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.

• What is your definition of $>$? Commented Jan 11, 2018 at 6:56
• ah, I think this is the right idea I'll try to work that out Commented Jan 11, 2018 at 7:00
• $\sqrt{x^2} = |x|$ not $x$. Commented Jan 11, 2018 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. Commented Jan 11, 2018 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 $x^2\leq y^2$.
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 $\sqrt{xx} = |x|$.
Let $f(t)=t^2$ for $t\ge0$.
Then $f(t)$ is increasing function. Then if $y<x$ then $f(y)<f(x)$. Then $y^2<x^2$