Prove or disprove that if y=x, then 0=0?

Question

I got this wrong on my exam. Where did I go wrong?

Disproof: suppose $y=x$. Then $y-x=x-x$ is true. Dividing both sides by $y-x$, $$\frac{y-x}{y-x}=\frac{x-x}{y-x}$$The right hand side simplifies down to $1$, while the left hand side simplifies down to $0$ since $x-x=0$. Thus $$1=0$$ which is a contradiction.

EDIT: Contrapositive. Suppose $0≠0$ (for example $0.0000000001 ≠ 0.00000000000000000002$). Then there exists $k \in R$ such that $0=0+k$. Subtracting $x$ from both sides, $0-x=0-x+k$. Adding $y$ to both sides, $0-x+y=0-x+y+k$. Subtracting $0$ from both sides, $k-x+y=-x+y+k$. Subtracting $k$ from both sides, $-x+y=-x+y$. Thus $x=y$.

I'm at a loss as to how I should prove it...

Answer 1

You've divided by $0$.

Needless to say, this will give you garbage results.

Answer 2

when $y=x$, $y-x=0$ and hence, you can not divide by $(y-x)$ as division by zero is not allowed.