What is wrong with this equations?

Question

so our math teacher told us how 2 * 2 = 5 today and we were like :O

I thought really hard to disprove this but it seems correct, would someone please tell me how is this possible!

proving 2 * 2 = 5 :

we know that : 25 - 25 = 0
and : 20 - 20 = 0
so : 25 - 25 = 20 - 20
now we can factor this equation into : 5 ( 5 - 5 ) = 4 ( 5 - 5 )
now both ( 5 - 5 )s are identical right? so : 5x = 4x
therefore : 5 = 2 * 2 , 5 = 2 + 2 , 10 = 8 , 15 = 12 ,...

oh and another thing : proving 2 + 2 = 2

lets say : x = 2 and y = 2
so : y^2 = xy
and : x^2 - y^2 = x^2 - xy
so : ( x - y ) ( x + y ) = x ( x - y )
we divide both sides by ( x - y ) and we have : x + y = x
now if we plug in numbers we have : 2 + 2 = 2

Answer 1

You should know that:

$ax=bx \implies a =b$

is not valid unless we know that $x \neq 0$.

That is: $$ax = bx \Leftrightarrow ax - bx = 0$$ $$\Leftrightarrow x(a-b) = 0$$

what is the conclusion ? $x = 0$ or $a=b$

In both reasoning above he exploited this fact to drive such conclusions.

Answer 2

The formulation logically is that $(AB = 0) \Leftrightarrow (A=0) \text{ or }(B=0)$. One, both, or neither of the secondary equations may have a solution. The problem is that in the cited example, the valid secondary equation was ignored and only the one with no solution was kept.

Incidentally, there's nothing wrong with coming up with an equation that says $5=4$. It's perfectly fine, and in fact, you often want to end up with something like that. It's the whole idea behind proof by contradiction. The proper conclusion, though, is NOT that $5$ and $4$ are equal; instead, since it is obviously false, you should conclude that the secondary equation leading to this falsehood has no solution.