# Thread: absolute value proof

1. ## absolute value proof

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

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

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

Is this correct?

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

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

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

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

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