Given the definition for any real numbers a and b, the max function is

$\max\{a, b\} = \begin{cases} a \text{ if } a \geq b \\ b \text{ if } a < b \\ \end{cases} $

Lemma: for any two real numbers $a\text{ and }b, a \leq \max\{a,b\}$ and $ b \leq \max\{a,b\}$.

Prove that for real numbers, $a, b$, and $x$, if a $\leq x \leq$ b then $|x| \leq \max\{|a|,|b|\}$.

So far, I have

$x \geq 0$ or $x < 0$ by the law of trichotomy

Case 1: $x\geq 0$

$|x| = x$ by the definition of absolute value

and I'm not sure where to go from there. Any hints?


Hints Through a Conversation

I'm going to restate your lemma to make it simpler, so you can see later how this lemma is applicable to your problem.

Lemma. For all real numbers $\alpha$ and $\beta$ $$\alpha \leq \max\big(\alpha, \beta\big)$$

We don't need to state $\beta \leq \max\big(\alpha,\beta\big)$, because if the above lemma is true for all $\alpha$, then it is true for all $\beta$; that is $\beta \leq \max\big(\alpha,\beta\big)$.


Next we need to observe the rather trivial inequality $|\alpha| \leq |\alpha|$. Obviously, they are equal, but it will be more useful at the moment to right it as an inequality, because, by the definition of absolute value, we obtain $$-|\alpha| \leq \alpha \leq |\alpha|$$ You can use this in your proof by saying something like "Since $|b| \leq |b|$ we have $-|b| \leq b \leq |b|$ by definition of absolute value." This is a valid move, because we are only using the definition of the absolute value and the trivial inequality $|b|\leq|b|$.

Now, notice that by definition of the absolute value, the inequality $|x|\leq\max\big(|a|,|b|\big)$ is equivalent to $$-\max\big(|a|,|b|\big)\leq x \leq \max\big(|a|,|b|\big)$$

Let's focus on the right inequality $x \leq \max\big(|a|,|b|\big)$. From the above lemma if we set $\alpha = |b|$ and $\beta = |a|$, then we get $$|b| \leq \max\big(|a|, |b|\big)$$ What remains to be shown is that $x \leq |b|$. Can you use our observation to establish the right inequality?

You now have enough information to work on a proof. I have provided my proof but concealed it. Hover over the yellowish area with your mouse to reveal my solution.


Let $a,b$ and $x$ denote real numbers such that $a \leq x \leq b$. Since $b \leq |b|$ we have $$x \leq b \leq |b| \leq \max\big(|a|,|b|\big)$$ Further $-\max\big(|a|, |b|\big) \leq -|a| \leq a$, so we deduce $$-\max\big(|a|, |b|\big) \leq -|a| \leq a \leq x \leq b \leq |b| \leq \max\big(|a|, |b|\big)$$ or $$-\max\big(|a|, |b| \big) \leq x \leq \max\big(|a|, |b|\big)$$ Therefore $$|x| \leq \max\big(|a|, |b|\big)$$


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