Calculus I presents this definition of absolute value:

$$f(x)=y=|x|=\left\{\begin{array}{}\;\;\;x&\text{ if}\,\,\,x\geq 0\\-x&\text{ if}\,\,\,x<0\end{array}\right.$$

But you can also define absolute value as $$|x|= \sqrt{(\pm x)^2} $$

I can't think of any other ways to define it.
What are some other ways of defining the Absolute Value?
Please include what I might not have learned yet, i.e., definitions encountered in advanced classes. Thank you.

  • $\begingroup$ But what is $\sqrt{x^2}$ for negative numbers (you can not say that it is the absolute value since you hadn't defined it). $\endgroup$ – Sigur Nov 2 '13 at 18:15
  • $\begingroup$ @Sigur I see your point, I edited the square root part. $\endgroup$ – Emi Matro Nov 2 '13 at 18:17
  • $\begingroup$ But you can not take the square root of a negative number (in your case, $-x^2$ is negative). $\endgroup$ – Sigur Nov 2 '13 at 18:17
  • $\begingroup$ But you square it first and it becomes positive, then you take the square root, no? $\endgroup$ – Emi Matro Nov 2 '13 at 18:18
  • 1
    $\begingroup$ So you should write $(-x)^2$. $\endgroup$ – Sigur Nov 2 '13 at 18:20

The supremum of two elements $x,y$ is usually written as $x \uparrow y$ and is characterized as follows: $$\boxed{\forall z :: x \uparrow y \leq z \ \equiv \ x \leq z \land y \leq z}$$ That is, $z$ is an upper-bound precicely when it's greater than the sup (ie the least-upper-bound).

Now the definition of absolute-value is $$\boxed{|x| = x \uparrow -x}$$

This definition shows its worth in calculations.

Here are some examples...

Theorem: $x \leq |x|$, (and $-x \leq |x|$)

Proof: In the characterization of $\uparrow$, take $y := -x$ and take $z := |x|$. That's a pretty-easy proof :)

Theorem: $|-x| = |x|$

Proof: $|-x| = -x \uparrow (--x) = -x \uparrow x = x \uparrow -x = |x|$, where we used the easily-proven property $a \uparrow b = b \uparrow a$.

Theorem: $|x+y| \leq |x| + |y|$ (The Triangle-Inequality)

Proof: Exercise! Try using this definition of absolute-value!

Theorem: $|x \cdot y| = |x| \cdot |y|$

Proof: Exercise! Try using this definition of absolute-value!

Hope this helps!

  • $\begingroup$ Also note the lack of case-analysis! $\endgroup$ – Musa Al-hassy Nov 13 '13 at 5:30

$|x|=x\text{sgn}(x) $,

where $\text{sgn}(x) = 1$ if $x>0$, $0$ if $x=0$, $-1$ if $x<0$.


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