Examples are not proofs.

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Consider the following problem.

For any \(x \in \mathbb{R}\), prove that

\[-|x| \leq x \leq |x|.\]

In case you forgot, the absolute value of a real number \(x\) is defined as follows:

\[|x| = \left\{\begin{array}{cc}x & \text{if $x\geq 0$},\\-x & \text{if $x < 0$}.\end{array}\right.\]

A notable property of the absolute value is that it is always non-negative (positive or zero).

Proof (wrong!): Let \(x = 3\). Then \(|x| = 3\). Thus, we have 

\[-3 \leq 3 \leq 3.\] 

So the proposition is true. Let \(x = -3\). Then \(|x| = |-3| = 3\). Thus, we have

\[-3 \leq -3 \leq 3.\]

So the proposition is also true. In either case, the proposition is true. Proved! ■

What is exactly wrong with this "proof"? It is that it only proves the cases when \(x = 3\) or \(x = -3\). But we need to prove that for any real number \(x\), and there are infinitely many (in fact, uncountably many) real numbers! What about \(x = 2.3\), or \(x = \pi^{-\sqrt{2}}\), or ...? Unless we prove the proposition for the general case, there is a possibility that the proposition might fail to hold for some strange real numbers.

Proof (correct): Let \(x\) be an arbitrary real number. If \(x \geq 0\), then \(|x| = x\). Thus,

\[-|x| \leq 0 \leq x \leq |x|.\]

(Recall that \(a \leq b\) means \(a < b\) or \(a = b\).)

If \(x < 0\), then \(|x| = -x\). Thus,

\[-|x| \leq x < 0 \leq |x|.\]

In either case, the proposition is true. ■

In this proof, \(x\) is an arbitrary real number, and what we show here applies to any real number. Therefore, the proof is general and correct.


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