The absolute value function

Main Question or Discussion Point

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

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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
1. If $x>0$ then $f(x) = x$ so $f(x) > 0$,
2. If $x=0$ then $f(x) = x = 0$,
3. If $x < 0$ then $f(x) = -x$ and so $f(x) > 0$,