A calculus of the absurd

4.9.5 Proof that a value is not greater than its modulus

This is not the most exciting proof, but it helps to illustrate a general principle when proving things about the modulus function: consider the positive and negative cases separately.

  • Example 4.9.3 Let \(a \in \mathbb {R}\) (that is, \(a\) is a real number). Prove that \(a \leqq |a|\).

We would like to show that \(a\) is less than or equal to its modulus. Here we can find two exhaustive cases by the law of the excluded middle (see Section 20.4 for details, but it just means that “either something is true, or it is false”), that is that either \(a \geqq 0\) or \(a < 0\). If these cases suspiciously mirror the definition3030 See Section 4.9.2 if you can’t remember the definition. of the modulus function, that’s because they do.

Why are these cases useful? Because they allow us to assume more about the structure of the modulus function. This comes at a cost - that we have lost generality (that is, we now know more about \(a\) but only in specific cases, so we must proceed case-by-case to prove this in general).

We can begin with the first case - the one where \(a \geqq 0\). First, we can consider what we know about \(|a|\) in this case, which is that \(|a| = a\). This is enough to prove the inequality as

\begin{align} a &\leqq a & \text {As $a = a$ it is certainly less than \textit {or equal to} the value of $a$} \\ &= |a| & \text {Which is what we wanted to prove!} \end{align}

We must then consider the other case, i.e. when \(a < 0\) in this case \(|a| = -a\), that is \(|a| > 0\), therefore in this case \(a < 0 < |a|\) which implies that \(a < |a|\) and thus that \(a \leqq |a|\)