Prove some properties of the absolute value

Prove the following:

$|x| = 0$ if and only if $x = 0$ .
$|-x| = x$ .
$|x-y| = |y-x|$ .
$|x|^2 = x^2$ .
$|x| = \sqrt{x^2}$ .
$|xy| = |x||y|$ .
If $y \neq 0$ then $|x/y| = |x|/|y|$ .
$|x-y| \leq |x| + |y|$ .
$|x| - |y| \leq |x-y|$ .
$||x|-|y|| \leq |x-y|$ .

Proof. $(\Rightarrow)$ If $|x| = 0$ then $-|x| \leq x \leq |x| \implies 0 \leq x \leq 0 \implies x = 0$ .
$(\Leftarrow)$ If $x = 0$ , then $|x| = x = 0$ by definition of $|x|. \qquad \blacksquare$
Proof. If $x \geq 0$ , then $|x| = x$ and $|-x| = -(-x) = x$ .
If $x < 0$ , then $|x| = -x$ and $|-x| = -x$ (since $x < 0 \implies -x > 0). \qquad \blacksquare$
Proof. If $x - y > 0$ , then $y - x < 0$ , so $|x-y| = x-y$ and $|y-x| = -(y-x) = x-y$ .
If $y-x > 0$ , then $x-y < 0$ , so $|x-y| = -(x-y) = y-x$ and $|y-x| = y-x$ .
If $x-y=0$ , then $y-x= 0$ so $|x-y| = |y-x| = 0. \qquad \blacksquare$
Proof. First, $|x|^2 = |x|\cdot |x|$ . So, if $x \geq 0$ then $|x|=x \implies |x| \cdot |x| = x^2$ .
If $x < 0$ then $|x| = -x \implies |x| \cdot |x| = (-x)\cdot(-x) = x^2. \qquad \blacksquare$
Proof. Since $\sqrt{x^2}$ is defined to be the unique non-negative square root of $x^2$ , from part (d) we have $|x|^2 = x^2$ , and $|x|$ is non-negative by definition; hence, $|x| = \sqrt{x^2}. \qquad \blacksquare$
Proof. If $x \geq 0, y \geq 0$ , then $xy \geq 0$ , so $|xy| = xy.$ Further, $|x| = x, \ |y| = y$ , so $|x||y| = xy$ .
If $x < 0$ and $y \geq 0$ , then $xy < 0$ so $|xy| = -(xy)$ . Then, $|x| = -x$ and $|y| = y$ , so $|x||y| = (-x)y = -(xy).$
If $x \geq 0$ and $y < 0$ , then $xy < 0$ so $|xy| = -(xy)$ . Then $|x| = x$ and $|y| = -y$ , so $|x||y| = x(-y) = -(xy).$
Finally, if $x < 0$ and $y < 0$ , then $xy > 0$ so $|xy| = xy$ . Then $|x| = -x$ and $|y| = -y$ , so $|x||y| = (-x)(-y) = xy. \qquad \blacksquare$
Proof. Using the definition of $x/y$ we have

$\left| \frac{x}{y} \right| = |xy^{-1}| = |x||y^{-1}|,$

by part (f). But then since $|y||y^{-1}| = |yy^{-1}| = |1| = 1$ , we have $|y^{-1}| = |y|^{-1}$ . So,

$|x||y^{-1}| = |x|(|y|^{-1}) = \frac{|x|}{|y|}. \qquad \blacksquare$
Proof. Using the triangle inequality and the fact that $|-y| = |y|$ ,

$|x-y| = |x+(-y)| \leq |x| + |-y| = |x|+|y|. \qquad \blacksquare$
Proof. Here we use the trick of adding and subtracting $y$ , and then the triangle inequality,

$|x| = |(x-y)+y| \leq |x-y| + |y| \implies |x| - |y| \leq |x-y|. \qquad \blacksquare$
Proof. Using a similar trick to part (i), we have

$|y| = |x+(y-x)| \leq |x| + |y-x| = |x| + |x-y| \implies |y| - |x| \leq |x-y| \implies |x| - |y| \geq -|x-y|.$

Then, combining this inequality with the one in part (i) and applying the definition of the absolute value,

$-|x-y| \leq |x| - |y| \leq |x-y| \implies ||x|-|y|| \leq |x-y|.$

6 comments

November 17, 2021 at 1:30 am

Anonymous says:

b) |-(-5)|=-5…

- December 5, 2023 at 6:18 am
  
  Anonymous says:
  
  |-(-5)| = -(-5)
  
July 10, 2021 at 10:41 am

Muhammad marwan iliyasu says:

What a wonderful site keep it up

March 20, 2020 at 9:04 am

Tiago says:

Actually I would make a small alteration: in part (f), you should have that if $x \geq 0$ and $y$ < 0, then $xy \leq 0$ (just to be more precise)

April 18, 2017 at 9:25 pm

Anonymous says:

hmm good….

June 24, 2016 at 11:23 am

GENTIL A. ESTEVEZ says:

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Stumbling Robot

Prove some properties of the absolute value

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