1. ## absolute value proof

Prove $|x| \leq y \Longleftrightarrow -y \leq x \leq y$.

If $x \geq 0$, then $|x| = x$. So $x \leq y$. Then $|x| = x \geq -y$ and so $-y \leq x \leq y$. If $x < 0$, then $|x| = -x$. So $|x| = -x \leq y$ and $|x| = -x \geq -y$. Thus $-y \leq x \leq y$.

If $-y \leq x \leq y$ and $x \geq 0$, then $|x| = x \leq y$ and $|x| = x \geq -y$. Similar process if $x < 0$.

Is this correct?

2. Looks correct to me, though there are some weird things...

This is of course assuming that $y \geq 0.$

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Prove $|x| \leq y \Longleftrightarrow -y \leq x \leq y$.

If $x \geq 0$, then $|x| = x$. So $x \leq y$. Then (<< then ? it's not a consequence of $x \leq y$ but of $x \geq 0$) $\underbrace{{\color{red}|x| =}}_{\text{useless}} x {\color{red}\geq 0} \geq -y$ and so $-y \leq x \leq y$. If $x < 0$, then $|x| = -x$. So ${\color{red}\dots}-x \leq y$ and ${\color{red}\dots} -x {\color{red} \geq 0} \geq -y$. Thus $-y \leq x \leq y$ (<< by multiplying each inequality by -1)

If $-y \leq x \leq y$ and $x \geq 0$, then $|x| = x \leq y$ and $|x| = x \geq -y$. Similar process if $x < 0$. (<< this one is good)