I can't spot the mistake. What's happening?

Question

Let x and y be equal integers, x or y isn't equal to 0 or 1. We have the following:

$x=y$ -subtract x to both sides

$x-x=y-x$

$0=y-x$ -divide y²-x to both sides

$\frac{0}{y^2-x}=\frac{y-x}{y^2-x}$

$0=y$ -divide y to both sides

$0=1$

What's happening?

$\frac{y-x}{y^2-x} \ne y$. Perhaps you are thinking of $\frac{y-x}{y^2-xy}=\frac{y-x}{y(y-x)} = y$ for $y-x\ne0$ — zahbaz, Apr 17 '18 at 05:44
Write it down step-by-step for $,x=y=2,$, and it will become obvious where it goes wrong. — dxiv, Apr 17 '18 at 05:47
Well, not (number-theory) either; if you hover with the mouse over the tag you will see a short description, which should make it clear why it's not a suitable tag in this case. Perhaps (algebra-precalculus) and/or (fake-proofs)? — Hans Lundmark, Apr 17 '18 at 07:08

score 0 · Accepted Answer · answered Apr 17 '18 at 08:00

Your first error is that $\frac{y-x}{y^2-x} \ne y$, but similar (fake) proof could avoid this. The most important thing to note is that you divide $y^2-x$ or $y$ on both side. There is the problem.

When you have an equality $a=b$, you must $\textbf{always}$ check that $b\ne 0$ if you want to conclude that $\frac{a}{b}=1$. Indeed, $\textit{dividing b on both side}$ actually means $\textit{multiply by} \ \frac{1}{b}$ on both side, which is only possible if $b\ne0$.

Now, as said @dxiv, try to do your reasoning with $x=2, y=2$ to see where you cheated.

But i said x or y isn't equal to 0 or 1. Anyways, i will accept your comment. Thank you. — Vincent William Rodriguez, Apr 17 '18 at 10:30

I can't spot the mistake. What's happening?

1 Answers1