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When one begins to study real analysis, the absolute value function quickly enters and a large number of exercises involve manipulations with it.

What are the basic properties of absolute value that one should know? (Similar to how there are basic properties of $+,-,\cdot,/$ that we use all the time.) I'd like to know the theory of absolute value and the demonstration of theorems about it, not just how to solve equations with it.

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closed as not a real question by Did, Andrés Caicedo, Isaac, Fabian, Pete L. Clark Jan 17 '12 at 2:56

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Try wikipedia. The part for real numbers should get you started: – 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
up vote 8 down vote accepted

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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