I don't fully understand this example the book gives. I understand the section where they examine when $x\gt0$ but when $x\lt0$ things get murky for me.
Prove that for every real number $x$, $x\le\lvert x \rvert$.
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4If $x$ is negative then $x\lt 0\lt |x|$. More concretely, let $x=-17$. Then $|x|=17$, so certainly $x\lt |x|$. – André Nicolas Sep 21 '14 at 15:32
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There are 2 cases: $x < 0$ or $x \ge 0$.
Case 1: $x \ge 0$
Now, when $x \ge 0,\ |x| = x$. Since $x \ge x,\ |x| \ge x \implies x \le |x|$.
e.g. when $x = 5,\ |x| = 5.\text{ So}\ 5 \le 5$
Case 2: $x < 0$
When $x < 0,\ |x|=-x>0$.
Now, a positive number $-x$ is always greater than a negative number $x$.
Thus, $-x = |x| > x \implies x \le |x|$.
e.g. when $x = -4,\ |x| = 4 \ge -4.$
From 1 and 2, $x \le |x|,\ \forall x \in \mathbb R$
Aldo
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taninamdar
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Ah this makes more sense, I think the book wrote your second line in Case 2 a little differently from what you said, but that makes it pretty clear. – secondubly Sep 21 '14 at 16:09
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The most used definition of $|x|$ is
$$|x|=\max(x,-x)$$ and knowing this it's clear that
$$|x|=\max(x,-x)\ge x$$
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$$-1\le|-1|=+1\\+1\le|+1|=+1$$ You can replace $1$ by any (positive) number.