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### Proof: 2 = 1?

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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

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.
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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 :)
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

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.

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!  http://mathforum.org/dr.math/
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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.
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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
yet.

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.

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
http://mathforum.org/dr.math/
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