False Proofs: Are All Numbers Equal?

One of the ways to show the need for careful proof, which I discussed last time, is the existence of apparent proofs of false “facts”. These are also useful for training us to think carefully. In this and the next couple posts, I will look at a few categories of false proofs, focusing on what they teach us. This time, we have some proofs that unequal numbers are equal, based on restrictions in algebra that are easily overlooked.

A proven fact?

We’ll start with a question from 1998 that serves largely to introduce why we care about this:

1 + 1 Doesn't Equal 2

It's a proven fact that 1 + 1 does not equal 2. What is the math equation that proves this? My math teacher taught me this, but I forget.

Most likely Sara didn’t really believe the “proven” claim, but meant only to ask if we knew the “proof” she had seen. But there are many of them! I began with a short list of references to “proofs” that could be described as showing that 1 + 2 is not 2: two that I’ll look at below, a now-broken link, and this:

Classic Fallacies (University of Toronto)

Then I added:

Of course, these aren't really proofs, because they all have some error in them. What's important about these examples is that they show ways in which you can make a mistake in using math if you aren't careful enough. If you can understand where the error is, then you can look for the same kinds of errors in your own work, whether it's a proof for school or a calculation you make when you're designing a bridge. It also explains why mathematicians and scientists don't publish their results without first having others check them to make sure there isn't some subtle error in their calculations.

The first step in writing correct proofs is to be fully aware that you are fallible, and to know what kinds of mistakes you are likely to make, so you know where to look for errors.

Fooled by zero

Here is one of the false proofs I referred to, from 1997:

Proof: 2 = 1?

What is wrong with this problem?

Okay, here is how it goes.

                       a = b
               a squared = ab
   a squared - b squared = ab-b squared
              (a-b)(a+b) = b(a-b)
                     a+b = b
                     b+b = b
                      2b = b
                       2 = 1

I asked my Geometry teacher about this problem and she said it is mathematically sound but we cannot figure out how 2 = 1. a = b is factored to a squared = ab by multiplying both sides by a. Please help me!

Doctor Luis answered:

Actually, the fact that you got 2 = 1 is PROOF that your method to get the solution is NOT mathematically sound. Think about it :) Why should a process that leads to a contradiction be sound? This means that somewhere, in some line, you have a false statement, one that does NOT follow from the previous one. Where? That's for me to know, and for you to find out... however, I am going to give you a hint: remember the first line ( a = b )? Given that, what can you say about a - b ?

The “proofs” we are discussing are not actual proofs of false things, but false proofs! It’s easy to see that the result is false; from that we already know the proof itself contains an error. The fun part is to find where the error is. (We’ll get to that below, but give it some thought while we continue …)

Let me leave you with one warning about division. Whenever you have two numbers x and y, you can always divide x by y and get another number x/y, ONLY when y is not zero. This is because you don't get a unique answer when you divide by zero. Since you can't get a solution when you do that, mathematicians say that division by zero is undefined.

We have seen previously why division by zero is undefined. Doctor Luis provides an elementary explanation:

I'm going to elaborate a bit more,

    When you divide 6 by 2 what do you get? 3, of course.
    Why? Because 2 times 3 is 6.

    What if I divide 20 by 5? I get 4. Right?
    That's because 5*4 = 20.

    In general, x/y = z  if and only if y*z = x

    Every time we divided, we got a unique answer, that is,
    we only got one number.

    6/2 does not equal 4 because 2*4 = 8 does not equal 6    
    The only number that satisfies the equation 2*x = 6,
    is x = 3 ( x is unique).

    Up to this point everything seems to be fine. However,
    weird things happen when you start dividing by zero.

    What is 1 / 0 ?  Is it 1? No. Because 0 * 1 = 1 is false.  
                     Is it 2? No. Because 0 * 2 = 1 is false.   
    In fact, if you say 1/0 is the number x, then x has to
    satisfy the equation 0*x = 1, which tells you that 0 = 1,
    and that's a false statement. This means that there is
    no such number x, but x was 1/0, and so there is no
    number 1/0, and we say it's undefined. Similarly, you can
    show that 2/0 is undefined, 3/0 is undefined, 4/0 is
    undefined, and so on...

So, dividing a non-zero number by zero yields no answer, because no number would work.

    Now, what about 0/0 ? What is it?

    Is it 0? yes, because 0*0 = 0

    There's only one problem. You can show that 0/0 is also 1 !

    Because 0*1 = 0.

    You can also show that 0/0 is also 2 !      

    Because 0*2 = 0.

    You can show that 0/0 is any number!

    Because 0*(any number) = 0

    Since you can show that 0/0 can be any number, we say that
    0/0 is not unique, and for that reason it is also undefined.

Dividing zero by zero yields no unique answer, because any number would work – so we can’t decide what answer to give.

Note that Doctor Luis has explained the background, but didn’t point out the actual error. That was left for the reader to find. Where are we dividing by zero? There’s no visible zero in the proof! We’ll get there momentarily …

Finding the error step by step

In 2003, Josh found this page and asked a very detailed question about it:

I was browsing the Math Forum and saw the "proof" that 2=1. I read how, since two does not equal one, somewhere along the line an error was made. It was then concluded that the source of error was division by zero. Here is the proof I am discussing:

a=b               Given
a^2=ab            Multiplication Property of Equality
a^2-b^2=ab-b^2    Subtraction Property of Equality
(a+b)(a-b)=b(a-b) Sum and Difference Pattern/Distributive Property
a+b=b             Division Property of Equality
b+b=b             Substitution Property
2b=b              Combination of like terms
2=1               Division Property of Equality

In case you didn’t follow all of this, here is what Josh has done: He copied the “proof” and gave a reason for each step, showing that they are valid (well, he omitted the conditions under which each property is valid, which we’ll be looking at). He agrees that division by zero is an issue, though he didn’t explicitly point out where that is happening, namely that \(a-b = 0\), so at the “Division Property of Equality”, where we divide by \((a-b)\), we are dividing by zero, which is forbidden by that property. I have omitted his alternative suggestion, which turned out, inĀ  further discussion that was not archived, to be a “crank” claim that multiplication by zero does not follow the rules of algebra.

I replied with my own analysis:

No matter how you look at it, the problem in the "proof" is that zero behaves differently, and no account is taken of that. We're agreed there, right?

I showed an orderly way to locate where the error is in any such “proof”:

You are suggesting that it is in the step I would simply call "factoring" [the fourth line] that something "goes wrong." But let's look at the whole proof with an eye on the fact that a=b, so that we are aware of what is zero:

  written           reality            judgment
  -------           -------            --------
  a=b               a=a                true
  a^2=ab            a^2=a*a            true
  a^2-b^2=ab-b^2    a^2-a^2=a*a-a^2    true (=0)
  (a+b)(a-b)=b(a-b) (a+a)(a-a)=a(a-a)  true (=0)
  a+b=b             a+a=a              false (unless a=0!)
  b+b=b             a+a=a              false
  2b=b              2a=a               false
  2=1               2=1                false

Go through that carefully; you see that the factored step is the last one that is true. It has not introduced the error, but it sets things up for an error, by putting the equation into a form in which we are multiplying by zero. It also introduces the 2, in a hidden form; so you are right in saying that this is not a coincidence. It's sort of like a burglar entering your house. When he walks up your street, he has not yet broken in, but it is no coincidence that he is there. At this point in the "proof," the burglar is on the scene, but isn't in yet.

So the factoring step (his “Sum and Difference Pattern/Distributive Property”, which others would call “Factoring a Difference of Squares” and “Factoring Out a Common Factor”) rewrites what had been, in the case \(a=b\), merely \(0 = 0\), into the form \((2a)(0) = a(0)\); an unthinking division in the next step turns out to be a division by zero.

Now, you are right that at this step we have, formally,

  2a * 0 = a * 0

and it already shows signs of the problem to come; but it is not wrong yet, just as

  2 * 0 = 1 * 0

is not wrong. It is only if you forget that 0 times anything is 0, and that zero is special at this point, that there is a problem, and that is exactly what is forgotten in the next step.

Here we have to introduce the full form of the Division Property of Equality:

The problem, of course, is that what you call the "division property of equality" has a condition:

  if ab = ac, AND a is not zero, then b = c

If we omit that condition, then it is not true, because

  0b = 0c = 0 for ALL b and c, regardless of whether b = c

The error in the proof is that this condition has not been checked, so that a false "property" was applied.

For a similar false proof based on division by zero (from the first week of operation of Ask Dr. Math), see


And here is another:

Proof that 1 + 1 = 1?

Fooled by square roots

Let’s look at one more, this one from 2005:

Proof That All Numbers Are Equal?

Theorem: All numbers are equal.

Proof: Choose arbitrary a and b, and let t = a + b.  Then
             a + b = t
    (a + b)(a - b) = t(a - b)
         a^2 - b^2 = ta - tb
          a^2 - ta = b^2 - tb
a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4
       (a - t/2)^2 = (b - t/2)^2
           a - t/2 = b - t/2
                 a = b

So all numbers are the same, and math is pointless.  What is the error in this proof?  I think that if a = b then a - b = 0 and we cannot multiply both sides of the equation by (a - b) because when a number is multiplied by zero the answer is always zero.

RV is familiar with false proofs like those above where the issue is a hidden zero; but in this case, we aren’t assuming that a = b, only concluding that (falsely) at the end. Something else is involved.

Before we continue, let’s analyze this as I did above, by taking some specific numbers for a and b, and seeing where things go wrong. I’ll take a = 3 and b = 2, so t is 5:

             a + b = t                           3 + 2 = 5
    (a + b)(a - b) = t(a - b)                   (5)(1) = 5(1)     [= 5]
         a^2 - b^2 = ta - tb                     9 - 4 = 15 - 10  [= 5]
          a^2 - ta = b^2 - tb                   9 - 15 = 4 - 10   [= -6]
a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4  9 - 15 + 25/4 = 4 - 10 + 25/4
                                                                  [= 1/4]
       (a - t/2)^2 = (b - t/2)^2               (1/2)^2 = (-1/2)^2 [= 1/4]
           a - t/2 = b - t/2                   3 - 5/2 = 2 - 5/2  [FALSE]
                 a = b                               3 = 2

So the 7th line is the first false statement; the error must be in what was done after the 6th. Can you see the error?

Doctor Minter began with a tongue-in-cheek reply, as if RV seriously said that math is all wrong (as some others do):

You pose a VERY interesting point.  But before I give up my career as a mathematician and concede that all numbers are equal, let me demonstrate that there are two paths to be taken from the sixth line of your proof.

This, again, is the last correct line.

On one side, you have (a - t/2)^2, and on the other, (b - t/2)^2.  If you want to eliminate the exponent by taking the square root of both sides, remember that the square root of the square of a number is the "absolute value" of that number, since we cannot take the square root of a negative number without delving into the realm of complex (a.k.a. "imaginary") numbers.

For example, if we know that

  sqrt(x^2) = 4

we can simplify this to 

  |x| = 2

and then x can be equal to +/- 2.  Either value makes the equations true.  x does not have to be BOTH positive AND negative 2 simultaneously.  It can be either.

Do you see that the “proof” didn’t take an absolute value?

In regards to your example (which I may use on my friends, regardless of this reply), the aforementioned line of your proof would simplify to 

  |a - t/2| = |b - t/2|

Having absolute value signs on both sides of the equation is redundant, so we can simplify again to

  |a - t/2| = b - t/2 (or a - t/2 = |b - t/2|, it makes no difference)

So then,

  +(a - t/2) = b - t/2


  -(a - t/2) = b - t/2

The first indeed does show your result.  The second, however, simply shows that a + b = t, which was the original assumption, anyway.  To know if these results are correct, you simply have to plug in values for a and b.  If they are equal, then you can show it with this method.  If you want to show that a + b = t, you can show that also.  But it's not required that a = b, because there are two "paths" that you can take after eliminating the exponents.  It is not necessary that both of these "paths" SIMULTANEOUSLY lead to logical results.

So, line 6 doesn’t necessarily imply that a = b, but rather that either a = b or a + b = t. Since we started with the latter, we know that is true, and the former need not be true. The error in the proof lies in neglecting the possibility that at/2 and bt/2 have opposite signs.

I hope that this has been of some help.  I did thoroughly enjoy the proof that you provided, and I will likely show others for laughs.  Please feel free to write again if you need further assistance, or if you have any other questions.  Thanks for using Dr. Math!

Next time, we’ll look at some false proofs based in part on another aspect of square roots, as we expand to complex numbers.

1 thought on “False Proofs: Are All Numbers Equal?”

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