Prove that |-a| = |a| I'm struggling to understand the basic proof for the following theorem: 
$\lvert -a\rvert = \lvert a \rvert$
The provided solution goes on to describe the following: 
$(i)$ For $a \ge 0 $, since $-a \le 0$   
        $\lvert -a \rvert = -(-a)=a, \lvert a \rvert = a$
My question: Why does $\lvert -a \rvert = -(-a)$ hold true? 
 A: The definition of $|a|$ is:
$|a| := \begin{cases} a \quad a \geq 0\\ -a \quad a\leq0 \end{cases}$
So, suppose $a \geq 0$
Then, $|a| = a$
$|-a| = -(-a) = a$. Therefore, $|-a| = |a|$
Now, suppose $a \leq 0$
Then, $|a| = -a$
$|-a| = -a$
Hence, $|a| = |-a|$
So, for all $a \in \mathbb{R}$, we find:
$$|a| = |-a|$$
A: The definition of $|\cdot|$ is:
$$|x| := \begin{cases}x & x \ge 0\\-x & x<0\end{cases}$$
So, for $a \ge 0$, since $-a\le 0$, we have $|-a| = -(-a)$, by the definition of $|\cdot|$.
A: Provided we are working the a subring $A$, of the reals (i.e. $\mathbb{Z, Q, R}$), then the absolute value function is defined for $a \in A$ as:
\begin{equation}
    |a|=
    \begin{cases}
      -a, & \text{if}\ a<0\\
      0, & \text{if}\ a=0 \\
      a, & \text{if}\ a>0
    \end{cases}
  \end{equation}
So to answer your question, in the case $a \geq 0$, $|-a|=-(-a)$ comes straight from the definition, as $-a \leq 0$ (obviously noting $0=-0$). 
A: Let's take -a≥0 then |-a|=-a ........(1)
        0<a then |a |=a  ........(2)

by (1) and (2) |-a|=|a |
Let's take -a<0 then |-a|=-(-a)=a .........(3)
        0≤a then |a |=a ...............(4)

by (3) and (4) |-a|=|a |
conclusion= |-a|=|a |
NOTE:The absolute value function always gives the positive value of some number.That is why the negative number -a gives the positive number -(-a) as the absolute value.
