# Thread: Where does this problem fail?

1. ## Where does this problem fail?

So, someone was showing me a math problem where he said he could get 1 = 2. Here's what he did:

$\displaystyle x = 1$

Thus
$\displaystyle x^2 = x^2$

So
$\displaystyle x^2 - x = x^2 - 1$
Right?

Then, he factored:
$\displaystyle x(x -1) = (x - 1)(x + 1)$

Divided by (x - 1):
$\displaystyle [x(x-1)]/(x-1) = [(x-1)(x + 1)]/(x - 1)$
to cancel out (x - 1)

Resulted in:
$\displaystyle x = x + 1$
From where he plugged in x = 1:
$\displaystyle 1 = 1 + 1$
Therefore:
$\displaystyle 1 = 2$

Please explain to me why this fails, because it seems like it does in so many places, but I want to know why exactly! For instance, dividing by $\displaystyle (x - 1)$ would be the same as dividing by zero, which would fail, but he thinks he's pretty much found where math has a weakness. I need expert verification!

2. There is a division per zero while simplifying, which invalidates all the following, sorry. You cannot divide per zero, because it would mean that $\displaystyle 2 = 1$, crumbling the whole theory of mathematics. Look at this :

$\displaystyle a = b$

$\displaystyle a^2 = ab$ (multiply by $\displaystyle a$)

$\displaystyle 2a^2 - 2ab = a^2 - ab$ (add $\displaystyle a^2 - 2ab$)

$\displaystyle 2(a^2 - ab) = a^2 - ab$ (factorize on left side)

$\displaystyle 2 = 1$ (simplify by $\displaystyle a^2 - ab$)

Which is impossible, because $\displaystyle a^2 - ab = 0$ (division per zero).

3. Ray, you are awesome! Thanks!

Quick edit: When you say (add $\displaystyle a^2 - 2ab$), do you mean $\displaystyle 2a^2 - 2ab$?

4. Originally Posted by audax
Ray, you are awesome! Thanks!

Quick edit: When you say (add $\displaystyle a^2 - 2ab$), do you mean $\displaystyle 2a^2 - 2ab$?
No, I mean $\displaystyle a^2 - 2ab$

5. ## can't divide by 0

note that x-1=1-1=0, so you can't divide by 0, as you have done.