This will blow your mind, well, at least it blew my mind. Some of you other math nerds will probably be able to blow down my card house, but until someone does...
Anyway, one does equal two, and I can prove it.
So, without any further ado:
a = b
a2 = ab
a2 + a2 = a2 + ab
2a2 - 2ab = a2 + ab -2ab
2a2 - 2ab = a2 - ab
2(a2 - ab) = 1(a2 - ab)
2 = 1
Well, there's a catch in it, if you can find it, you're doing better than I did (I had to ask my pre-calculus professor)!
a^2 + a^2 = a^2 + ab
is not necessarily equivalent to
2a^2 - 2ab = a^2 + ab - 2ab
You added a^2 - 2ab on the left side but added only -2ab on the right side. The above two expressions are only equivalent if a^2 - 2ab = -2ab
a^2 - 2ab = -2ab
a^2 = 0
a = 0
Since a = b = 0, the last step in the argument, where you divided both sides by a^2 - ab, is invalid, since a^2 - ab = 0 and it would be a division by zero. More correctly,
2(a^2 - ab) = a^2 - ab
2 * 0 = 0
0 = 0
Oops. My bad, I must have misread. The step where you add -2ab on both sides is not erroneous.
The more proper correction would be
2(a^2 - ab)/(a^2 - ab) = 1(a^2 - ab)/(a^2 - ab)
implies that a^2 - ab != 0
a^2 - ab != 0
a^2 != ab
aa != aa
0 != 0
Since 0 != 0 is not true, it's not true that 2 = 1.
Any division is only valid if the divisor is non-zero. Every time you cancel out some multiplication in an equation you're implying that something is not zero. Forgetting to follow up on those implications is how you arrive at contractions such as the one above, or how you miss solutions to a system of equations. It's quite common to do something like
x^2 = x
x = 1
And forget that x = 0 is also a solution. I'm simplifying, of course; not every example is this obvious.