Show that if x $\neq$ 0, then $x^2$ > 0 
Show that if $x \neq 0$, then $x^2 > 0$
The author gives the hint that:
"By the trichotomy rule, if $x \neq 0$, then either $x > 0$ or $x < 0$. Consider the two cases separately."

I have no idea how to go about doing proofs myself but here's an attempt at doing one by contradiction.
Proof:
Let's say that $x^2\ngtr0$ for $x\neq 0$
By the law of trichotomy, the two possible cases are that either:
i. $x^2 = 0$
ii. $x^2 < 0$
Now case ii cannot be true because the square of $x$ such that $x \in\mathbb{R}$ will never be a negative number.
The case i will be True $\iff$ $x=0$, which by our initial supposition is false.
Hence for $x \neq 0$  we have that $x^2 > 0$
Now as I said, I'm a total amateur. Is my proof fine? Please let me know how I can go about making it better or doing it right in case it isn't right.
I would really appreciate any help with this.
 A: As suggested in the comments, at a certain point you assume what you're trying to prove. Let me write down what I think could be a solution:
The two cases the hint is talking about are $x<0$ and $x>0$, because any real number is positive, negative or null (this is the trichonomy is talking about). Let us analyze them separately.


*

*$x<0$: then $x^2=x\cdot x>0$, as the product of two negative numbers is positive.

*$x>0$: then $x^2>0$ as the product of two positive number is positive.
Since we are assuming $x\neq 0$, this ends all possible cases, and we are done.
A: Let $x\in \mathbb R$.
Suppose $x\neq0.$ Then by trichotomy rule either $x\gt0 \ \text{or} \ x\lt0$.
Case i: $x\gt0$. 
multiply both sides of the inequality by $x$. then $x\cdot x\gt 0\cdot x\implies x^2\gt0$
Case ii: $x\lt0$. 
multiply both sides of the inequality by $x$. then $x\cdot x\gt 0\cdot x\implies x^2\gt0$ (inequality reveserd since $x\lt0$)
thus in both cases, $x^2\gt0.$
A: "Now case $ii$ cannot be true because the square of $x$ such that $x\in\mathbb{R}$ will never be a negative number." - as JMoravitz has pointed out, this is essentially what you have to show. Follow the line hinted in the book.
Let $x>0$. Then we know, product of two strictly positive numbers is positive. Thus $x^2>0$.
Let $x<0$. Then $-x$ is positive. Then, again since product of two strictly positive numbers is positive, $x^2=(-x)\cdot(-x)>0$.
If you want a proof of the fact that product of two strictly positive numbers is positive, you can use the order axioms. Let $x$ and $y$ be two positive reals. If for some $z\in\mathbb{R}$, $x>z$, then, since $y>0$, so from the order axioms, we have, $x\cdot y>z\cdot y$. In particular since $x>0$, we can choose $z=0$ in this scenario, which in turn gives $x\cdot y>0$.
