I'm in precalculus and totally new to this 'proofs with axioms' concept, and it's stumping me beyond belief. I get the concept, I just don't understand how to start or where to implement the proofs.
The problem states: Give a careful proof of $|a-b| = |b-a|$. Justify each step.
This is how I attempted…
assume $a,b ∈ \mathbb{R}$
$(a,b) \geq 0$, only $0$ when $a=b$ -absolute value property 1
Lets say $|b-a| = c$, so
$|a-b|=c$
$|a||-b|=c$ (absolute value property 3)
$|a||+b|=c$ (absolute value property 2)
$|b||a|=c$ (commutative for multiplication axiom //would this be addition?)
$|b||-a|=c$ (absolute value property 2)
$|b-a|=c$ (absolute value property 3)
$|b-a|=|b-a|$ ??
I'm confused because the hint for the homework says I will need to use one of the distributive rules, the definition of absolute value, and some of the absolute value properties.
..I have no idea what the point would be to use the distributive rule for this. Add another factor? For the sake of what?
Here are the absolute value properties…
- $|a| \geq 0$
- $|a| = |-a|$
- $|ab| = |a||b|$
- $|a/b| = |a|/|b|$
- $|a+b|<= |a|+|b|$
Best Answer
Think. Why does $|a-b| = |b-a|$. Is it because $a - b$ and $b-a$ are equal? No because they aren't. So when else can absolute values be equal? Well, what's your casual definition of absolute value? Something like $|a| = a$ if $a \ge 0$ and $|a| = -a$ otherwise? Or maybe some casual "the absolute value without regard to sign?
In any event, $|a - b| = |b -a|$ because $a - b$ and $b-a$ are the "same size" but opposite signs.
So which properties will we use?
Let's see:
|a|≥0
Not really. But are non-negative.
|a|=|−a|
Absolutely $(a - b) = -(b - a)$ so $|a-b| = |-(a-b)| = |b-a|$
So here is our proof in !!!ONE!!! line:
$|a-b| = |-(a-b)| = |b-a|$ (property 2)
That's it. We are done. That is EVERYTHING we need to do.
|ab|=|a||b|
Useless, we don't have any multiplication.
|a/b|=|a|/|b|
or division.
|a+b|<=|a|+|b|
This tells us $|a - b| \le |a| + |-b| = |a| + |b|$.
And $|b - a| \le |b| + |-a| = |b| + |a| = |a| +|b| $ .. which seems to need more work if we are going to get it to help us.