The question:

What is wrong with the following "proof"? Let x=y. Then

$\displaystyle x^2 = xy$
$\displaystyle x^2-y^2= xy-y^2$
$\displaystyle (x+y)(x-y)= y(x-y)$
$\displaystyle x+y= y$
$\displaystyle 2y= y$
$\displaystyle 2=1$

Solution: I know that 2 does not equal 1 and my answer was an incorrect substitution in step 5 however that makes no sense as x=y. The book says that in step 3 'they' incorrectly divided by (x-y)=0. I do not understand, if (x-y) is a whole number then would it not be valid to perform the division?
i.e.$\displaystyle \frac {y(x-y)}{(x-y)} = y$?

Now, perhaps this post would be better suited in the set theory/logic section but as the concept under discussion is of elementary character I felt that would be an unnecessary aggrandizement. Also I am unsure as to the formal system of 'proof' being employed, is it deduction?

2. Originally Posted by Foxlion
The question:

What is wrong with the following "proof"? Let x=y. Then

$\displaystyle x^2 = xy$
$\displaystyle x^2-y^2= xy-y^2$
$\displaystyle (x+y)(x-y)= y(x-y)$
$\displaystyle x+y= y$
$\displaystyle 2y= y$
$\displaystyle 2=1$

Solution: I know that 2 does not equal 1 and my answer was an incorrect substitution in step 5 however that makes no sense as x=y. The book says that in step 3 'they' incorrectly divided by (x-y)=0. I do not understand, if (x-y) is a whole number then would it not be valid to perform the division?
i.e.$\displaystyle \frac {y(x-y)}{(x-y)} = y$?
since x=y then clearly x-y=o

3. It is clear isn't it, and yet here I am. Thank you

4. And it is NOT valid to divide by the whole number 0.

5. haha no, it isn't, though it is tempting to a layman such as I. It's an English thinker's problem in attempting to reconcile the irreconcilable.