Negative of Absolute Value

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[edit] Theorem

Let x \in \mathbb{R} be a real number.

Let \left|{x}\right| be the absolute value of x.


Then - \left|{x}\right| \le x \le \left|{x}\right|.


[edit] Corollary

  • \left|{x}\right| < y \iff -y < x < y;
  • \left|{x}\right| \le y \iff -y \le x \le y.


[edit] Proof

Either x \ge 0 or x < 0.

  • If x \ge 0, then - \left|{x}\right| \le 0 \le x = \left|{x}\right|.
  • If x < 0, then - \left|{x}\right| = x < 0 < \left|{x}\right|.


[edit] Proof of Corollary

  • First we show that \left|{x}\right| < y \Longrightarrow -y < x < y.

Suppose \left|{x}\right| < y.

Then from the above, x \le \left|{x}\right| and \left|{x}\right| \ge -x.

So x < y and -x < y, and so x > -y from Ordering of Inverses.

It follows that -y < x < y.


Now suppose that \left|{x}\right| \le y.

If \left|{x}\right| < y then -y < x < y and so -y \le x \le y.

Otherwise, if \left|{x}\right| = y then either x = y or -x = y and hence the result.


  • Next we show that -y < x < y \Longrightarrow \left|{x}\right| < y.

Suppose -y < x < y.

Then x < y and -x < y.

For all x, \left|{x}\right| = x or \left|{x}\right| = -x.

Thus it follows that \left|{x}\right| < y.


Now suppose that -y \le x \le y.

If -y < x < y then \left|{x}\right| < y and hence \left|{x}\right| \le y.

Else, if either -y = x or x = y then \left|{x}\right| = y and hence the result.


[edit] Sources

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