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.
Best Answer
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!