Originally Posted by

**Syrus** I seem to think my proof of the following theorem has been made more complicated/lengthy than necessary. Any thoughts will be highly appreciated =)

__Prove that for all real numbers a and b, |a|≤ b iff -b ≤ a ≤ b.__

ADDENDUM! I went forward and completed part b) of this exercise before any responses were posted, so I figured I attach that as well (they are related).

Prove that for any real number x, -|x|≤ x ≤ |x|.

__Proof:__

Let a and b be arbitrary real numbers.

(-->) Suppose |a|≤ b.

**Case I.** a ≥ 0. Then |a|= a and we have 0 ≤ |a| = a ≤ b. This also means that 0 ≤ b, and so -b ≤ 0. Thus, -b ≤ 0 ≤ |a| = a ≤ b. In particular, then, -b ≤ a ≤ b.

**Case II.** a < 0. Then 0 < |a| = -a ≤ b. But since |a| = -a ≤ b, -|a| = a ≥ -b. Thus, -b ≤ a. Also, though, we have that 0 < -a and 0 < b. Adding these two inequalities yields 0 < b - a. Adding a to each side gives a < b. Clearly, then, a < b or a = b, so we may state that a ≤ b. Finally, together: -b ≤ a ≤ b.

Since these cases are exhaustive and each results in -b ≤ a ≤ b, it follows that if|a|≤ b, then -b ≤ a ≤ b.

(<--) Suppose -b ≤ a ≤ b. This means -b ≤ a and a ≤ b.

**Case I.** a ≥ 0. Then |a| = a ≤ b. So in particular, |a|≤ b.

**Case II.** a < 0. Then |a| = -a ≤ b. So in particular, |a| ≤ b.

Since these cases are exhaustive and each results in |a| ≤ b, it follows that if -b ≤ a ≤ b, then |a| ≤ b.

Lastly, since we have proven both directions of the biconditional for arbitrary real numbers a and b, we may conclude that for all real numbers a and b, |a|≤ b iff -b ≤ a ≤ b.