A troll can also prove that 2 == 1 by continuously folding the peaks of an equilateral triangle down to the baseline.
Another fun one:
$1 = 100¢
$.1 = 10¢
$.1^2 = 10¢ ^ 2
$.01 = 100¢
thus
$1 = 1¢
Trollface
Understanding you to be a distinguished algebraist (that is, distinguished from other algebraists by different face, different height, etc.), I beg to submit to you a difficulty which distresses me much.
If x and y are each equal to 1, it is plain that
2 * (x^2 - y^2) = 0, and also that 5 * (x - y) = 0.
Hence 2 * (x^2 - y^2) = 5 * (x - y).
Now divide each side of this equation by (x - y).
Then 2 * (x + y) = 5.
But (x + y) = (1 + 1), i.e. = 2. So that 2 * 2 = 5.
Ever since this painful fact has been forced upon me, I have not slept more than 8 hours a night, and have not been able to eat more than 3 meals a day.
I trust you will pity me and will kindly explain the difficulty to Your obliged,
Lewis Carroll.
You can't divide by 0 you just get nonsense.
So, the actual problem is dividing by zero. Your assumption that "we are dealing with algebra and not numerical values" is false because it completely ignores the "if x and y = 1" part.
The problem really is that you can't divide by zero, even in an algebraic expression.
A simpler example of this phenomenon (which blew my mind when I first encountered it) occurs with the equation x = x^2. If you divide by x, you get x = 1, which is a solution to the equation, but where did the other solution x = 0 go??
Whenever you divide an equation by an algebraic expression, you need to consider the possibility of that expression being zero and treat it as a special case. So in the case of x = x^2, you can reason as follows: maybe x = 0, in which case … what … ah yes, that's a solution! Or maybe x ≠ 0, in which case we can divide by it and get x = 1. That doesn't contradict the assumption x ≠ 0, so it's okay, and x = 1 is the other solution.
0.01 (dollars)^2 = 100 (cents)^2
0.01 (dollars)^2 = 100 (1/100 dollars)^2