# Absolute Value theory [closed]

Could someone explain me the basics about absolute value? Not only how to solve it, but also the theory and the demonstration of theorems about it.

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## closed as not a real question by Did, Andres Caicedo, Isaac, Fabian, Pete L. ClarkJan 17 '12 at 2:56

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Try wikipedia. The part for real numbers should get you started: en.wikipedia.org/wiki/Absolute_value#Definition_and_properties –  Mikko Korhonen Jan 16 '12 at 17:48
@m.k. I tried wikipedia, my book, and notes. But I have my trouble yet... –  Overflowh Jan 16 '12 at 17:52
What does "how to solve [absolute value]" mean? Absolute value is a function; you don't solve functions, you evaluate them. –  Arturo Magidin Jan 16 '12 at 20:14

The most important thing is, of course, the definition: $$|x|=\left\{\begin{array}{ll} x &\text{if }x\geq 0\\ -x &\text{if }x\lt 0. \end{array}\right.$$

The absolute value of a real number $x$ is its distance from $0$.

Basic properties:

1. $|x|\geq 0$ for all $x\in\mathbb{R}$.
2. $|-x| = |x|$ for all $x\in\mathbb{R}$.
3. More generally, $|xy|=|x|\,|y|$ for all $x,y\in\mathbb{R}$.
4. $|x+y|\leq |x|+|y|$ for all $x,y\in\mathbb{R}$. Equality holds if and only if the sign of $x$ and $y$ is the same (both nonnegative or both nonpositive).
5. Let $a$ be a nonnegative real number. Then $|x|=a$ if and only if $x=a$ or $x=-a$.
6. Let $a$ be a nonnegative real number. Then $|x|\leq a$ if and only if $-a\leq x\leq a$ (that is, $-a\leq x$ and $x\leq a$).
7. Let $a$ be a nonnegative real number. Then $|x|\geq a$ if and only if $x\geq a$ or $x\leq -a$.

Proof. 1 follows from the definition. If $x\geq 0$ then $|x|=x\geq 0$. If $x\lt 0$, then $-x\gt 0$, and $|x|=-x\gt 0$.

Two follows by figuring out what $|-x|$ is: $$|-x| = \left\{\begin{array}{ll} -x & \text{if }-x\geq 0\\ -(-x) & \text{if }-x\lt 0 \end{array}\right. = \left\{\begin{array}{ll} -x &\text{if }x\leq 0\\ x &\text{if }x\gt 0 \end{array}\right. = |x|;$$ the only possible issue is that we get a different formula at $x=0$ (in $|x|$ we get $x$, here we have $-x$) but $-0 = 0$, so the two functions are equal.

For 3, consider the four possibilities: if $x,y\geq 0$, then $xy\geq 0$, so $xy = |xy|$ and $|x|\,|y|=xy$. If $x\geq 0$, $y\lt 0$, then $xy\leq 0$, so $|xy|=-xy$, and $|x||y|=x(-y) = -xy$. If $x\lt 0$ and $y\geq 0$, same thing. And if $x,y\lt 0$, then $xy\gt 0$, so $|xy|=xy$, and $|x||y|=(-x)(-y) = xy$.

Point 4, the triangle inequality, is the more interesting one. There are many ways of verifying the inequality; for example, if $x$ and $y$ have the same sign, then either $x$, $y$, and $x+y$ are all positive, or $x$, $y$, $x+y$ are all negative. Then $|x+y|=x+y = |x|+|y|$ if all are positive, $|x+y| = -(x+y) = -x-y = |x|+|y|$ if all are negative (this also establiishes the equality in the case where they both have the same sign.)

Now assume that $x$ and $y$ have opposite signs; by switching $x$ and $y$, we may assume that $x\gt 0$ and $y\lt 0$. If $x+y\gt 0$, then we have $|x+y| = x+y \lt x-y = |x|+|y|$ (since $y\lt 0$, we have $y\lt 0 \lt -y$, so $x+y \lt x-y$). In summary, $|x+y|\lt |x|+|y|$. On the other hand, if $x+y\lt 0$, then $|x+y| = -(x+y) = -x-y \lt x-y = |x|+|y|$ (since $x\gt 0$, then $x\gt -x$, so $x-y \gt -x-y$). Either way, $|x+y|\lt|x|+|y|$, with equality if and only if $xy\geq 0$.

For 5, note that $|x|=a$ if and only if $x\geq 0$ and $x=a$, or $x\lt 0$ and $-x=a$; if and only if $x=a$ or $x=-a$.

For 6, if $x\geq 0$, then $|x|\leq a$ if and only if $0\leq x\leq a$; if this holds, then $-a\leq x\leq a$. If $x\lt 0$, then $|x|\leq a$ if and only if $-x\leq a$, if and only if $x\geq -a$. If this holds, then $-a\leq x\leq a$. Either way, if $|x|\leq a$, then $-a\leq x\leq a$. Conversely, if $-a\leq x\leq a$, then either $x\geq 0$ and $|x|=x\leq a$; or $x\lt 0$ and $|x|=-x\leq a$ (since $-a\leq x$, so $a\geq -x$).

And 7 is similar: assume $|x|\geq a$. If $x\geq 0$, then this means $x\geq a$; if $x\lt 0$, then this means $-x\geq a$, or $x\leq -a$. Conversely, if $x\geq a$, then $x\geq 0$ so $|x|=x\geq a$; if $x\leq -a$, then $x\lt 0$, so $|x|=-x\geq a$. $\Box$

Other properties are derived from the above properties. For example, $$|x-y| \leq |x|+|y|$$ follows from 2 and 3: $$|x-y| = |x+(-y)| \leq |x|+|-y| = |x|+|y|.$$

The "other" triangle inequality, $|x+y|\geq |x|-|y|$ follows by taking $y+x-y$ and using 4: $$|x| = |y+x-y| \leq |y+x|+|-y| = |y+x|+|y|.$$ Subtracting $|y|$ from both sides we get $$|x|-|y| \leq |x+y|.$$ By switching the roles of $x$ and $y$ you also get $|y|-|x| \leq |x+y|$, hence $$|x+y|\geq \Bigl| |x|-|y|\Bigr|.$$

These are the basics; everything else follows from them. I have no idea what it means to "solve [absolute value]"; equations that involve absolute values are solved either by using 5, or by considering cases. Inequalities involving absolute value are solved either by using 6 and 7, or by considering cases.

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