Thread: Confusion with absolute value proof

1. Confusion with absolute value proof

Hi, I hope this is the correct forum, if not my apologies

I've been confused with this for a while and have decided I am really stuck.

I have to prove that |-a| = |a| using the absolute value definition which was given to us as:

$\displaystyle |a|=\left\{\begin{array}{cc}a,&\mbox{ if } x\geq 0\\-a, & \mbox{ if } x<0\end{array}\right$

Now I understand the following by definition:

if a>0 then

$\displaystyle |a| = a$

if a<0 then

$\displaystyle |a| = -(-a) = a$

But with $\displaystyle |-a| = |a|$ I get really confused! My lecturer explained it as follows:

if $\displaystyle a > 0$, then,

$\displaystyle LHS = |-a| = -(-a) = a$
$\displaystyle RHS = |a| = a = LHS$

and if $\displaystyle a < 0$, then

$\displaystyle LHS = |-a| = -a$
$\displaystyle RHS = |a| = -a = LHS$

Firstly, $\displaystyle |-a| = -(-a) = a$ Is this because originally we treat a as positive, but because it is made negative within the absolute value, then we treat it as if it is negative, thus -(-a)?

Then with a < 0, is it so because, we originally treat it as a<0, thus -(-a) but then as a negative, thus -a? Also with the RHS, -(-a) then -a?

Some clarification on this would be really appreciated, here is my attempt at proving it:

if $\displaystyle a>0$, then

$\displaystyle |-a| = |a| \Rightarrow -(-a) = a \Rightarrow a = a, therefore |-a| = |a|$

if $\displaystyle a<0$, then

$\displaystyle |-a| = |a| \Rightarrow -(-(-a)) = -(-a) \Rightarrow -a = a$

As you can see, I'm missing some small fact to make this true or I am making a really silly mistake somewhere as this seems like it should be very easy!

2. Do three cases.
If $\displaystyle a=0$ then $\displaystyle -a=0$ so $\displaystyle 0=|a|=|-a|$

If $\displaystyle a>0$ then $\displaystyle -a<0$ so $\displaystyle |a|=a~\&~|-a|=-(-a)=a$. Again $\displaystyle |a|=|-a|$

If $\displaystyle a<0$ then $\displaystyle -a>0$ so $\displaystyle |a|=-a~\&~|-a|=-a$. Again $\displaystyle |a|=|-a|$