The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Proof: 2 = 1?

Date: 10/03/97 at 21:58:06
From: Wall, Mark
Subject: 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!

Date: 10/11/97 at 03:42:32
From: Doctor Luis
Subject: Re: What is wrong with this problem?

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 ? 

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 

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...

    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.

    If you have any questions feel free to reply :)

-Doctor Luis,  The Math Forum
 Check out our web site!   

Date: 05/30/2003 at 21:38:40
From: Josh
Subject: Analyzing false proofs to determine the cause of error

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

I suppose division by zero makes sense as a valid reason for fallacy, 
but how do we know for sure that it is the cause?  The Sum and 
Difference Pattern also makes sense, but when a=b I found that the 
solution actually changes:

(a+b)(a-b)=a^2-b^2    Sum and Difference Pattern
(a+a)(a-a)=a^2-a^2    Substitution Property
a(1+1)a(1-1)=a^2(1-1) Distributive Property
2a^2*0=a^2*0          Combination of Like Terms, Inverse Axiom of 

Notice that here the LHS is exactly twice the RHS if you disregard 
the Property of Zero in Multiplication, which states that 0a=0.  If 
you momentarily assume that 0a may not equal zero for all a, then the 
fact that the LHS is twice the RHS accounts for the fact that the 
solution to the false proof has an LHS twice the RHS. Therefore, if 
you assume that 0x=0y is only true when x=y, the Sum and Difference 
Pattern is revised to (a+b)(a-b)=2(a^2-b^2) when a=b. If you 
substitute this into the equation in step four, the solution is a 
true identity.

Is it possible that this is the reason that the proof is false?  
Otherwise it would seem hard to believe that the fact that the Sum 
and Difference Pattern's "identity" when a=b and the "identity" of 
the false proof are proportional is just a coincidence.  Perhaps the 
difference between 0x and 0y where x does not equal y is so minute 
that it cannot be perceived from our perspective and is impossible to 
tell apart, but can be viewed from our perspective and made apparent 
by cancelling out, or dividing, by zero?  Is it possible to assume 
that the error in this false proof may have resulted from something 
other than division by zero?

Thank you.

Date: 05/30/2003 at 23:15:43
From: Doctor Peterson
Subject: Re: Analyzing false proofs to determine the cause of error

Hi, Josh. Great question!

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?

You are suggesting that it is in the step I would simply call 
"factoring" 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 

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.

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.

Your reasoning depends entirely on your condition:

  If you momentarily assume that 0a may not equal zero for all a, ...

That is just what I am saying, except that you seem to be considering 
it possible that this assumption could be true. Since it is not, all 
the rest of what you say really doesn't mean much. I think it's a lot 
cleaner to say, as I have, that factoring just sets us up to make the 
mistake, by unwittingly making this assumption, rather than that 
there is something in the factored equation itself that implies that 
2=1. You are suggesting that there should be a law against people 
walking up the street, rather than against breaking and entering.

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
Associated Topics:
Elementary Number Sense/About Numbers
Middle School Number Sense/About Numbers

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.