Absolute value in real number system

Question

I got a theorem from my lecture's note. Proof that \begin{align*} |x|≤|y| \leftrightarrow x^2≤y^2 \end{align*}

Then, I try to proof that theorem. \begin{align*} |x|≤|y| \leftrightarrow \sqrt{x²}≤\sqrt{y²}\\ \leftrightarrow x²≤y² \end{align*}

But, i am not sure with my answer. Is it proven enough like that? Thanks for any help

Answer 1

The idea behind your proof is OK. Whether it is correct depends on what your definitions are and how much you are allowed to assume or have already proved.

In your first step, can you explain why $|x| = \sqrt{x^2}$?

In your second step, can you explain why squaring each side of an inequality preservers the inequality? (It's not always true.)

Answer 2

Since $a\geq b, c>0\Rightarrow ac\geq bc$, we have $|y|\geq|x|\Rightarrow|y||y|\geq|y||x|\geq|x||x|$, so $y^2\geq x^2$ for $x,y\neq0$ and it is trivially true for either $x=0$ or $y=0$.

Answer 3

You can make this preliminary observation: for any $x\in\mathbf R$, $\;x^2=|x|^2$, so, as we have only non-negative numbers for the first equivalence, $$|x|\le |y| \iff |x|^2\le |y^2|\iff x^2=y^2.$$

Answer 4

I have: $ \left | x \right | \leq \left | y \right | $ and $ \left | x\right | - \left |y \right | \leq0$. Now I multiply by $( \left | x\right | + \left | y\right | )$ that is always positive. So $ \left | x\right |^2\leq \left | y \right |^2$. Because of $ \left | x \right |^2 = x^2$, I end up with: $$x^2\leq y^2$$