Exercise 11.1

Let \(a, b \in \mathbb{Z}\), with both nonzero. We want to show that \(\lvert a \rvert \leq \lvert ab \rvert\).

There are a few ways to do this. For the way I’m going to use here, our first step is to show that \(\lvert ab \rvert = \lvert a \rvert \lvert b \rvert\) for any integers \(a, b.\) In fact, this always holds, even if \(a\) or \(b\) is zero. This can be proved by cases. We’ll consider four cases:

  • \(a \geq 0, \; b \geq 0\)
  • \(a \geq 0, \; b < 0\)
  • \(a < 0, \; b \geq 0\)
  • \(a < 0, \; b < 0\)

In the first case, since both \(a\) and \(b\) are nonnegative, we have that \(\lvert a \rvert = a\) and \(\lvert b \rvert = b\), so it follows that \(\lvert a \rvert \lvert b \rvert\) is equal to \(ab\). Also, since \(a\) and \(b\) are both nonnegative, their product \(ab\) is also nonnegative, so \(\lvert ab \rvert = ab\) and we are done.

In the second case, we have that \(\lvert a \rvert = a\) as before, but \(\lvert b \rvert = -b\), since \(b\) is negative, and so the right hand side is equal to \(-ab\). Also, in this case, the product \(ab \leq 0\), so on the left hand side, we have \(\lvert ab \rvert = -ab\). So both sides are equal to \(-ab\) and we are done.

The remaining two cases play out similarly, so I won’t bother to write them out.

Now we know that \(\lvert ab \rvert = \lvert a \rvert \lvert b \rvert\), we can return to our original question. Using our new knowledge, we can rewrite the statement we are trying to prove as \(\lvert a \rvert \leq \lvert a \rvert \lvert b \rvert\). One thing we can do with inequalities is divide both sides by a positive number. Since we have by assumption that \(a\) is nonzero, it follows that \(\lvert a \rvert > 0\) and so we can divide both sides by \(\lvert a \rvert\), leaving us with \(1 \leq \lvert b \rvert\). And since \(b\) is a nonzero integer, \(\lvert b \rvert\) must be a positive integer, so \(1 \leq \lvert b \rvert\) is necessarily true, and we are done.