# Spivak's Calculus: Ch. 1, Problem 14 (absolute value inequality)

Still getting used to the level of rigor in this text, question 14 is where my confidence in my proofs starts to degrade. I am looking for critiques of my proofs--not limited to what I got flat-out wrong, but ways the proofs could be more concise or elegant.

14a) Prove that $|a| = |-a|$.

My proof:

$a \ge 0 \to |a| = a$ $\land$ $-a \le 0$

$-a \le 0 \to |-a| = -(-a) = a$

$\therefore |a| = a = |-a|$

So by transitivity of equality, $|a| = |-a|$

b) Prove that $-b \le a \le b \iff |a| \le b$.

I decided to go for a contradiction first to get the 'only if' part down.

Given $b \ge -b$, we know that $b \ge 0$

$\therefore |a| \gt b \to (a \gt b)\land(-a \lt -b )$

$\therefore |a| \gt b \to -b \not\le a \not\le b$

This completes the contradiction. Now to demonstrate that the inequality we want does follow from the condition provided.

$|a| \le b \to (a \le b)\land(-a \ge -b)$

This is the first bit of frustration, because I understand intuitively that the magnitude of a being less than b means that it must be greater than -b and less than b, but I'm not sure how to get from the above, which only compares a to b and -a to -b, to a unified inequality of -b, a, and b.

$\therefore -b \le a \le b$

The question also mentions that it follows that $-|a| \le a \le |a|$ I'm not sure I understand how this follows from what I just demonstrated, though I do understand the inequality itself. Is it just that when you substitute |a| in for where b was, you get that result? Finally,

14c) Use this fact to give a new proof that $|a + b| \le |a| + |b|$

This is where it gets very ugly for me, but here's what I've tried:

Using 14b, $-|a| - |b| \le a + b \le |a| + |b| \iff |a+b| \le |a| + |b|$

By the definition of absolute value, we know that the left-hand statement is true, therefore the right-hand statement must also be true.

That seems overly simplistic, even primitive, but it's the best thing I could come up with.

Would a direct proof be alright?

For example, result (b) follows directly from the definition of the absolute value:

By definition:

$$|a|=\left\{\begin{array}{rl}-a,&a<0\\a,&a\ge 0\end{array}\right.$$

So if $a<0$ then $|a|\le b$ means (i.e. bidirectional implication) $-a\le b \equiv a\ge -b$. With the assumption included: if $a<0$ then $-b\le a<0$. Similarly, if $a\ge0$ then $0\le a\le b$. Since exactly one of $a<0$ and $a\ge 0$ is true (they are contradictory predicates of $a$), we can assert that $a<0$ or $a\ge 0$. Thus $|a|\le b\equiv (-b\le a<0)\lor (0\le a\le b)\equiv -b \le a \le b$.

For (c) I don't see a problem with your proof, assuming result (b), but we might show how the left is true based on the definition of absolute value (you have just mentioned that it is) if it hasn't already come up.

Can't you use $|x|=\sqrt{x^2}$ for part (a)?