\beginsection{3.5}

(a) Show that $|b|\le a$ if and only if $-a\le b\le a$.

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First prove the implication.
Let $|b|\le a$.
Then by Theorem 3.2, $-a\le-|b|$.
By $-|b|\le b\le|b|$ we have
$$-a\le-|b|\le b\le|b|\le a$$
Hence $|b|\le a$ implies $-a\le b\le a$.

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Now prove the converse.
By hypothesis we have
$$b\le a\eqno(1)$$
Also by hypothesis we have $-a\le b$ which implies
$$-b\le a\eqno(2)$$
Therefore by (1) and (2) we have
$$|b|\le a$$

