Prove |a|=|b| $\Rightarrow$ a=b or a=-b

[ver_mathstats]

Homework Statement

Let a, b ∈ ℝ. Prove that:

If |a| = |b|, then a = b or a = -b.

Homework Equations

The Attempt at a Solution

[/B]
I am having difficulties with beginning this proof.

Would it make sense to have: Case 1: b≥0 and Case 2: b≤0?

 

[jack476]

You don't need to break this into cases. A direct solution is best here.

Here's a hint to help you get started: |x| = |-x| = √(x²)

 

[member 587159]

Homework Statement

Let a, b ∈ ℝ. Prove that:

If |a| = |b|, then a = b or a = -b.

Homework Equations

The Attempt at a Solution

[/B]
I am having difficulties with beginning this proof.

Would it make sense to have: Case 1: b≥0 and Case 2: b≤0?

Indeed, it is a very good idea to split in cases (better 4 instead of 2 though), depending on the sign of $a$ and $b$.

(1) $a \geq 0, b \geq 0$
(2) ...
(3) ...
(4) ...

 

[member 587159]

You don't need to break this into cases. A direct solution is best here.

Here's a hint to help you get started: |x| = |-x| = √(x²)

"No need to break this into cases". Maybe, if you have already proven that $|x| = |-x|$ and $|x| = \sqrt{x^2}$. Proving this will ultimately rely on breaking up in cases... I think introducing unnecessary square roots obscures things, but that's just my opinion.

 

[jack476]

"No need to break this into cases". Maybe, if you have already proven that $|x| = |-x|$ and $|x| = \sqrt{x^2}$. Proving this will ultimately rely on breaking up in cases... I think introducing unnecessary square roots obscures things, but that's just my opinion.

Yes, if you're not already given that $|x| = |-x|$. Actually, I kind of thought that that was true by definition.