How to prove $x \leq 0, 0 \leq y \implies x.y \leq 0$?

Question

In real analysis course, I need to prove $x \leq 0$ and $0 \leq y \implies x.y \leq 0$. I applied following approach but am not sure whether it is correct and acceptable proof.

1. Assume $x \leq 0$
2. Adding $-x$ to both sides yields $0 \leq -x$
3. Assume $0 \leq y$
4. We obtain $0 \leq -x.y$ from (2) and (3). (Recall the axiom $0 \leq x$ and $0 \leq y \implies 0 \leq x.y$)
5. $0 \leq (-1).x.y$ and put $z = x.y$
6. $0 \leq (-1).z$
7. Add $z$ to both sides
8. So, $z \leq (-1).z + z$ which is $z \leq -z + z$
9. $z \leq 0$ which is $x.y \leq 0$

Answer 1

Every thing is good except you are taking it for granted that $(-x)y = -(xy)$ and that $-x = (-1)x$.

That must be proven.

And before you can prove that you must prove $0\cdot x = 0$.

You can not take either of those as a given.

Claim: $0x = 0$. Pf: $0= 0\cdot x + (-(0\cdot x) = (0 + 0)x + (-(0\cdot x))= 0\cdot x + 0\cdot x + (-(0\cdot x) = 0\cdot x + (0\cdot x + (-(0\cdot x)) = 0\cdot x + 0 = 0\cdot x$.

And

Claim: $(-x)y = -(xy)$. Pf: $0 = 0*y = (-x + x)*y= (-x)y + xy$ So $0 + (-xy) = (-x)y + xy + (-xy)$ so $-xy = (-x)y + 0 = (-x)y$.