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


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?

  • 5
    $\begingroup$ This isn't obvious and is not just a matter of writing your thoughts down formally. You have to actually come up with an algebraic justification for it using only the facts you know. Exactly how you can do that depends on what the facts you know are. $\endgroup$ Commented Jul 26, 2019 at 18:36
  • $\begingroup$ You need to show that the function $x^2$ is strictly increasing on $\mathbb{R}_+$. One way to go is to observe the derivative. $\endgroup$ Commented Jul 26, 2019 at 18:38
  • 3
    $\begingroup$ @HamidMohammad It's a bit overkill to use calculus on this problem. $\endgroup$
    – Jam
    Commented Jul 26, 2019 at 18:45

3 Answers 3


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).

  • $\begingroup$ Thanks, I understand...! $\endgroup$
    – Analysis
    Commented Jul 26, 2019 at 19:27
  • $\begingroup$ @ParabolicAlcoholic Awesome! $\endgroup$
    – cmk
    Commented Jul 26, 2019 at 19:29

You could start with the difference:

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

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

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


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


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,

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


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