$|x_i|$ is the smallest number $z_i$ that satisfies $x_i \leq z_i$ and $-x_i \leq z_i$

Question

I'm a bit ashamed that I have a math degree and can't figure out this seemingly simple problem, but that's what the situation is. From Introduction to Linear Optimization by Bertsimas and Tsitsiklis, they mention on p. 18 that

$|x_i|$ is the smallest number $z_i$ that satisfies $x_i \leq z_i$ and $-x_i \leq z_i$

How does one prove this?

I wish I could provide some more context to this, but this statement seems to have been mentioned as a side comment. We don't know anything other than that $x_i \in \mathbb{R}$.

I do know that by definition (and I have to admit, it is confusing that they use $z_i$ to represent $|x_i|$ as well):

$$|x_i| = \begin{cases} x_i, & x_i \geq 0 \\ -x_i, & x_i < 0\text{.} \end{cases}$$ I believe what we're trying to prove is that $$|x_i| = \min\{z_i \mid x_i \leq z_i, -x_i \leq z_i\}\text{.}$$ However, I'm not sure how to derive this from the definition of $|x_i|$ given above.

Answer 1

Let $z=\min\{z_i \mid x_i \leq z_i, -x_i \leq z_i\}\text{.}$ Let us also get rid of the index $i$ of $x_i$: I will denote it $x$.

First, we have $x\leq |x|$ and $-x\leq |x|$, which can be established simply by a disjonction of cases (depending on whether $x$ is negative or non-negative). Hence, by definition, $z\leq |x|$.

Second, we know that $z\geq \max(x,-x)$ by definition. But we have the identity $\max(x,-x)=|x|$ which again may be shown by disjonction of cases.

We conclude that $|x|=z$.