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 
of your proof.  

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 
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

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.