they need to charge 3.09%, not 3%. $100 + .0309 (100) = $103.09 Now when the credit card company takes their 3%, 3% of $103.09 is $3.09, which was the full amount passed on to the customer, who paid $103.09, leaving the store with the full purchase price.

This is why the first method didn’t work.

The division calculation you mentioned does have a name. It is called themargin, not the markup. These are easy to confuse.Markup is the percentage gain based on COST, whilemargin is the percentage gain based on PRICE.

(I have corrected a typo, where “profit” was written instead of “price”.)

For example, returning to the example of an item that usually retails for $100: If you mark it up 30%, you would calculate 30% of 100, which is $30, so the selling price is $130. The markup is 30/100 = 30%. The MARGIN, however, is 30/130 = 23%. This is because selling the item for $130 results in a $30 profit, and 30/130 means that 23% of the money the store took in was profit. We say their margin was 23%. In fact, a 30% markup will always result in a 23% profit margin. To calculate the selling price at a given margin, you do what you said: divide the cost by (1 - margin percent). So if you want a 30% profit on selling this item, then your cost will be 70% of the sales price: 70% of sales price = $100 sales price = $100/.70 = $142.86 This means that you will make $42.86 profit, giving you a margin of 42.86/142.86 = 0.30, or 30% Your cost ($100) is 100/142.86 = 0.70, or 70% of the sales price.

This last calculation confirmed that the sales price did lead to the required margin. And I imagine 30% rather than 3% was used in this example in order to make differences more visible. Now back to the actual question:

So when the store divides the purchase price by 0.97, it is a quick way to calculate the sales price (not the markup) in a way that ensures that they get a margin of 3% (extra) profit, all of which will be passed on to the credit card company, leaving the store with no loss on the sale. The hard thing is to explain to a customer how this all works, and why they are getting charged $3.09 instead of $3.00 for a 3% fee!

I imagine the customer wouldn’t normally see how much the fee is …

]]>We can start with this question from 1996:

Price before Discount A calculator is discounted 20 percent. After adding 7 percent tax, the total cost is $20.07.What was the list priceof the calculator, without tax, before discount?

Here we have to back off both a tax (which amounts to a percent increase) and a discount (a percent decrease). So this question covers both.

Doctor Brian answered:

When20 percent is discountedfrom a price, you still have to pay 80 percent of the original price. When7 percent tax is addedto a price, you have to pay that plus 100 percent of what you normally would, for a total of 107 percent of the actual price. We'll use these two ideas in the problem. If x is the original price, then we are paying 107 percent of (the sale price), which is 107 percent of (80 percent of the original price). And since the word "of" means "times" in percent problems, we get 1.07 * .80 * x = 20.07 .856 * x = 20.07 x = 23.45

Recall from last time that 80% is called the **complement** of the 20% decrease, and represents the remaining percentage. The 107% is similar (but unnamed, to my knowledge): It represents what percentage the new value is of the old value.

What he has done here is first to **combine** the two changes (7% increase and 20% decrease) by multiplying 107% and 80% to get a net multiplier of 85.6%. To undo this, we just **divide** by the multiplier. We could instead have merely divided **separately** by each multiplier (1.07 and 0.80) to get the answer.

Turning these into formulas, we can reverse a P% decrease using $$\text{original} = \frac{100}{100-P}\cdot \text{new},$$ and we can reverse a P% increase using $$\text{original} = \frac{100}{100+P}\cdot \text{new}.$$

We can also **check** the answer. From a list price of $23.45, a 20% discount would be $4.69, reducing the price to $18.76. A 7% tax on this would be $1.31 (rounded from 1.3132); adding this, the total price is $20.07, which is just what we want. (I recommend always doing this check in real-life problems, because rounding can influence the results.)

The next question, from two months later, picks up where that left off:

Restoring an Original Price I know there are formulas for restoring a price to its original value after taking a discount on it. How do you determine what themultiplieris forgetting back to your original number? Example: I start with a cost of $48,695. I discount it by 30 percent, bringing it to $34,087. What do I multiply $34,087 by to get back to $48,695? Your help in this matter would be very much appreciated. Please explain the method of getting the formula.

Doctor Mike answered, first answering the specific question before broadening it:

If you discount by a certain percentage, like 30 percent, thediscounted price is 70/100(or 70 percent) of the original. You thenmultiply the discounted price by 100/70(about 1.4286) to get the original price back. This multiplier is equivalent to 142.86 percent of the discounted price.

All we’ve done is to take the 70% (complement) and stand it on its head, so to speak: Rather than divide by it, as we did in the first answer, we can **multiply by its reciprocal**.

We can also turn this into a percent increase:

The 142.86 breaks down to 100 + 42.86, which means that to get back up to the original price you have to take 100 percent of the discounted price and add to that 42.86 percent of the discounted price. The 42.86 percent part of it is the "percent of increase" over and above the discounted price. This can be surprising at first; that youdecrease by 30 percentbut then mustincrease by about 43 percentto get back up to where you started. This is just another way of saying that "30 percent of the original price" is about the same as "43 percent of the discounted price."

Here we are reversing the process we used to change a 7% increase to a 107% multiple, by subtracting 100% from 142.86%.

And it turns out that the discount (30% of $48,695 = $14,608.50) is about 43% of the original price (42.86% of $34,087 = $14,609.69, a little off due to rounding).

This can be turned into a formula:

What we did for 30 percent we can do for any percentage "P". The discounted price is (100-P)/100 of the original. You multiply the discounted price by 100/(100-P) to get back up to the original price. Multiply this by 100 to get the equivalent percentage (like 142.86 above), and then subtract 100 to get the percent of increase (like 42.86 above). / 100 \ 100*P | ------- * 100 | - 100 = ------- = Grossing Percentage \ 100-P / 100-P Checking for P = 30% , (100*30)/(100-30) = 3000/70 = 42.85714.... If you want to work with multipliers rather than percentages of increase, just use "100/(100-P)" for the "grossing up" multiplier.

“Grossing up” means determining the “gross” (before discount or tax) amount required to obtain a given “net” (after discount) amount. The formula to reverse a P% decrease is $$\text{% increase} = \frac{100P}{100-P}.$$

The next question, from 2002, is similar, but about a percent increase:

Undoing Percentage Changes If 10000 x 102% = 10200, then how do I figure thepercentage to get backto my base number (10000), e.g., 10200 * 1.96% - 10200 = 10000.08?

Here we have a 2% increase, and want to find out that it will be undone by a 1.96% decrease. (Elias had his work backward, and meant 10200 – 10200 * 1.96% = 10000.08. And, of course, this is not exactly 10000 because of rounding.)

Doctor Ian started with the formula rather than the specific example:

It's probably easier to see how this works if you use variables instead of numbers. If your original amount is A, and the percent increase is p, then the new amount is A' = A(1+p) You want to decrease it by some percentage q, to get back to A. That is, you want to find q such that A = A(1+p)(1-q) 1 = (1+p)(1-q) 1/(1+p) = 1 - q q = 1 - 1/(1+p) = (1 + p - 1)/(1+p) = p/(1+p)

This is much like the formula in the previous answer, except that (a) the sign is changed because it was a percent increase, and (b) Doctor Ian, like most mathematicians, is working with the decimal forms rather than the actual percent number, so that his *p* is not 2, but 2% = 0.02. Using the same notations as we saw above for grossing up, the formula to reverse a P% increase is $$\text{% decrease} = \frac{100P}{100+P}.$$

Let's check this with a simple example. If weincrease something by 100%, we should have todecrease it by 50%to get back to where we started: q = 1.0 / 2.0 = 0.5 If we increase something by 1/3, we should have to decrease it by 1/4: q = (1/3) / (4/3) = 1/4 So this seems to work okay. So if p is 2 percent, q would be q = 0.02 / 1.02 Does this help?

And this *q* turns out to be 0.0196078…, which rounds to Elias’ 1.96%. It could also have been calculated as 2/102.

Here’s a 2003 question that focused on distinguishing what we’ve been doing here from a straight percentage calculation:

Percentage Increase vs. Percentage If there were 10,000 claims in 2001, and that is a300 percent increasesince 1999, how many claims were there in 1999? My colleagues are giving me all different answers. I think the answer is 2,500. My colleagues say 3,333.

Very likely the large percentage was part of the reason for the dispute, as it can confuse people. (We’ll be looking at that soon!) Doctor Ian replied:

A300% increasemeans that (claims in 2001) - (claims in 1999) 300 ----------------------------------- = --- (claims in 1999) 100 That is, we're saying that the _increase_ is 300% of the original value. Since the numbers are so nice and round, we can do this: (claims in 2001) (claims in 1999) 300 ----------------- - ---------------- = --- (claims in 1999) (claims in 1999) 100 (claims in 2001) ----------------- - 1 = 3 (claims in 1999) (claims in 2001) ----------------- = 4 (claims in 1999) So the number of claims in 2001 is 4 times whatever it was in 1999, which means there were 2,500 claims in 1999.

Another way to say this would be that a 300% increase means that the new value is 100% + 300% = 400% of the original (that is, 4 times the original); so to get back to the original, we divide by 4.

So how did the others get their answer? They were missing the word “increase”:

Note that if we change the wording slightly, we can come up with the other answer. That is, if we say that the number of claims in 2001 is300% of the number of claimsin 1999, then we're saying (claims in 2001) 300 ---------------- = --- (claims in 1999) 100 which we can rearrange to get 100 * (claims in 2001) (claims in 1999) = ---------------------- 300 100 * 10,000 = ------------ 300 10,000 = ------ 3 = 3333 1/3

That is, if the new is just 300% of the old (3 times as much), we just divide by 3. Doctor Ian didn’t mention it, but their misreading of the problem may very well have been due to the thinking discussed in this post: Three Times Larger: Idiom or Error?

Let's put the two cases together, so you can compare them more easily: 1) Claims in 2001 are anincrease of 300% overclaims in 1999. (The increase is 300% of the old value.) (claims in 2001) - (claims in 1999) 300 ----------------------------------- = --- (claims in 1999) 100 2) Claims in 2001 are300% ofthe claims in 1999. (The new value is 300% of the old value.) (claims in 2001) 300 ---------------- = --- (claims in 1999) 100 Does this make sense? I find that it's useful to keep a few simple examples in my head. One of my favorites is this: If I start with $1, an increase of 100% is an increase of $1, which gives me $2, but $2 is twice as much as $1, which means it's 200% of $1.

The fact that a percent increase is reversed by a *different* percent decrease leads us to a common confusion that is, as we’ll see, complicated by calculators. This question is from 2002:

Dueling Percentages Can you tell me why, for instance, when youadd 15%to 100, you get 115, but if yousubtract 15%, you get 86.95...instead of 85?

Pete’s calculation is wrong, so it needs to be diagnosed: What is he really doing? Doctor Ian answered:

Are you sure about that? 100 + (15% of 100) = 100 + (0.15 * 100) = 100 + 15 = 115 100 - (15% of 100) = 100 - (0.15 * 100) = 100 - 15 = 85

Increasing or decreasing by 15% mean adding or subtracting 15% of the same number.

I think what you're asking is why, when you reduce something by 15%, increasing the reduced figure by the same percentage doesn't get you back to where you started, e.g., 85 + (15% of 85) < 100 86.95 + (15% of 86.95) = 100

We get 86.95 by dividing 100 by 1.15, so that increasing *this*, not the 85, by 15% gets us to 100.

So what Pete called **decreasing** by 15% was really **undoing an increase** by 15%. We’ve already seen that these are different things. Why?

The clearest way to show why it works this way is to use symbols instead of concrete values. Suppose we start with some amount, A, and we reduce it by some percentage, P. Then the new value is A' = A(1 - P) Now if we increase the reduced amount, by the same percentage, we get A'' = A'(1 + P) = A(1 - P)(1 + P) = A(1 - P^2) which is smaller than what we started with!

In our example, decreasing by 15% and then increasing by 15% results in a number that is only \(1 – 0.15^2 = 97.75\%\) of what we started with. And this is because what we added is 15% of a smaller number than the 100.

It can also be easier to grasp if you use extreme examples. If you start with $100, and I take 99% of it, that leaves you with $1. If I increase your dollar by 99 percent, will that get you anywhere near $100?

Subtracting 99% of $100 ($99), then adding back on 99% of $1 ($0.99) gets you only to $1.99, nothing near $100. (We’ve seen this idea of going to extremes several times recently.)

And here is a third way to look at it. Suppose we start with some amount, +---+---+---+---+---+ | | | | | | +---+---+---+---+---+ | | | | | | +---+---+---+---+---+ | | | | | | +---+---+---+---+---+ | | | | | | +---+---+---+---+---+ | | | | | | +---+---+---+---+---+ and reduce it by some percentage (say, removing one item out of every 5): +---+---+---+---+---+ | | | | | - | +---+---+---+---+---+ | | | | | - | +---+---+---+---+---+ | | | | | - | +---+---+---+---+---+ | | | | | - | +---+---+---+---+---+ | | | | | - | +---+---+---+---+---+ Now we want to increase it by the same percentage (i.e., adding one item for every 5): +---+---+---+---+---+ | | | | | - | +---+---+---+---+---+ | | | | | - | +---+---+---+---+---+ | | | | | - | +---+---+---+---+---+ | | | | | - | +---+---+---+---+---+ | | | | | - | +---+---+---+---+---+ | + | + | + | + | +---+---+---+---+ The increase, being the SAME fraction of a SMALLER number, will be smaller than the reduction, so we end up replacing only SOME of the amount that was removed: +---+---+---+---+---+ | | | | |-/+| +---+---+---+---+---+ | | | | |-/+| +---+---+---+---+---+ | | | | |-/+| +---+---+---+---+---+ | | | | |-/+| +---+---+---+---+---+ | | | | | - | +---+---+---+---+---+ The difference between the increase and the reduction is the square of the percentage.

That little square at the lower right is the difference.

Pete responded, explaining what his question really meant:

Thanks very much for replying yesterday to my question. The problem lay in the fact that I was using a desktop calculator, andthe -TAX buttonfor some unknown reason does not giveminus 10% of 100as 90. Any ideas why this is so, as it happens on all models?

Not all calculators have +TAX and -TAX buttons; in fact, those used by mathematicians or in math classes don’t:

I've never seen a -TAX button on a calculator, but it sounds to me as though it might be assuming that you know the price WITH tax, and the tax rate, and you want tofigure out the price WITHOUT tax. For example, if the price with tax is $100, and the tax is 10%, and the price without tax is P, then P(1 + 0.10) = $100 P = $100 / 1.10 = 90.91 In other words, a retail price of $90.91, with 10% tax added to it, comes to $100.

So the button is for **reversing a percent increase** (removing a known tax), not for **making a percent decrease** (as if there were such a thing as a negative tax). Pete was misunderstanding the label on the button.

The same issue comes up with % buttons that many (non-scientific) calculators have, as this 2006 question shows:

Finding Percentages on a Calculator How do I use a calculator to check percentages? For example, how would I check 6% of 180,000 using a calculator? Would I multiply 180,000 x 6.00%?

It’s encouraging that Donna doesn’t use the calculator to do her work, but to check; that’s what I recommend to students. (Or did she just say that to avoid offending us?)

I took this one, having a lot of experience with it:

Probably! Unfortunately, I've seen some variation in how calculators handle percents, so you'll want to check the instructions that came with yours to be sure. Buttry it out and see!What I'd do, if I had an unknown calculator with a % key, is to enter something like 200 x 5 % = and see if it gave the right answer, namely 10. (Also note whether it gives the answer even before you press "="; some work that way, and then you should learn NOT to use "=" here. That is true, for example, of the calculator that comes in Windows.)

Some calculators will immediately replace 5% with 0.05 in their display, so the multiplication above amounts to \(200\times 0.05 = 10\), which is correct. This mirrors how I think about such calculations. On this type, the “=” is needed. (The current Windows calculator, in Standard mode, works this way, not as I described.)

Then I might also try 5 % x 200 = and see if that worked too. (On many calculators, it won't, so I'd have learned not to do it in that order!)

The Windows calculator, when I tried this, replaced 5% with 5% of the number that I’d left in the display from a previous calculation, which would be a disaster! Two calculator apps on my phone both handled this correctly.

There's much more variation in how they handle percentage increases, like 200 + 5 % which many calculators, but not all, would take as "increase 200 by 5%", giving 210. If you need to do things like that, you'll want to be sure to check the manual.

The Windows calculator, and my two calculator apps, all replace 5% here with 10, implicitly calculating 5% of the number being added to, so that pressing “=” results in 210, the correct answer.

On the other hand, two scientific calculators with a % button (a TI and a Casio) both do what I would expect of such a calculator, converting 5% to 0.05, then adding to get 200.05, which is what the expression literally means.

In other words, calculators meant to be used in financial calculations take “+ 5%” to mean “add 5% of the preceding number”, twisting the notation to fit the application. Calculators meant to be used in contexts where math notation is interpreted carefully don’t do that, in part because it would make it impossible to be sure how to interpret more complicated expressions.

Let’s close with a summary of how to apply and reverse percent increases and decreases, in response to this 2002 question:

Markups and Discounts I have been selling products over the past 30 years, and I still run into this question. What is the correct way to markup or discount a product? $100 markup (x) 1.1 = $110 = 10% $100 markup (/) .9 = $111.11 = 10% $100 discount (/) 1.1 = $90.90 = 10% $100 discount (.9) = $90 = 10% I find it easier to mark up (/) .9 .8, which equal 10% 20% etc., and then mark down (x) .9, .8, .7, which equal 10 to 30%. It seems faster if there is no other explanation. I have yet to find anyone who can explain it to me. Thanks, Gary B.

His methods are a mix of the forward and backward methods. I replied, explaining what each calculation really does:

If youmark up by 10%, you are adding 0.10 to the price of an item: new price = old price + 0.10 * old price = 1 * old price + 0.10 * old price = old price * (1.10) (I am using "*" for multiplication.) To determine theoriginal pricegiven the marked up price, you have to undo this by dividing: old price = new price / 1.10 If youdiscount by 10%, you are subtracting 0.10 of the price of the item: new price = old price - 0.10 * old price = old price * (0.90) To determine theoriginal pricegiven the discounted price, you have to undo this by dividing: old price = new price / 0.90

To emphasize that these are different, I added:

Note that if you mark up by 10% and then discount by 10%, you don't get back the original value, because you are taking off 10% of a larger amount: $100 * 1.10 * 0.9 = $110 * 0.9 = $99 Here we took off $11 for the discount, not the $10 we added in the markup. I think this differs from what you wrote, if I understand it correctly. You always use 1.10 (1 plus the percentage) for markups, and 0.90 (1 minus the percentage) for discounts. Youmultiply to applythe markup or discount, anddivide to undoit.

Next time, we’ll look at another perspective on this: Gary may in part have been confusing **markup** and **margin**.

Let’s start with a basic question from 2002:

Finding Sale Price

We are studying how tofind the sale price by subtracting the discountfrom the regular price. Can you help me?

I answered:

We have a number of answers to questions in this area, which you can find by searching our site. Most of these deal with the reverse problem, finding the original price when you know the sale price. Here's a sample problem in the forward direction: If the regular price is $8.99, and the item is being sold at 30% off, what is the sale price? Well,30% of $8.99means 30/100, or 0.30, times $8.99: 0.30 * 8.99 = 2.697 That much is beingtaken off the price, so the sale price is $8.99 - $2.697 = $6.293 which would be rounded down to $6.29.

So, the basic idea of a percent decrease, such as a discount, is to find that percent of the given number, and subtract it. The formula for this would be $$\text{new value} = \text{old} – \frac{\text{percent change}}{100}\times \text{old}.$$ In the example, I directly converted 30% to the decimal 30/100 = 0.30, and multiplied by that.

Another way to do this would be to recognize that if 30% is being taken off, then what's left is 70%; so you can just multiply $8.99 by 70% to get the sale price: 0.70 * $8.99 = $6.293

This method looks like $$\text{new value} = \left(1 – \frac{\text{percent decrease}}{100}\right) \times \text{old}.$$

That second method is the subject of this 2001 question:

Opposite of Percentage

I am beginning to wonder if there *is* an answer to this. I am trying to write a document explaining a common procedure, namely,calculating a discounted price. To calculate the discount, you multiply the "base" by the "percent discount" to yield "percentage." At least, these were the terms I was taught. So, if the price is $32 and the discount is 15%, then you multiply 32 times .15 to get the amount that is to be subtracted from the initial price ($4.80) to yield the discounted price ($27.20). I have been taught that the "initial price" ($32) is the "base," the .15 is the "percent" or "discount rate," and the resulting discount amount ($4.80), is the "percentage." Another way to calculate the final price would be to subtract the percentage from 100 (100% - 15% = 85%) and then multiply that result by the original price (.85 times $32 = 27.20). Or, as it appears in many computer programs, net = price * (1-discount_rate), where it is understood that discount_rate is a decimal fraction. My question is one of terminology.What is the name given to the term "(1-discount_rate)" in this example? I have looked up reciprocal, complement, inverse, etc., but none of those fits. I need to be able to define this word once and then use it repeatedly in my document, rather that explaining it every time I use it, which will be many times.

One of the terms Marty looked up turns out to be the answer, as I explained, using what is now a dead link:

What you want is "thecomplement of the discount rate." Here is one site I've found that defines it: Trade Discounts - Business Mathematics, Confederation College Trade discounts are used in the retail sector.

They are used to calculate how much a retailer

will pay a manufacturer for a given product... Complement of a Trade Discount: This is the

difference between the discount rate and 100%.

The complement can be any percent that results

when you subtract the trade discount rate from 100%. In general mathematicians use the word "complement" to mean "all except ..." or "the whole minus a given part"; here it means 1 minus a fraction - though I haven't found any dictionary that gives this definition.

So we can apply a discount (percent decrease) using the complement: $$\text{new value} = \text{old}\times (1 – \text{percent decrease}).$$

Similarly, we can apply a percent increase (markup – we’ll be discussing this later) this way: $$\text{new value} = \text{old}\times (1 + \text{percent increase}).$$

Here’s an interesting question from 2002:

Order of Discounts and Taxes

I'm wondering if you could tell mewhy it doesn't matterin determining the cost of an itemwhether you take off the tax before or after you take off the saving rate. For instance, if I buy an item that originally costs 100 dollars during a 20% off sale and the tax is 10% (for sake of simplicity), intuitively I think that I want to take the 20% off before I calculate the tax, so I will be taxed on a lower amount, but it doesn't seem to matter whether I take the tax off the 100 dollars first and then calculate the savings, or I take the savings first and then calculate the tax.Am I right that order doesn't matter, and is there an explanation?

Since the question was about why it works rather than how to do it, Doctor Ian replied with an algebraic view:

As with most things, it's often easier to see what's going on if you use variables instead of actual numbers. Suppose the normal price of an item is P, and the tax rate is T. Then the amount you'd normally pay for the item is P(1 + T) Now, suppose you discount the price by some percentage, D. If you do it before the tax, you end up paying [P(1 - D)](1 + T) But if you apply the discount to the after-tax amount, you end up paying [P(1 + T)](1 - D) Since you're just multiplying things together, the order doesn't matter; so you end up paying the same thing in both cases.

He is using the complement to find the discount, and its equivalent to apply the tax. As an example, take that 20% discount and 10% tax. The discount results in paying 80% of the full price; the tax leads to paying 110%. So the net price is \($100\times0.80\times1.10 = $88\). Taking them one at a time, the discount on $100 would be $20, leaving $80; the tax on $80 is $8, so you pay $88. In the other order, the tax on $100 is $10, so you owe $110; but the discount on $110 is $22, so you only pay $88, the same as before.

Here is a question from 2005 that focuses on finding the percentage, given the two numbers:

Formula to Calculate Percent Increase I want to compare two figures and show the percent increase that has been made. For example, if 55 has increased to 75,what percentage increase is that?Whenever I get a figure, I am never sure if there isa way of checkingback that it is correct.

Doctor Ian took this, explaining the appropriate formulas for both the **calculation** (finding the percent) and the **check**. First, what does a percent increase *mean*?

When we say Q increased by P percent what we _mean_ is that the new value of Q is Q' = Q + (P percent of Q) which is to say, Q' = Q + (P/100 * Q) So far, so good?

He is defining percent change in terms of how it is **applied**: If we increase *Q* by *P*%, we are adding *P*% **of Q** to

Now we reverse the calculation to **find the percentage** by which *Q* was increased to get *Q*‘:

If you have Q (the old value) and Q' (the new value), then you want to find P. If we solve the equation for P, we get Q' = Q + (P/100 * Q) Q' - Q = P/100 * Q Q' - Q ------ = P/100 Q Q' - Q ------ * 100 = P Q So yousubtract the old value from the new one, and divide the difference by the old value. Then multiply by 100 to get the percentage.

Having seen the algebra, you don’t need to repeat it each time; you can just use that formula: $$\text{percent change} = \frac{\text{new } – \text{ old}}{\text{old}}\times 100\%.$$

Another way to think of that is: amount of change ---------------- x 100 = percent change original amount

I find this easier to remember than an actual formula: We are simply asking, “**What percentage of the original amount is the change from that to the new amount?**” That, of course, is exactly what the phrase “percent change” means.

I also like to think of the multiplication by 100 as a unit conversion, where we multiply a number (typically a decimal) by 1, in the form of 100%, to get an answer measured in percent.

An example will make it clearer. First we calculate a percent change, then check by applying it:

If you want to check your work, you can go back to the definition, plug the numbers in, and see if what you get is true. For example, suppose I have an increase from 40 to 50. The formula gives me 50 - 40 10 P = ------- * 100 = -- * 100 = 25 40 40 So I think it's 25%. Let's go back to the definition: 50 = 40 + (25% of 40) = 40 + 10 = 50 This is true, so I have the right percentage increase.

The change divided by the old value was 0.25, which is 25%. And the check is to apply that percent change to the old value and see that we get the new value: \(40 + 0.25\times40 = 50\), the final value we started with. It’s good!

The same sort of work applies to finding a **percent decrease**; we just get a negative number if we subtract old from new. We still divide by the original amount.

Another version of the formula is obtained by dividing each term of the numerator by the denominator: $$\text{percent change} = \left(\frac{\text{new }}{\text{ old}} – 1\right)\cdot 100\%.$$ For example, in the case above, we have $$\left(\frac{50}{40} – 1\right)\cdot 100\% = (1.2 – 1)\cdot 100\% = 0.2\cdot 100\% = 20\%.$$

Here is a similar question from 2002, with a different context:

Percentage Error

Please help me find thepercentage errorfor the following numbers. 5.7 estimated findings 5.8 actual findings

Doctor Achilles took the question, starting with the formula:

Percentage error is just how much your guess was off from the actual value. The formula is: |estimate - actual|/actual * 100% [That is: the absolute value of (the estimate minus the actual) all divided by the actual, all multiplied by 100%.]

Note that this is the same formula we saw above, except that, by taking the **absolute value**, we are ignoring the direction of the error. We are treating the **actual** value as the “original”, from which the **estimate** deviates by some percentage, called the percentage error.

But he goes beyond the mere answer, focusing on the reason for the calculation:

Let's think aboutwhy we use this formula. If you want to know how close your estimate is, the first thing to do is just to askhow much you missed by, that is the absolute value of the difference between the two numbers (the absolute value is used because you're only concerned with how much you missed by, not whether you were too high or too low). So in this case, you missed by 0.1. 0.1 is a small number, so it sounds like your guess is pretty good. So why do we bother with this business of dividing by the actual amount?

Let's take a couple of other examples: Example 1: 100.0 (estimated) 105.3 (actual) Example 2: 10.0 (estimated) 15.3 (actual) Example 3: 1.0 (estimated) 6.3 (actual) Example 4: 0.1 (estimated) 5.4 (actual) In all cases, you missed by the same amount (5.3). But in the first example, it seems as if your guess was a lot better. Even though it's off by just as much as the last example, missing by 5.3 out of 105.3 isn't too bad, but missing by 5.3 out of 5.4 seems pretty darn bad.So we're concerned here not just with how much you missed by, but with what percent of the actual value you missed by.So in example 1, you missed by 5.3 out of a total of 105.3: 5.3/105.3 * 100% equals 0.0503 * 100% equals 5.03% And in example 4, you missed by 5.3 out of total of 5.4: 5.3/5.4 * 100% equals .9815 * 100% equals 98.15% This is a good way to represent the intuition thatmissing by 5.3 out of 105.3 is a pretty good guess, but missing by 5.3 out of 5.4 is a pretty rotten guess. In one case, your percent error is only about 5% (small error), while in the other case, your percent error is a big 98% (huge error).

So the point of the percent error is to put the amount of error into context; it’s also called relative error.

The discussion above related to a percent increase or decrease **from one number to another**. The next question, from 1997, gets us into another kind of percent change, through variations in wording.

Percent Change, Increase, Difference

I haven't had math in years and I need a little help here. Say you have two numbers 5 and 7. You want to know what is the difference in percent between the two numbers. Below is an real example of the problem I'm working on. 1991 = $2346.80 1992 = $3608.29 Percent change 91-92 = ?

The two examples are potentially quite different, because the first just has two numbers with no context, while the second is a practical type of problem with a specific meaning. Doctor Rob answered, starting with the latter, which is the same type we did above:

First you need to compute thedifference between the two amounts. Then you make afraction of that difference over the first of the two amounts. Finally you convert the fraction to a percentage by dividing the denominator into 100 times the numerator. In your example, dropping the dollar signs, the difference is 3608.29 - 2346.80 = 1261.49. The fraction is 1261.49/2346.80, and the percentage comes from dividing 2346.80 into 126149. The result is 53.75, and since the sign is positive, this is a 53.75 percent increase from 1991 to 1992.

Here we are specifically asked for the change from 1991 to 1992, with a definite “old” and “new”, so we do just what was done above.

But the first example is different, and a little ambiguous. Which number comes first?

In the other example,5 and 7, the difference is 2, the fraction is 2/5, and the percentage is 5 divided into 200, or 40. Thus 7 is40 percent largerthan 5. On the other hand, if you take7 and 5, the difference is -2, the fraction is -2/7, and the percentage is 7 divided into -200, or -28.57. Thus 5 is28.57 percent smallerthan 7.

So the percent change from 5 to 7 is \(\frac{7 – 5}{5}\cdot 100% = \frac{2}{5}\cdot 100% = 40%\), while the percent change from 7 to 5 is \(\frac{5 – 7}{7}\cdot 100% = -\frac{2}{7}\cdot 100% = -28.57%\).

It may seem paradoxical that if we start with 5 andadd 40 percent, we get 7, but to get back to 5 we have tosubtract only 28.57 percent. The reason these percentages are different is that they are percentages of different amounts: 40 percent of 5, and 28.57 percent of 7 - both are equal to 2.

We’ll be coming back to this issue (increase and decrease being asymmetrical because of the different bases) next time.

But there’s another issue, which was brought up by a reader, Jerone, in 2003 (and edited into the same page):

The above explanation definespercent differenceas ((q1-q2)/q2)*100, which is used in many calculations forpercent errorrather than percent difference. Percent difference is defined as (|q1-q2|/((1/2)*(q1+q2)))*100 for comparing values. I am confused, as the terms percent difference and percent error are not consistent with what is expected in many classes.

There is both a contextual and a semantic issue here; technically, no one above used the exact term “percent difference”, and what Allistair was asking for (using the inaccurate term “difference in percent”) is indeed a **percent change**, equivalent to what Jerone calls a **percent error** (as we saw above under that title). The **percent difference** is a case we haven’t seen yet.

I replied:

The above is really about "percent change" or "percent increase," rather than "percent difference," sincethere is an "old" and a "new" value. In that case, you take the percentage of the old value.When there is no directionalityto the difference, so you can't distinguish an "old" value from a "new" value, you have to use the average of the two as the standard of comparison, as in your formula. This is one of many cases where terms are used differently in various contexts, and what is a minor error of terminology in one question can lead to major confusion for people coming at it from a different direction.

“Percent difference” properly refers to a comparison between two numbers, neither of which takes priority as “original” or “correct”.

To sum up, when we are describing a **change** (**increase** or **decrease**) from one value **to** another, or the difference from a correct value to an estimate (**percent error**), the formula is $$\text{percent change} = \frac{\text{new } – \text{ old}}{\text{old}}\cdot 100%;$$ but when we just have two numbers, treated as equals, we use $$\text{percent difference} = \frac{a – b}{\frac{a + b}{2}}\cdot 100% = \frac{2(a – b)}{a + b}\cdot 100%.$$

This is the context of the recent question, which came from George in March:

If I’m baking a clay brick in an oven and know that it shrinks by 25% as a result of this process, and I need the final size to be one meter, I’m looking for the best way to determine the initial size.

I’m thinking the percentage difference equation probably comes into play:

x% = (a-b) / ((a+b)/2)

So, if x=25 and b=1, what’s the equation we use to solve for a? Let me know.

Here George is using the percent *difference* formula, but he is really doing a percent *decrease*. I replied:

Your equation is for the

percentage difference between two numbersthat arenot distinguished(e.g. as original and new, or true and measured, for example). SeePercent Change, Increase, Difference.

For a

percent change, you just add or subtract the percentage of that value. in your case, you have an (unknown) initial size x that is decreasing by 25%, so 25% of x is subtracted from x. That leavesfrom a given valuex – 0.25x = 1x – 0.25x = 0.75x

To reverse this, you can divide by 0.75. So if the final size is to be 1, we have to solve0.75x = 1

so the solution is

x = 1/0.75 = 1.33… = 4/3.

In other words, the new size is 100% – 25% = 75% of the original, so the original is the new divided by 0.75. This is a division by 3/4, which is the same as multiplying by 4/3. The item has to be 1/3 larger to offset a decrease of 1/4.

We’ll be getting to this reversal process next time. But the question raises the interesting question: How do you reverse a *percent difference*, to find the “other” value when you know one value and the percent difference. Give it some thought!

Let’s begin with the paragraph in our FAQ on infinity that makes the claim we’ll be looking at:

FAQ: Large Numbers and Infinity Now for the fun part! Even though infinity is not a number,it. Mathematicians divide infinite sets into two categories, countable and uncountable sets. In a countably infinite set you can 'number' the things you are counting. You can think of the set of natural numbers (numbers like 1,2,3,4,5,...) as countably infinite. The other type of infinity is uncountable, which means there are so many you can't 'number' them. An example of something that is uncountably infinite would be all the real numbers (including numbers like 2.34... and the square root of 2, as well as all the integers and rational numbers). In fact,ispossible for one infinite set to contain more things than another infinite setthere are more real numbers between 0 and 1 than there are natural numbers (1,2,3,4,...) in the whole number line!

The “numbers” that are used to identify the “size” (cardinality) of infinite sets are called **transfinite numbers**. The claim here is that the cardinality of the real numbers (or of any interval) is **greater** than the cardinality of the integers — it is **uncountable**.

For some initial ideas on how to “count” infinite sets (such as the rational numbers), see:

One-to-One Correspondence and Transfinite Numbers Infinite Sets Counting Rationals and Integers

Here is a 1997 question about the claim that there are more real numbers than integers:

Size of Infinity How can one infinity be bigger than another infinity? For example, the number of real numbers between 0 and 1 is infinite. The number of integers greater than 1 is also infinite. Yetthe first infinity mentioned is bigger than the second one. I don't get it.

Doctor Rob replied:

This has puzzled many people over the years. It has to do with the way we count. When you count objects, often you touch (or look at) one of them and say "One." Then you touch another and say "Two." You continue touching previously untouched objects until you run out of them. Then the last number you said is the number of objects. In the abstract, what you are doing is setting up aone-to-one correspondencebetween the objects and a finite set of natural numbers of the form {n | 1 <= n <= N}. It is one-to-one because no number is used twice, and it is a correspondence because no object is used twice.

That is, one object corresponds to one number. We say that the cardinality of the set (that is, the cardinal number for it) is N, the last number used.

An extension of this can be used to "count" infinite sets. Again a one-to-one correspondence is set up between the set ofall natural numbersand the set of objects. If all the objects correspond to a natural number, and all natural numbers correspond to an object, then the set of objects is said to have the same _cardinality_ as the set of natural numbers. The same procedure can be used for any two sets. If they can be put into one-to-one correspondence, then they are said to have thesame cardinality. In the case of a finite set, the cardinality is just the number of objects. In the case of infinite sets,one might naively think that all of them would have the same cardinality, and call that "infinity."

That is, it feels natural to think of infinity (∞) as a “number” that counts *any* infinite set, even though we know that infinity can’t really be used as a number, because it doesn’t follow the usual laws. (For instance, ∞ + 1 = ∞, which is impossible for actual numbers.)

In fact, there are infinite sets A and B whichcannotbe put into one-to-one correspondence with each other, but A can be put into one-to-one correspondence with a subset of B. In this case, the cardinality of A is the same as that of the subset of B, but not the same as the cardinality of B. B is said to havelarger cardinalitythan A. Notice that in the case of finite sets, the cardinality of a proper subset of a finite set is always smaller than the cardinality of the whole set, so the analogy is apt. Your example is like that: the natural numberscannotbe put into one-to-one correspondence with the real numbers between 0 and 1, although they can be put into correspondence with a subset (via n <-> 1/(n+1), for example).

We always have to be careful with infinity, though; the cardinality of an infinite subset of an infinite set can be the same as the cardinality of the whole set. An easy example to prove (though not to grasp) is that there are “just as many” even integers as there are integers. (To prove this, just let each integer *n* correspond to the even integer 2*n*. This matches …, -2, -1, 0, 1, 2, … with …, -4, -2, 0, 2, 4, … .)

Doctor Rob’s last paragraph matches the natural numbers 1, 2, 3, .. with the real numbers 1/2, 1/3, 1/4, … in the interval (0, 1).

The hard part is to*prove*that constructing a one-to-one correspondence isimpossiblein certain cases. That is done bycontradiction, by making the assumption that one exists, and from that deducing a false statement. This means that the assumption had to be false to begin with, and the existence is therefore impossible. In this case,suppose there were such a correspondence. Make a table with the first column containing the natural numbers 1, 2, 3, 4, ..., and the second column containing the corresponding real numbers between 0 and 1, written out in decimal expansions. It would take the form: 1 0.*********... 2 0.*********... 3 0.*********... 4 0.*********... ... ...

Note that we are not claiming there actually **is** such a list; we are supposing that there was, and showing that something would go wrong. So don’t try too hard to imagine that list.

According to the assumption,all real numbers between 0 and 1will appear in some row in the second column. Nowwe will construct one which doesn't appear, as follows. Consider the number D beginning "0." and continuing with the digits appearing on the *diagonal* of the array of digits in column 2 of the table. Nowpick any number X whose digits *disagree* with D in every decimal place, but which contains no 0 or 9.(There are 7 or 8 choices for each digit of X, so this is easy to do.) Where could it appear on the list? For any natural number k, it cannot be in row k, because its k-th digit differs from the k-th digit of the number appearing in row k. This number is not on the list in any row. As a result, the assumed one-to-one correspondence doesn't exist. Notice that it did not matter in what order we put the numbers in column 2 above, the same construction works. Notice also that there are huge numbers of X's which can be constructed this way: at least 7^k possibilities for just the first k digits to the right of the decimal place; and *none* of them is on the list. (The funny business with 0 and 9 is to avoid the fact that some numbers have two decimal expansions: for example, .500000000... = .499999999... . The constructed X misses both of these representations whenever they exist.) I hope I haven't boggled your mind with this. Mostly it wasn't known until the time of Georg Cantor, in the 19th century.

There are a couple tricky points in the proof, as a result of which it is very easy to give a form of the proof that leaves some loopholes (such as the 9’s). Here is another question about it, from 2001:

Orders of Infinity I recently read a book about infinity that set forth several arguments for why there are different sizes or orders of infinity.None of them seems convincing to me, and, since my ignorance of math is vast, I was wondering if you might fill me in on where my reasoning has gone wrong or provide more convincing lines of argument to me.

Doctor Luis answered; I’ll extract just the central part of his answer, and modify it according to a suggestion from a Professor Loewen, who wrote to us in 2010 to point out a typo that made the proof as originally given incorrect, and an additional modification (following Doctor Rob’s mention of 0 and 9 above) that would make it complete. Somehow this correction never made it into the archive. Here is the proof, as amended:

Cantor showed (via his famous "diagonalization" proof) that the set of real numbers is uncountable. He actually found a much more complicated proof first and he didn't convince many people, but years after having found that solution, he found a much more elegant proof. In fact, it's so simple that I'll reproduce it here, in case your book didn't. First I'll start by noting that there are as many elements in the set of real numbers as there are elements in the set (0,1) = {x | 0<x<1 }. To see this, you need only find a bijection (one-to-one function) that takes (0,1) to R. For example: f(x) = Arctan(Pi*(2x-1)) Now assume that we can set up a one-to-one correspondence between the natural numbers and (0,1). Naturally, I'd expect to see an explicit listing of such a correspondence: 1 -> 0.0011213213111... 2 -> 0.8199191988110... 3 -> 0.8878181888108... 4 -> 0.3996771726507... ...... And so on, where each real number in (0,1) appears exactly once to the right of a unique natural number on the left. (Here, I'm implicitly asking for a single decimal representation for each real number, with no repeating 9's so that you don't get any duplicates on the right. Example: 0.499999... = 0.500000... This is clearly not much to ask of our listing, but you'll see why this point is important later on.) So far so good? It was Cantor's genius to realize that such a listing is impossible. "Why?" you ask. Well, we can start by observing that we've required every number to be on the list. Now, let's define a new real number y with the following property: The n-th decimal digit of y is either 7 or 8. We choose 7 whenever it is different from the n-th decimal digit of the n-th number in our list, and 8 whenever this doesn't work. So, given the list above, we would define y = 0.7787... Our real number y is clearly well-defined. You provide me with a listing of all real numbers in (0,1) and I'll be able to produce a number y. (That's why I required a single decimal representation for each real number, so that y could be defined properly. It's also why I chose to make y without ever using the digit 9.) Now, here comes the key step. y is itself in (0,1), so it must have a corresponding entry in our listing. We did after all, list all real numbers. The problem is, what natural number m corresponds to y? Well, m can't be 1, since by definition y differs from the first number in the first decimal digit. m can't be 2, since by definition y differs from the second number in the second decimal digit. In fact, if n is a natural number, then m can't be n since by definition y differs from the n-th number in our list by precisely the n-th decimal digit. (y is defined by the diagonal entries in our list, hence the term "diagonalization proof.") We have produced a number that's not in our list! This is a glaring contradiction. A one-to-one correspondence between the natural numbers and (0,1) is impossible. Therefore the set (0,1), and by extension, the set of real numbers, is innumerable.

Now let’s move from the “how can this be?” questions to the “you’re wrong” responses.

Many claim to be able to list all the real numbers. The following came to us in 1997, not actually telling us we are wrong, but humbly asking where he is wrong:

Counting Infinities I was re-re-rereading Gamow's _1,2,3...Infinity_ andthought I found a flaw in the uncountability proof. My method of listing decimals was to enter the numbers inbinary, writing them backwards: .0, .1, .01, .11, .001, .101, . . . I was thinking I would get all possible decimals. The counterproof of changing the numbers in the diagonal would produce .111111... = 1, which shouldn't be included anyway. Since I don't really think that aleph 1 > aleph 0 just because we have ten fingers, I decided that I must not really be including all decimals, but I still can't see why. For instance,I suspect that I don't have 1/3 in my table, but since I include the first 10, first 1000, and first googol digits of 1/3 somewhere (I could tell you which lines, but you don't really care, do you?) in my table, it seems, (by induction?) thatI should get exactly 1/3 eventually, but I get myopic around a googolplex.

He has taken the binary numbers 0, 1, 10, 11, 100, 101, … and reversed them to get .0, .1, .01, .11, .001, .101, … . Others have suggested doing the same thing in decimal, turning the list 1, 2, 3, …, 10, 11, 12, 13, … into 0.1, 0.2, 0.3, …, 0.01, 0.11, 0.21, 0.31, … . Like Gene, Cantor used binary rather than decimal representations, which is more natural, mathematically (though doing so forced him to use a different approach than avoiding 9 and 0 in order to get past the double-representation problem).

Doctor Rob took this:

Even when the proof is given writing the numbers indecimal, you have to avoid .9999999.... = 1.0000000...., an artifact of all such positional numeral systems. The usual "diagonal construction" contains the provision that you pick for the n-th digit a digit other than the one in row n, column n, *and also different from 9*. That avoids the ambiguity. Inbinary, if you pick a digit other than the one in row n, column n, and also different from 1, you may have no choices left, as in your example. You could modify this slightly to use *pairs* of bits, avoiding the pair 11 and the pair occurring in row n and columns 2n-1 and 2n. This would fix up your construction. Since 1/3 = .01010101...., you can use this method to show that it cannot appear.

The important thing is that his list contains only terminating representations, so it doesn’t even include all rational numbers, not to mention irrationals:

As to the matter of 1/3, it is not in your list. The numbers in your list are just those of the form a/2^b, where a is 0 or odd, and b is positive. 1/3 cannot be written in this form, because of the Fundamental Theorem of Arithmetic (unique factorization into primes). If it could, 1/3 = a/2^b <==> 2^b = 3*a = n, and 2 and 3 prime, violates unique factorization of n. You do have excellentapproximationsto 1/3, as good as you like, butyou don't have 1/3 itself. This is somewhat like the situation where the rational numbers do not contain sqrt(2), although they do contain excellent approximations, as good as you like.

Others have claimed that the *method of proof *fails, by showing that it produces nonsensical results when applied to other sets than the reals. Here is one from 2005, which was not archived:

Dear Dr.Math! There is apopular alleged proofof the fact that the irrational numbers are non-denumerable, which seems to me to allow for aparallel proof that the rational numbers are non-denumerable. ... What I simply DO NOT get is where the mistake is, if we replace all references to irrational numbers with references to rational numbers and then follow through with a parallel "proof" that if we assume that the rational numbers are denumerable, then we can create a new rational number not already in our list in the same way as before. I mean, the argument cannot be that we have independent proofs that the rationals are denumerable and that they are not (for then we might as well claim that the irrationals could some day be proven to be denumerable, which is a kind of reasoning that would demolish any idea that mathematical proofs proof something), nor that the rationals cannot be written as decimal expansions (which they obviously can!)?!

Doctor Jacques replied:

The proof does not work for rational numbers, becauseevery decimal expansion does not necessarily represent a rational number. The decimal expansion of a rational number is either finite or periodic. If you try to use Cantor's argument with the rational numbers, you will end up with a decimal expansion that does not appear on the list, butyou cannot derive a contradiction from this, because you cannot prove that the new string of digits is finite or periodic, i.e., that it corresponds to a rational number.

The proof for real numbers depends on the fact that *any* (non-terminating) decimal corresponds to a real number.

In fact, youcanmake a list of the rational numbers between 0 and 1 (see below). If you apply Cantor's argument, you will end up with a decimal expansion that is not on the list -this simply means that the corresponding real number is irrational. In the case of real numbers, you get a contradiction, because you assumed that all real numbers were on the list. To make a list of the rational numbers between 0 and 1, you can, for example, order them by increasing denominators first, and leaving out fractions that are not in lowest terms: 0, 1, 1/2, 1/3, 2/3, 1/4, 3/4, ...

Christian wrote back:

Thank You very much! I am a bit embarrassed that I did not see the central point - that no contradiction arises in the case of trying to create a new rational number since the newly created number cannot be proved to be rational - but I am also very thankful to you, for your swift and clear response. Thank You again so much!

Some others have tried to apply the diagonalization proof to **integers**; in that case, the new number is an “integer” with infinitely many digits, which (unlike numbers with infinitely many decimal places) is nonsense.

We can start with our first question about it, from 1995:

Getting All the Utilities to Each House I've been stumped-here's the problem: Draw 3 one inch squares representing houses horizontally across the page. Draw three circles, one under each square. Number your squares and put a G for gas, E for electric, and W for water in each circle.Your job is to connect each utility to each home WITHOUT crossing any lines.You may not pass a line through a house or other utility circle. Leave an inch and a half between each house. Put each circle an inch below each square.

Here is a picture (borrowed from the FAQ):

So we have to draw nine (curved) lines, connecting each dot at the bottom to each of the houses. Here is an attempt:

We are missing one line, from Electric to House 1. The line we started is trapped between the green and red lines, one of which it would have to cross to get to the house.

It’s important to keep in mind that this is a pencil-and-paper puzzle, not a real-world problem; real electric lines can go in the air and pipes can be at different levels in the ground, so crossing is not a problem. Also, we are thinking of each square or dot as a single entity, effectively a point, so we can’t pass a pipe through a building as we might in real life; nor can we join wires into a loop from house to house. (All these are things people have written to us to suggest.) We are thinking of it as an abstract problem in graph theory, which deals with points joined by curves.

If you keep trying, you will keep failing the same way, until you should wonder, as Mr. Borstein probably did above, whether you are just missing something, or there is something wrong with the problem itself.

Doctor Ken replied briefly:

I hope I can make you feel a little better about the fact that you couldn't solve it, becausethere is no solution possible. What's neat (and you might want to pass this along as a challenge to your students) is that while the problem can't be solved in the Euclidean plane,it CAN be done on the surface of a torus(a doughnut). See if your students can figure out why.

This is the first level of answer: Don’t worry, it’s not just you!

When a mathematician finds that something feels impossible, he has two next thoughts: Can I **prove** it really is impossible? And, can I **modify** the conditions to make it possible? Those are the two points Doctor Ken made.

A similar question came in in 1996:

Graph Without Crossing Lines There are three houses 1 2 3 and there are three utilities gas water electric How can you connect the three houses individually to the three utilities without crossing your lines? (You have to draw a line from each house to each of the utilities without the lines ever crossing.)

This time Doctor Patrick answered, adding one more piece to his answer:

Hi, sorry to tell you this, butthat problem is impossible on most normal surfaces. The only one I know of where it will work is what is called a torus--which is basically a doughnut-shaped figure with a hole in the center. If you can, try the problem again on a surface like that and see what happens. Since a torus might be hard to come by, you cancut a hole in the center of the paperand allow lines to go onto both sides and through the hole onto the other side. Why do you think it works in this shape?

A piece of paper with a hole in it, when you have permission to go around to the other side both at the hole and at the edges of the paper, is equivalent to a torus. Here is an example solution, where we have fixed the problem in the attempt I showed above by taking that last line through the hole:

(Here the light green represents a green line on the back of the paper.)

By the way, the hole is necessary, not just the permission to wrap around; without the hole, wrapping around just turns the paper into the equivalent of a sphere, which doesn’t help.

The FAQ shows such a solution on an actual torus. Now, let’s look at one of the proofs that it can’t be done on a plane.+

Here is a question from 2002:

Water, Gas, Electricity to Three Homes Puzzle This is a puzzle that I have tried over and over and over to try to get. You have the three letters W G E arranged in any position and you also have 3 circles (homes) arranged in any position. You have to draw a line from each of the W G and E to each of the 3 circles without crossing lines. You can put the letter and shapes in any shape or positions to do this.I have always got to one line left over and over and over.I believe that this is an impossible one to solve. If you figure it out, can you tell me what order you put down the lines since that's half of how you do the problem.I HATE THIS PUZZLE!

Doctor Anthony provided not only the answer, but a proof (similar to one of the two proofs in the FAQ):

We have 3 utilities A, B, C and 3 houses 1, 2, 3 as shown A B C 1 2 3 and we must join each letter to each number without any of the lines crossing each other. This problem comes under the heading of graph theory. You have probably heard ofEuler's formula for plane graphs. If p = number of vertices, q = number of edges, r = number of regions then p - q + r = 2 A plane graph has none of its edges intersecting except at a vertex. For example, in a square, you have p = 4, q = 4, r = 2 (one region inside the square, one region outside the square) and p - q + r = 4 - 4 + 2 = 2 as required. The proof of Euler's formula is straightforward and can be found in any textbook on basic graph theory. We now use it to show that it is impossible to join the three letters to the three numbers without any intersections of the edges.

We previously discussed Euler’s Formula. Commonly we use the letters v, e, and f for the number of vertices, edges, and faces (regions).

We prove the result by contradiction. Suppose that it is possible to draw the figure without any intersections of the edges. We sum the number of edges lying on the boundary of each region over all r regions and denote this number by N. Since no edges join letters to each other or numbers to each other, the graph contains no triangles, so it follows that N >= 4r [To illustrate this, think again of a square. The inner region has 4 edges on its boundary. The outer region has 4 edges on its boundary and so N = 8. If we had a triangle then N would be 3+3 = 6 but as stated above there are NO TRIANGLES.]

Here is an example of a region with four edges:

On the other hand N counts each edge AT MOST twice (an edge can only have 1 or 2 regions on either side), so N <= 2q (= 2 x 9 = 18). We had N >= 4r so 4r <= N <= 18 r <= 9/2 (=4.5) ......(1) However by Euler's formula, p - q + r = 2 6 - 9 + r = 2 r = 5 ..................(2) Comparing (1) and (2) we see that there is a contradiction and our assumption that the figure is a plane graph is clearly impossible. It follows that the letters and numbers cannot be joined without at least one intersection of two lines.

This also provides an explanation for the fact that the puzzle can be solved on a torus: the equivalent of Euler’s formula there is p – q + r = 0, from which we get r = 3, and there is no contradiction.

Asking whether the problem can be solved on other surfaces is interesting math. But many people claim to be able to solve the problem by some “trick”. Here is someone in 1999 who claimed to do it without “tricks”:

Three Houses, Three Utilities I know that you have answered this before: the question about the three houses and the three utilities (gas, electricity, water). Well, the guy who gave me this puzzlesays there is a way of solving it in 2D, without any tricks. He says that it is simple, once you figure it out. I don't get it. Everywhere, it says that it can only be done using 3 dimensions. Can you solve it using 2 dimensions? How?

Hmmm …

Doctor Rob answered with a clarification of the problem:

You can only solve this if you allow one of the utility lines to runthrough someone else's house, orthrough one of the other utility companies, which I suppose is possible, but is usually forbidden by the conditions of the puzzle.

Such evasions are generally considered tricks. Here is such a “solution”:

I added a comment, listing the other basic tricks we’ve seen:

He may not call it a trick, but any solution that's really 2D (that is, done justby drawing non-intersecting curves on a flat sheet of paper) has to twist the rules somehow. He might, for example, draw the houses as rectangles and say that it's legal to open the front and back doors of one house andpass a pipe through. I call that a trick. Another trick is to solve iton the surface of a donut(a torus) and point out thatany surface is itself 2-dimensional, even though it exists in a 3-dimensional space. Or you can allow going around to the other side of the paperthrough a hole, which is essentially the same thing.

But *why* are the rules what they are? How do we “know” that these tricks aren’t allowed? Because we know **the genre of the problem**, which is mathematical rather than a riddle. I continued:

When the problem is stated carefully in mathematical terms (continuous non-intersecting curves from each of three points to each of three other points), there's no solution; but presented in terms of houses and utilities (which are inherently three-dimensional), there are lots of ways to get around it.

So this problem, like some of the others (Boy or Girl? being the best example), is commonly stated as if it were a real-life problem to make it more relevant and impressive, but in doing so it can miss the point, running the risk of making math appear pedantic.

Here is what another person (Bruce, unarchived) said in 2007 in arguing against our answer, claiming he had a solution:

How my solution works isn't based on mathematics or tricks - it simply takes advantage of the way the puzzle is stated (the rules).If there is no rule barring the action, then it is "legal" to do it.The reason nobody has "figured out" the puzzle is A) Not paying attention to the rules. B) People keep re-writing the puzzle and adding their own (new) rules to it. And C) People are missing the very simple solution because they are paying more attention to what IS being stated than what ISN'T being stated in the rules. For instance - Howard Dudeney did not say you could not "leave the plane" as you have asked me here. So while the puzzle as stated probably before you were born was solvable due to that opening - you have changed the rules in order to "make" it unsolvable. I have seen other variations and people trying to make it stricter (hence "unsolvable"), but even so I still have come up with several methods of solving the puzzle, becausethere's always a "loophole" in the rules. Let me know if you would like to see - I bet you'll be unhappy about it, because it isn't some "high math" solution... just plain observance.

*Henry* Dudeney was a popular puzzle creator in the early 20th century, who put this puzzle in its current form, in 1913. (For more information, see Wikipedia.) The notes there lead to this copy of the original puzzle, in Strand Magazine:

He doesn’t present it in the abstract mathematical form; and in fact, his solution is the same trick answer we’ve seen:

Quite likely, this is also Bruce’s solution. How do we justify disallowing it, when Dudeney accepts it? Doctor Ian responded:

I think you might be missing the point of the problem.What's interesting to mathematicians is the abstract problem, defined purely in terms of mathematical concepts, and there are no loopholes there. Of course, there's also very little there to interest anyone who isn't a mathematician. It just gets expressed in terms of things like houses and utilities to make it more understandable to the general public. Of course, that requires setting up analogies, and anytime you set up an analogy, it creates openings for pushing it so hard that it breaks. That's inherent in the concept of an analogy, so it's not terribly interesting when it happens. Now, if you'd like to forget about houses and utilities, and show that it's possible to construct a planar graph from each of three nodes to each of three other nodes, then that's interesting. Can you do that?

That is, to a mathematician the puzzle is really an opportunity to prove something abstract about points and lines (that is, about graph theory); and this puzzle, in that form, predates Dudeney. Dudeney was interested in puzzles, so in his context the trick is what makes it fun. To each his own!

By the way, someone wrote this in 2005 (unarchived, and in fact unanswered):

Dudeney wrote the utilities problem long before running water, gas, and electricity were found in houses. What was his original problem back in 1917, since it is impossible to believe he wrote it the way it is found today?

The assumption is wrong, of course; but it would be interesting to see the older forms Dudeney referred to. Possibly they didn’t allow his trick.

Some people write to us having seen the puzzle long ago without, apparently, being told the point (that it can be proved to be impossible) or allowed to use a trick. That can become unintentionally sadistic! Here is one, from Medora in 2003:

My question is mostly rhetorical. When I was 12, this question was given to my math class by Mr. Ostrander with the words, "See if you can solve it." He didn't know how stubborn I was. I worked on this problem for years, until it dawned on me that I had been duped--the problem was insoluble in two dimensions. (Note that I was always a "humanities" type student: literature, music, art, history, philosophy, and now theology are my areas.) Thank you for providing an answer for me 43 years after I began struggling with this. I came across your site by accident while looking for the secret of the universe.I just wonder why math teachers feel the need to torment people like me for decades!

Doctor Jaffee answered:

I know that your question is rhetorical; nonetheless, I would like to suggest an answer to you that may make you feel a bit more positively about Mr. Ostrander. I believe, as perhaps Mr. Ostrander did, that oftentimes finding the answer to a problem is not as important as the process of seeking the answer. This process requires tenacity (which apparently you have), creativity, the willingness to try new and different ideas, and oftentimes the process opens doors to new ways of thinking. I suggest that your success in the areas you listed above may have been enhanced by the time you spent struggling with the "3 utilities to 3 houses" problem. I hope this makes some sense to you. At any rate, thank you for your letter and best wishes in your future studies. Write back if you would like to discuss this some more.

But I have to say, it seems inappropriate not to tell students the point! We need an “out” – one of those “open doors”, which might be either a lateral-thinking trick, or a way to prove impossibility.

In closing, here is something I wrote in 2011:

Although this puzzle is stated as a real-world problem, its interest to mathematicians is in its abstract form: Given two sets of three points on a plane, it is impossible to make continuous non-intersecting curves joining each of the first three points to each of other three points. What makes this interesting is that]]>we can prove that it is impossible-- not just give up after trying, and justbelievethat it is impossible. I think this is the important point that many people miss, because they don't see the problem from a mathematician's perspective.Trick solutions are fun, but proofs of unexpected facts are amazing!

We’ll start by looking at an early statement of the issue, from 1996:

Trisecting an Angle I've recently been pondering the possibility of trisecting an angle. I wasn't sure if it was possible and found a FAQ submitted to you that stated it was indeed impossible. Has it beenproved to be impossibleor is it thatno one has proved it possible?The reason I ask is that I think (but am not sure) thatI have found a way to trisect an angle. I am a seventh grade math teacher, so I am not completely ignorant of mathematical proofs. I would appreciate some information on who would be the best person to contact with my "proof".I hope you believe me because I think this actually works.

This is the benign species of trisector, one who just isn’t sure what it is that was proved, and isn’t insisting that everyone else is wrong. Doctor Tom answered, presenting the key ideas one by one:

There is, in fact, a proof that atrisection is impossible. Many people think they have found trisections, but they eitherdon't understandexactly what the problem is or theirmethod is flawed. The problem requires the construction of the trisection using an(unmarked) straightedge and a compass. If, for example, you were allowed to make 2 marks on the straight-edge, there is a trisection. The straight-edge can only connectpoints already constructed, or use arbitrary points, and the compass can only be used similarly. An exact trisection in afinite number of stepsis also required since it's easy to do a trisection that doesn't require too much accuracy. As a limiting case, it might be possible to perfectly trisect an angle in an infinite number of steps.

So what was proved impossible is a *specific task* using *specific methods*; if you break the restrictive rule on what tools are available and how they are to be used, trisection is *not* impossible. For discussions of two such alternative tools, see

Trisecting an Angle: Proof Trisecting an Angle Using the Conchoid of Nicomedes

(The first of these has a broken link that should now be http://www.faqs.org/faqs/sci-math-faq/trisection/. The second has a link that is now http://www-history.mcs.st-and.ac.uk/history/HistTopics/Trisecting_an_angle.html.)

So how was it proved to be impossible?

The proof that it's impossible basically considers the arithmetic form of the points that can be obtained using a compass and straight-edge. In a sense, they are "quadratic" devices -the new points generated are solutions of quadratic equations of previously constructed points. So you can clearly get things like square roots, fourth roots, eighth roots, and so on (and it's a bit more complicated than that - what you can get are field extensions of degree 2 over whatever you begin with). Butyou CANNOT solve irreducible cubic equations this way- such solutions require extensions of degree 3, and no combination of multiples of 2 gives a multiple of 3 (to put it in an inexact, but hopefully clear, way). You can, beginning from nothing, construct a 60 degree angle. So if you can trisect anything, you can trisect the 60 degree angle which you produce. So if you have a trisection, you can construct, from scratch, a 20 degree angle. Hence you can construct the sine and cosine of 20 degrees. But it's easy to show thatthe sine and cosine of 20 degrees are roots of an irreducible cubic equationover the rational numbers. This is a contradiction, so a trisection is impossible. To get the details, read about fields and field extensions in an abstract algebra book.

One excuse many trisectors give for doubting the proof is that they think algebra can say nothing about geometry. But it can!

As for this teacher’s claimed trisection,

So the bottom line is, there's probably something wrong with your trisection, not because I've seen it, but because I've studied field extensions and irreducibility.

That idea, “I know it’s wrong without needing to see it,” angers a lot of people; but that’s how math works!

Here’s another trisection attempt of the relatively benign variety, from 1999:

Attempt at Trisecting an Angle We have a bisector of an angle of 30 degrees (or any degree) that extends into the angle 1" and extends outside the angle 2". Then we have bisector line 3" long. From the vertex of the angle, a circle with a 1" radius is drawn. Then 2" outside the angle, and on the bisector, a circle with a 3" radius is drawn. The two circles meet on the bisector of the angle 1" inside the angle. Question:Are the arcs of the two circles within the angle the same length?Arc a-b of small circle = arc A-B of larger circle. Both arcs are 1" within the angle.

Hollis doesn’t come right out and say he’s doing the impossible; here is a picture of his work, so you can see what he is doing (using A_{1}, B_{1}, A_{2}, B_{2} for his a, b, A, B):

I responded first by noting that the arcs are not, in fact, the same length, and observing that if they were, he would have trisected angle A_{1}OB_{1}:

I could show you the trig to work out the actual arc lengths, but it's pretty ugly and probably not worth the effort. I'll just tell youthey are not the same. Your construction is reminiscent ofsome attempts I've seen to trisect an angle. In particular, the figure looks much likeArchimedes' method, which requires a marked straightedge, and that in itself is enough to tell me that the answer will be no. If the arcs were the same, then the angles would be in a 3:1 ratio (since they have the same arc length and the radii are in 3:1 ratio), and you would have trisected the angle; since I know that's impossible with a construction that can be done with compass and straightedge, I don't really have to do the extra work.

I tried to encourage him to do math, without the aim of trisecting an angle:

If that's what you're trying to do, don't feel it's foolish to try. As the alchemists discovered useful things in the process of trying to do the impossible,you may learn a lot about what does work in trying to do what won't. There are some fascinating ways to trisect an angle usingspecial tools, as our FAQ tells you if you dig deep enough, ... and the attempt to prove that any given method might work can help you get a better feel for geometry, and stretch your mind considerably. What you'll want to do, though, is todevelop the skills to prove for yourself whether something is true, rather than make a conjecture and have to ask whether it is true. That's the fun part of geometry, and what makes it more challenging than any other part of elementary math.

Hollis wrote to us periodically over the following years, sometimes with questions that I answered with comments like, “This looks suspiciously like an attempt to trisect an angle, which of course can’t be done (exactly) with compass and straightedge. Can you tell me what you are doing?” Eventually, he explicitly became a serial trisector, immune to whatever we tried to say.

Here is a trisection attempt that turned out to be interesting:

Trisecting an Angle I've been looking for someone to talk to about the trisecting an angle problem from ancient Greece. I understand the whole modern algebra proof thing butit doesn't disprove the possibility of angle trisection. At least not from the way I have read it and understood it. I came up with a method when I was 16 and in geometry class andso far no one can disprove it.I tried proving it geometrically for a while but gave up. All I know is that with best drawing programs using only circles and unmeasured lines, it works at least to the naked eye for any angle from 5 to 175 degrees. I've done numerous examples trying to find one where it is off by more than the error when using a compass andso far everything I've done is a perfect trisection. Who can I show this to? I think I have something here...

Eric clearly doesn’t understand that algebra *did* prove that trisection (by the rules) is impossible; and that his attempt can’t be called exact until it is *proved* to be exact — the burden of proof is on him to *prove* it, not on others to *disprove* it! But I asked him to show his work, and (in a discussion too long to include here) discovered that it could be made exact if only you could adjust a center point so that other points coincide with a circle. This makes it like a marked-straightedge construction.

Unfortunately, this page got me started on a side career of making similar checks of approximate trisections.

A 2001 question from Joe provided a perfect opportunity to discuss what people are misunderstanding when they think they have a trisection. It started innocently, as many do:

Angle Trisection: Construction vs. Drawing Has anyone ever divided an angle into three equal parts by construction?I have been told it has not been accomplished.

Note that Joe interprets the claim that it *can’t* be done as a mere challenge, that no one *has yet* done it. This is what drives trisection mania.

Doctor Tom replied:

Using the standard methods of construction with a straightedge and compass, it can be proven that it isimpossibleto trisect anarbitrary angle. People who think they have solved the problem usually make one of two mistakes: 1) They trisect aparticular anglethat happens to allow a trisection. For example, anyone can trisect a 90-degree angle. 2) Theydo not understand the "rules"of straightedge and compass construction. For example, if you are allowed to make two marks on the straightedge to turn it into a sort of ruler, you can trisect any angle. But the offical rules do not allow this.

We have had a number of claims related to specific angles; here is one:

Trisecting a Right Angle

Others start with a known angle, such as Hollis’ 30°, just as an example, but apparently thinking that measuring their result as 10° will constitute proof.

But Joe turned out to be a serious trisector, as he replied:

I have trisected arbitrary anglesup to 90 degrees. Don't laugh until you see it.Only a straightedge and a compass. No measuring.The drawing must be precise for the angle to be correct.Where could I submit my effort for confirmation?

So he claimed to satisfy the requirements, but something he said suggested the need to add another clarification, so I replied:

I notice something in what you just said that indicates where you are misunderstanding the trisection problem. Dr. Tom listed two mistakes people commonly make (trisecting only a specific angle, or using the wrong tools). But there is another that is even more common: not recognizingwhat we mean by an exact trisection. When you say that "the drawing must be precise," you show thatit is the drawing itself that you have been focusing on. But to a mathematician,the drawing itself is nothing. It is only a representation of something that really happens in an ideal world where lines have no thickness, and so on. In that world,we can prove that a construction is ABSOLUTELY exact; either it meets the precise point you claim, or it is a false construction. Andsince this is the world of the mind, ONLY proofs count. It doesn't matter how good a drawing you make, it proves nothing.

Every trisector we have heard from has made a claim without proof, depending only on how things look. (Some, after reading this page, have made attempts at proof, but they generally have no idea what a real proof looks like; otherwise, they would have understood that what is proved impossible really can’t be done.) But no measurements of a physical drawing can be accurate enough; whereas when you do a proof, there is no need to have an accurate drawing.

So unless you can prove that your construction really works exactly, you have nothing to show anyone. And we know that you can't, because it has been PROVEN that such a construction can't be done. I've seen many constructions that come remarkably close, usually just because they are very complex;there is nothing at all impressive about a close approximation. Please don't waste your time on this, as so many people have.

Unfortunately, the discussion didn’t end here; the next month, Joe wrote back telling us about his constructions, each of which assumed something that is not true about chords and arcs (see below on that). And in 2004 he wrote again:

Trisecting an Angle Using Compass and Straightedge

There, he asked me to analyze his construction on Geometer’s Sketchpad, which I had used on Eric’s construction that I mentioned above; I had to tell him that not only physical measurements, but also calculations (necessarily rounded) by a program do not constitute proof. I did analyze it, to show him that it was not exact, but also demonstrated what a proof looks like, giving a proof of an angle bisection for comparison. At the end, I added, referring to the FAQ:

As that explains, it has been proved to be impossible to exactly trisect any arbitrary angle using only compass and unmarked straightedge; and as I pointed out, "exact" to a mathematician means something entirely different from "as close as you can measure"; it means "provably exact". "Impossible" is a common thing in math. For example, it is impossible to find two odd numbers whose sum is odd; it is easy to prove that the sum of two odd numbers is always even, so we just accept that.I've never heard of people who, when told this fact, spend their lives trying to find two odd numbers whose sum is odd; but there have been thousands of people who have wasted a lot of time trying to find a way to trisect an angle. Most of them probably have no idea that a valid trisection is not just a drawing but a proof! They just hear that it can't be done, don't understand how that could be proved, and decide to show that it doesn't apply to them.Since it's not as obvious as adding odd numbers, they never realize how silly what they are doing is.To keep your mind alert, I would suggest something more "constructive" than trying to do what has been proved impossible. Take some time to learn the basics of geometric construction and proof, then work through the exercises in a good book on the subject, which will ask you to do constructions that are hard but possible. That way you are giving yourself a major mental challenge, but one that you can really master with effort. It's a lot more rewarding!

I have said this sort of thing to several others I have corresponded with. This discussion continued a little further, and ended with his saying, “I will put your note about proof under the glass on my desk as a reminder to leave the trisect impossibility alone.”

But, alas, he wrote again in 2009, with yet another attempt. It turned out not to be close at all. He hadn’t learned a thing.

Let’s look at one more page: a proof that a common assumption of trisectors is false:

Trisecting an Angle and the Opposite Side in a Triangle Prove that it is impossible to have a triangle in which the trisectors of an angle also trisect the opposite side. I am unsure how to prove this. It seems that if i trisect the angles of an equilateral triangle so that each of the trisected angles is 20, it would indeed divide the opposite side into 3 equal pieces. I have completed geometry, and I have tried several things. I got started trying to use exterior angle theorem, but got confused. I think that may be a good way to do it, but I dead ended after extending two of the sides to create isosceles triangles. I think it may be all of the criss-crossing lines that is getting me confused.

I replied to this refreshing question:

I don't think I've ever tried proving this, but it's a very nice little theorem! It may seem as if trisecting an angle in an equilateral triangle would work, but it is not true. I think I'd approach it by contradiction. Suppose that you have a triangle ABC with trisectors BD and BE, D and E being on AC, and further suppose that D and E trisect AC, so that AD = DE = EC: Now focus on triangle ABE. Here BD bisects angle ABE, while D is the midpoint of AE; so BD is both an angle bisector and a median. What does that imply? Repeat with triangle DBC, and look for a contradiction. There may be many other ways to approach this, so if you see any ideas of your own while you try this out, go ahead and pursue them!

The implication I had in mind is that triangle ABE must be isosceles, because triangles ABD and EBD are congruent. It also implies that BD is perpendicular to AC. Since the same must be true of triangle DBC, this triangle can’t exist.

In summary, here is what I wrote to someone who wrote in 2009, starting out “Why do you print false information on your site?”:

It has been proved that you can't construct a]]>provably exacttrisection of anarbitrary angleusing theclassical rules(only a compass andunmarkedstraightedge, in afinitenumber of steps). People who claim they can, generally don't really have a provable trisection, or are bending the rules in some way without realizing it. ... If you have no proof to show me, then you have not met the requirements of a valid construction, and most likely it is just measurably close, not mathematically exact. The imprecision of physical tools has nothing to do with it, because we do not demonstrate the correctness of a construction by doing it physically and seeing that it looks as close as our tools can make it. The tools we really care about are all in the mind, and that is where the proof has to be done.