Prove $\lvert x\rvert$ = $\lvert-x\rvert$ for all real numbers $x$

Question

Been at this one for a long time. I'm trying to use the fact that $|x|$ = $x$ if $x$ is greater than or equal to 0, and $|x|$ = $-x$ if $x$ is less than 0. Then I want to split the proof into these 2 cases. I'm not sure if I'm supposed to use anything other than that definition for absolute value.

Answer 1

If $x \geq 0$ then $-x \leq 0$, so that $\lvert-x\rvert = -(-x) = x = \lvert x\rvert$.

If $x < 0$ then $-x > 0$, so that $\lvert-x\rvert = -x = \lvert x\rvert$.

Answer 2

The map $x \mapsto \sqrt{x}$ is bijective on $[0,\infty)$, hence it is sufficient to show that $|x|^2 = |-x|^2 $ for all $x$. No need to deal with cases here.

Answer 3

I propose an Algorithmic Proof in the hope it will appear as “trivial”

Assumptions

• H1: $ | x | = x $ if $ x \ge 0 $
• H2: $ | x | = -x $ if $ x < 0 $

Thesis

• T: $ | x | = | -x | \quad \forall x \in \mathbb{R} $

Proof

Step 1 :

• Let’s evaluate $ | x | $ so if $ x \ge 0 $ then goto Step 2.1 otherwise goto Step 3.1

Step 2.1 :

• Use H1 as conditions holds and get $ | x | = x $

Step 2.2:

• Evaluate $ | -x | $ but for sure $ -x \le 0 $ (as we are in the $ x \ge 0 $ branch) so immediately use H2 then $ | -x | = -(-x) = x $

Step 2.3:

• Finally observe that $ | x | = x = | -x | $ and this branch concludes

Step 3.1:

• Use H2 as condition holds and get $ | x | = -x $

Step 3.2:

• Evaluate $ | -x | $ but for sure $ -x > 0 $ (as we are in the $ x < 0 $ branch) so immediately use H1 then $ | -x | = -x $

Step 3.3:

• Finally observe that $ | x | = -x = | -x | $ and this branch concludes

Finally observe that whatever the branch the conclusion is always $ | x | = | -x | $ hence T holds

Answer 4

You don't really have much except the definition to go on.

If you can prove that if 1) if $x < 0$ then $-x > 0$ and 2) that $-(-x) = x \forall x$ you are done.

Proving those will depend on depend on the field definitions and axioms you are given to work with.

$-(-x) = 0$ is a direct result of the definition of additive inverse (and the assumption that $\mathbb R$ is an ordered field so every element has a unique additive inverse). [x + (-x) = 0 so x = -(-x)]

If x > 0 then -x < 0 and vice versa, is a basic proposition from the order field axioms (assuming $\mathbb R$ was presented as an ordered field). [ if x > 0 then 0 = -x + x > -x + 0 = -x]

So if $ x \ge 0$ then $-x \le 0$ so $|-x| = -(-x) = x = |x|$.

If $x < 0$ then $- x > 0$ so $|-x| = -x = |x|$.