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Proof by cases that $x \le |x|$?

May I have an example?

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"Example"? Of what? The proof? Or cases, like $\,2\le |2|\;\;,\;\;-2<2=|-2|\,$ ...? – DonAntonio Feb 25 '13 at 12:54


$$|x|=\begin{cases}-x&,\;x<0\\{}\\\;\;\;x&,\;x\ge 0\end{cases}$$

so for $\,x\in\Bbb R\,$ we have

$$x<0\Longrightarrow x<-x=|x|\Longrightarrow x<|x|\\x\ge 0\Longrightarrow x=x=|x|\le |x|\Longrightarrow x\le |x|$$

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$|x|$ is defined as $$|x|=\left\{ \begin{array}{rl} x & x\geq 0 \\ -x & x<0 \\ \end{array}\right.$$ So when $x\geq 0$ than we have $|x| = x \geq x$ and this is known to be true.

For $x<0$ we have $$x<0 < - (x)= |x|$$

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The answer depends on your definition of $|x|$.

For example, if your definition of $|x|=\max\{x,-x\}$ then there is no need for proof by cases, $x\leq\max\{x,-x\}$ by definition of $\max$.

Generally if you define $f(x)=\begin{cases} f_+(x) & x\geq 0\\ f_-(x) & x<0\end{cases}$ (in the case of $|x|$ we have $f_+(x)=x$ and $f_-(x)=-x$, but that's not to the point), then a proof by cases would be as follows:

Let $x$ be a real number. If $x\geq 0$ then $f(x)=f_+(x)$ [and derive the property wanted in that case]; otherwise $x<0$, and then $f(x)=f_-(x)$ [and derive the property wanted in this case as well].

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