[ILikeFriedman]

Assume a=b=1
then
(a+1)=(b+1), (a-1)=(b-1), and (a*a)-1=(b*b)-1
and ......

(a*a)-(b*b)=(a-b)
use this last equation and apply the difference of 2 squares
(a+b)*(a-b)=(a-b)
divide both sides of the equation by (a-b)
(a+b)=(a-b)/(a-b)=1
since a=b=1, and substituting for a and b
1+1=1 or
2=1

[maro]

You've divided by zero, which is why you reach an impossible result.

Repeat after: math is consistent, math is consistent, math is consistent. If I reach a nonsensical result, I did something wrong.

[ShadowAce]

I've seen that result from a slightly different proof:

Assume x=y
x+x² = x²+y
x-x²-y = x²
x-x²-y-xy = x²-xy
(x-y)(x+1) = x(x-y)
x+1=x for all x

[ozidar]

divide by zero error.

[MarkL]

Anybody else here remember doing a two page mathematical proof, getting something screwed around, and winding up with the original equation?

Mark

[CardCarryingMember.VastRightWC]

Here's a slightly less complex version of it:

Given a=b, prove that 1=2.

PROOF:

      a   =   b          Given
    a*b   =   b^2        Multiply both sides by b
a*b - a^2 = b^2 - a^2    Subtract a^2 from both sides
   a(b-a) = (b+a)(b-a)   Factor
      a   =   b+a        Divide both sides by (b-a)
      a   =   2*a        Since a=b as originally given
      1   =   2          Divide both sides by a

[Cronos]

Your point

(a+b)=(a-b)/(a-b)=1

Then means (by resubstituting a=b=1)

1+1=(1-1)/(1-1)=1
2 = 0/0 = 1

You can't divide by 0.