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### Proof That All Numbers Are Equal?

```Date: 12/06/2005 at 17:00:34
From: RV
Subject: 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.

```

```
Date: 12/06/2005 at 20:08:07
From: Doctor Minter
Subject: Re: All numbers are equal

Hi RV!

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

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.

In regards to your example (which I may use on my friends, regardless
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

OR

-(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.

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!

- Doctor Minter, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 12/07/2005 at 07:26:26
From: RV
Subject: Thank you (All numbers are equal)

Dr. Minter - Thanks a lot for the explanation, I emailed it to my
friends and they are all amazed.
```
Associated Topics:
High School Basic Algebra
Middle School Algebra

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