# Show that if $x^2>y^2$ then $x>y$ for all positive real numbers [closed]

Would the solution be something like: \begin{align*} x^2 &>y^2 \\ xx&>yy\\ \sqrt{xx}&>\sqrt{yy}\\ x&>y \end{align*}

I feel like I'm not proving something though.

## closed as unclear what you're asking by Namaste, Olivier, Leucippus, JonMark Perry, Claude LeiboviciJan 12 '18 at 8:10

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• What is your definition of $>$? – user7530 Jan 11 '18 at 6:56
• ah, I think this is the right idea I'll try to work that out – mathminutemaid Jan 11 '18 at 7:00
• $\sqrt{x^2} = |x|$ not $x$. – steven gregory Jan 11 '18 at 7:01
• You aren't. You can't claim $a > b \implies \sqrt a > \sqrt b$ unless you have some reason to believe that and you have utterly none. That is precisely what you are trying to prove! In essence, all you have done is assumed it is true and claim you have proved it simply because you assumed it. – fleablood Jan 11 '18 at 7:36

$x^2>y^2$, then, $x^2-y^2>0$ but this is equal to $(x-y)(x+y)>0$. We know that $x,y> 0$, then, $x+y>0$ and because $(x-y)(x+y)>0$, thus $x-y>0$. This is equivalent to $x>y$.

By contradiction/contrapositive. If $x\leq y$, then, since $0<x$ by assumption, we have $x^2\leq y^2$.

Your solution looks correct, but you must be aware that you could have thought incorrectly anyway.

The step from $xx>yy$ to $\sqrt{xx}>\sqrt{yy}$ for example uses the property that $\sqrt\cdot$ is strictly increasing. Also the last step you could have thought incorrectly since $\sqrt{xx}=x$ only for positive $x$ (which is where you use that $x$ is positive), in general you have $\sqrt{xx} = |x|$.

Note that

$$x^2>y^2\iff x^2-y^2>0\iff (x+y)(x-y)>0$$

since

$$x+y>0 \implies x-y=0 \implies x>y$$

Let $f(t)=t^2$ for $t\ge0$.

Then $f(t)$ is increasing function. Then if $y<x$ then $f(y)<f(x)$. Then $y^2<x^2$