Prove by cases that:
$$|x|≤R \iff -R≤x≤R$$
$R$ is defined as $R≥0$.
I consider the two relevant cases to be $x≥0$ and $x<0$. However, for the latter, if $x<0$ then $|x|=-x$. This yields $-x≤R \iff x≥-R$. Moreover, we have $0>x≥-R$, which implies that $0>-R \iff 0<R$. Thus we end up with $R>0>x≥-R$, or, if turned around and omitting the zero, $-R≤x<R$.
I don't see how this expression is equivalent to the above. According to my book the equivalence should hold for $x≥0$ and $x<0$. Am I missing something?