# Show that $a\le|a|$

Show that $$a\le|a|$$

Case 1: If $$a\ge0$$ then by definition $$a = |a|$$, thus $$a\le|a|.$$

Case 2: If $$a<0$$ then by definition $$- a = |a|$$, thus $$-a\le|a|$$ and $$a\lt0\lt-a\le|a|.$$

In both cases $$a\le|a|$$ thus concluding the proof.

• Yeah. I'd go into a bit more detail as to why if $a < 0$ then why does that mean $a$ can not exceed $-a$. (It's fairly trivial $|a| \ge 0$ always, and if $a < 0$ then $a < 0 \le |a|$.) Sep 23, 2022 at 4:33
• This looks like one of the examples where the student is supposed to demonstrate a "proof by distinguishing cases". Thus, as you present your proof, you need to be very clear about: (a) Which cases you have considered, and (b) How is each case handled. I would say you have the right idea, but your demonstration lacks a bit in the part (b) above. Spell each case separately and work it out, it will then sound much better.
– user700480
Sep 23, 2022 at 4:51
• If $a < 0$, then use transitivity. Sep 24, 2022 at 5:52
• just reduce case two to case one by noting that if $a<0$ then $-a>0$, hence $a<0<-a\leq|a|$. Sep 24, 2022 at 12:18

It is almost valid. In your second case, you state "Since $$a < 0$$, $$-a = a$$" which is wrong. If $$-a = a$$, then $$a = 0$$. You could just say that if $$a < 0$$, as $$0 \leq |a|$$, then $$a < |a|$$.

You proof is right, as an alternative way, we have that

• Case 1: $$a \ge 0$$ by definition $$|a|=a$$ then

$$a\le |a| \iff a\le a$$

which is true.

• Case 2: $$a < 0$$ by definition $$|a|=-a>0$$ then by $$b=-a>0$$

$$a\le |a| \iff a\le -a\iff -b\le b$$

which is true, then $$a\le |a|$$ always holds.

Let, $$|a|
Since, $$|a|≥0$$, we have $$a>0$$.
This implies that, $$a. A contradiction.