# Where does this problem fail?

• Nov 9th 2009, 07:26 PM
audax
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:

$x = 1$

Thus
$x^2 = x^2$

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

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

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

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

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 $(x - 1)$ 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!
• Nov 9th 2009, 07:40 PM
Bacterius
There is a division per zero while simplifying, which invalidates all the following, sorry. You cannot divide per zero, because it would mean that $2 = 1$, crumbling the whole theory of mathematics. Look at this :

$a = b$

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

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

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

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

Which is impossible, because $a^2 - ab = 0$ (division per zero).
• Nov 9th 2009, 07:46 PM
audax
Ray, you are awesome! Thanks!

Quick edit: When you say (add $a^2 - 2ab$), do you mean $2a^2 - 2ab$?
• Nov 9th 2009, 08:07 PM
Bacterius
Quote:

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

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

No, I mean $a^2 - 2ab$ (Happy)
• Nov 10th 2009, 07:46 AM
picozzi
can't divide by 0
note that x-1=1-1=0, so you can't divide by 0, as you have done.