# The absolute value function

The absolute value function $f(x)=|x|$ has a global minimum at x=0. How could we prove this rigorously? In other words, how could we prove that there is no point $c \ \epsilon \ ℝ$ such that $f(c)<f(0)$

(Obviously, the function is not differentiable at x=0 so we cannot apply Fermat's critical point theorem).

BiP

I think the rigorous proof just goes something like "by definition, the absolute value of a real number is non-negative, therefore no such c exists."

pwsnafu