Proving that $-|a| \leq a \leq |a|$ for all $a \in \mathbb R$

Question

I'm confused on the notation of the definition of an absolute value of a number. According to a book I'm reading, it says that for any real number $a \in \mathbb R$, $|a|:= \begin{cases} a, & \text{if $0 \leq a$} \\ -a, & \text{if $a \lt 0$} \end{cases}$.

So I've already been given a particular set, the real numbers $\mathbb R$, and a single binary operation "$\leq$" which in this case, represents inequality. Since we haven't been given what "$\lt$" is, I'm assuming that for any real numbers $a,b \in \mathbb R$, "$a \lt b$" means that the proposition "$b \leq a$" is false.

The reason I'm asking this is that I'm stuck on proving that $-|a| \leq a \leq |a|$ for all $a \in \mathbb R$. In the case that $0 \leq a$, we have that $|a| = a$, so using the axioms of the inequality "$\leq$", I end up getting $-|a| \leq a \leq |a|$. But in the case that $a \lt 0$, we have that $|a| = -a$, and now I'm confused on proving that $a \leq -a$ now since I can no longer be using the axioms for the binary operation "$\leq$". I'm now dealing with "$\lt$", in which I haven't been given any axioms for them yet. How do I approach this?

Thanks in advance.

Answer 1

While perhaps uncommon, it is possible to define $\lvert a \rvert$ and prove that $$-\lvert a \rvert \leq a \leq \lvert a \rvert$$ without resorting to the strict order $<$ or invoking trichotomy.

We need to already know that addition $+$ and the weak order $\leq$ form the structure of a lattice-ordered abelian group with meet $\wedge$ and join $\vee$. We also need to know that the inclusion-exclusion principle $$a\wedge b + a\vee b = a + b$$ holds as an identity.

To start, define $$\begin{align}a^+&=a\vee 0 \\ a^-&=-(a\wedge 0) \\ \lvert a \rvert &= a^+ + a^-\text{.} \end{align}$$ Then we can show that $$\begin{align} 0&\leq a^+ \\ 0&\leq a^- \\ a&=a^+-a^- \end{align}$$ from which it follows that $$-\lvert a \rvert \leq a \leq \lvert a \rvert\text{.}$$

An advantage of this treatment over others is that it motivates the theory of Riesz spaces, which unify many results in functional analysis.

Answer 2

1. If $a=0$, then the inequality is trivial as $ 0\leq0\leq0 $ is certainly true.

2. If $a>0$ ($a$ is positive) $\implies -a<0<a.$

   $|a|=a \implies a\leq|a|=a$ and $a\geq-|a|=-a$

3. If $a<0$ ($a$ is negative), then $a<0<-a$

   $|a|= -a$ (remember $a$ is negative so $-a$ is positive) $\implies a\leq|a|=-a$ and

   $a\geq -|a| =-(-a) = a$

Answer 3

As you have already proved the result for $a\geq 0$, you can use exactly this as follows:

• $a<0 \Rightarrow a' = -a > 0$ and note that by your definition of the absolute value you have $|a| = |a'|$.

Now, apply your result to $a'> 0$. So, you get $$-|a'| \leq a' \leq |a'| \stackrel{a' = -a}{\Longleftrightarrow}-|a| \leq -a \leq |a|\Leftrightarrow \boxed{|a| \geq a \geq -|a|}$$