Show that $a\le|a|$

Question

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.

Is this a valid answer?

Answer 1

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 $a < |a|$.

Answer 2

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.

Answer 3

Proof by Contradiction.

Let, $|a|<a$

Since, $|a|≥0$, we have $a>0$.

This implies that, $a<a$. A contradiction.