How to write simple proofs? "If $x,y\geq 0$, then: $x>y\Rightarrow x^2>y^2$"

Question

How to write simple proofs? "If $x,y\geq 0$, then: $x>y\Rightarrow x^2>y^2$"

Hello,

indeed, this is really obvious. But I am having trouble writing down a proof.

It is already known that if you multiply two numbers greater than zero, you end up getting another number that is greater than zero. So I can conclude that $x^2,y^2\geq 0$. And since $x>y$, it follows that $x^2>y^2$.

But how can I write my thoughts down formally?

Answer 1

If $y=0,$ then the result is obvious, so suppose $y>0$.

If $x>y$ and both are non-negative, then multiplying both sides of the inequality by $x$ preserves the ordering, giving that $$xx>xy.$$ Similarly, multiplying both sides of the original inequality by $y$ gives that $$xy>yy.$$ Putting this all together, $$x^2=xx>xy>yy=y^2.$$

The preservation of the inequality after multiplying by a positive number comes from $\mathbb{R}$ being an ordered field (which I assume you know for this problem).

Answer 2

You could start with the difference:

$x^2 - y^2 = (x + y)(x - y)$

Now, $$x, y > 0 => x + y > 0$$

And $$x > y => x - y > 0$$

Hence,

$x^2 - y^2 > 0$ $=> x^2 > y^2$

Answer 3

Recall the following order axioms:

If $a > b$ and $b > c$, then $a > c$.

If $a > b$ and $c > 0$, then $ac > bc$.

Since $x > y > 0$,

$$ x^2 = x \cdot x > x \cdot y $$

and also because $x > y > 0$,

$$ x \cdot y > y \cdot y = y^2$$

and so by the first order axiom listed above,

$$ x^2 > y^2.$$