2 = 1

If we write everything out as an equation it will look like this:

step 1) $a = b$

step 2) $a^2 = ab$

step 3) $a^2 + (a^2 - 2ab) = ab + (a^2 - 2ab)$

simplify

$2(a^2 - ab) = a^2 - ab$

step 4) $\frac{2(a^2 - ab)}{a^2 - ab}$ $=$ $\frac{a^2 - ab}{a^2 - ab}$

$2 = 1$

It sure looks like a proper proof without any holes in it, but there is actually something that's not quite right with it.

During the fourth step we divide $2(a^2 - ab)$ with $a^2 - ab$ and since we know that 2 times something is in fact two times bigger than that something we can easily see that 2 times something divided by the something will bring us a quotient 2. And since we also know that something divided by itself can only bring us the quotient 1, there isn't any reason to be doubtful of this proof. But if we look at the first step again we will find something that makes this proof invalid.

$a = b$ it says. Wait a minute. If a is equal to b doesn't that mean that $a^2 - ab$ could also be written $a^2 - a^2$ ? And something minus itself will give us 0. That means that the equation can be written like this

$\frac{2(a^2 - a^2)}{a^2 - a^2}$ $=$ $\frac{a^2 - a^2}{a^2 - a^2}$

$\frac{2 \times 0}{0}$ $=$ $\frac{0}{0}$

Since anything multiplied by 0 equals 0, $2 \times 0$ $=$ $0$

This results in that the equation can be written like this

$\frac{0}{0}$ $=$ $\frac{0}{0}$

$\frac{a}{b}$ means "the number when multiplied with $b$ gives $a$ " and in the case of $\frac{0}{0}$ every number satisfies that criteria. That's why $\frac{0}{0}$ $=$ $unidentified$

For short:

$2 ≠ 1$