Proofs in Cases

Doug Baldwin

Spring 2021

For an example of what a proof by cases might look like when fully written out, suppose \(f(x)\) is defined by the following:

\begin{equation} f(x) = \begin {cases} x^2 & \mathrm {if\ } x < 0 \\ x & \mathrm {if\ } 0 \le x \le 1 \\ 2x - 1 & \mathrm {if\ } 1 < x \end {cases} \end{equation}

Then we have

Conjecture 1. For all real numbers \(x\), \(f(x)\) as defined by (1) is greater than or equal to \(0\).

Proof. We will prove that \(f(x) \ge 0\) for all real numbers \(x\) by analyzing each case in the definition in turn.

For the first case, consider \(x < 0\). Then \(f(x) = x^2\), which is greater than or equal to \(0\) for all values of \(x\).

For the second case, consider \(0 \le x \le 1\). In this case \(f(x) = x\) which is greater than or equal to \(0\) from the constraint \(0 \le x \le 1\).

Finally, consider \(1 < x\). Here, \(f(x) = 2x - 1\) which is an increasing function, i.e., \(f(x) > f(1)\) for all \(x > 1\). Since \(f(1) = 1 \ge 0\), we see that \(f(x) \ge 0\) for all \(x > 1\).

We have now shown that \(f(x) \ge 0\) for all cases in the definition of \(f\), and thus that \(f(x) \ge 0\) for all real numbers \(x\). □