Answer
See proof below.
Work Step by Step
"If and only if" statetments are equivalence statements.
Proving an equivalence $A\Leftrightarrow B$ is done by proving $A\Rightarrow B$ and $B\Rightarrow A$
$1.\quad A\Rightarrow B$
Let $\quad |x|\gt a.$
Then, a positive number $k$ exists such that
$|x|=a+k$
By definition of absolute value, $|x|=\left\{\begin{array}{ll}
x, & x\geq 0\\
-x, & x \lt 0
\end{array}\right.$
If x is nonnegative, then $x=a+k$, that is $x \gt a.$
If x is negative, then $x=-(a+k)=-a-k$,
which is less than $-a$, because we are subtracting a positive number from $-a.$
The implication $ |x|\gt a\quad \Rightarrow \quad x\gt a \quad $or$\quad x \lt a.$
has been proved.
$2.\quad B\Rightarrow A$
Now, let $x \gt a$.
Since $a$ is positive, it follows that x is also positive.
By definition, $|x|=x, $and since $x \gt a$, it follows that $|x| \gt a.$
Finally, let $x \lt -a$.
x is negative, so by definition, $|x|=-x$, and
when we multiply the last inequality with $(-1)$, the inequality direction changes:
$-x\gt a$
$|x|\gt a$
This implication has also been proved.
Thus, $A\Leftrightarrow B$, and the statement has been proved.
