Difference between revisions of "Absolute value"

(Generalized Absolute Values: Added (Non-negative) and (Completely Multiplicative) for consistency with (The Triangle Inequality). Added Ultrametric Inequality as alt. name for Strong Tri. Ineq.)
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The '''absolute value''' of a [[real number]] <math>x</math>, denoted <math>|x|</math>, is the unsigned portion of <math>x</math>. Geometrically, <math>|x|</math> is the [[distance]] between <math>x</math> and [[zero]] on the real [[number line]].
  
The '''absolute value''' of a [[real number]] <math>x</math>, denoted <math>|x|</math>, is its distance from [[zero (constant) | zero]] on a [[number line]]. If <math>x\ge 0</math>, then <math>|x|=x</math>, and if <math>x<0</math>, then <math>\displaystyle |x|=-x</math>. This is equivalent to "dropping the minus sign."
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The absolute value function exists among other contexts as well, including [[complex numbers]].
  
Similarly, the absolute value of a [[complex number]] <math>z=x+iy</math>, where <math>x,y\in\mathbb{R}</math>, is <math>|z|=\sqrt{x^2+y^2}</math>, the distance of <math>z</math> from the origin.
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==Real numbers==
  
== Example Problems ==
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When <math>x</math> is real, <math>|x|</math> is defined as <cmath> |x| = \begin{cases} x & \text{for } x \ge 0,\\ -x & \text{for } x \le 0.\end{cases} </cmath> For all real numbers <math>x</math> and <math>y</math>, we have the following properties:
=== Simple Absolute Value Problems ===
 
<math>|x|=5</math>
 
  
Solution: You have to isolate the variable, and then make two equations; one negative, the other positive. The variable is already isolated, so we can make the two equations: <math>x=5</math> and <math>x=-5</math>. This works because x can be both positive and negative, and will still give the same result. The answer is <math>x=\{-5,\,5\}</math>.
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* (Alternative definition) <math>|x| = \sqrt{x^2}</math>
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* (Non-negativity) <math>|x| \ge 0</math>
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* (Positive-definiteness) <math>|x| = 0 \iff x=0</math>
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* (Multiplicativeness) <math>|xy| = |x| |y|</math>
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* ([[Triangle Inequality]]) <math>|x+y| \le |x|+|y|</math>
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* (Symmetry) <math>|x| = |-x|</math>
  
----
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Note that
  
Now, let's say that you have functions outside your absolute value: <math>4+3|7x|=151</math>.
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<cmath>|x| \le y \iff -y \le x \le y </cmath>
  
Just like in the other problem, you must isolate the variable. First, sutract 4 from both sides to get <math>3|7x|=147</math>. Then, divide by three to get <math>|7x|=49</math>.
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and
  
Now, try to solve it by yourself.
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<cmath> |x| \ge y \iff x \ge y \text{ or } x \le -y.</cmath>
  
Solution: We first get rid of the absolute value by making two equations: <math>7x=49</math> and <math>7x=-49</math>. Divide everything by 7 to get the answer: <math>x=\{-7,\,7\}</math>.
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==Complex numbers==
  
=== Practice Problems ===
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For [[complex number]]s <math>z</math>, the absolute value is defined as <math>|z| = \sqrt{x^2+y^2}</math>, where <math>x</math> and <math>y</math> are the real and imaginary parts of <math>z</math>, respectively. It is equivalent to the distance between <math>z</math> and the [[origin]], and is usually called the [[complex modulus]].
<math>-|x|=x-6</math>
 
  
<math>|7b|=21</math>
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Note that <math>|z| = |\overline{z}| = \sqrt{z\overline{z}}</math>, where <math>\overline{z}</math> is the [[complex conjugate]] of <math>z</math>.
  
<math>5+8|4x|=69</math>
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==Examples==
  
=== Word Problems ===
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# If <math>|x|=k</math>, for some real number <math>k</math>, then <math>x=k</math> or <math>x=-k</math>.
Absolute Value Functions are also very useful for solving problems.
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# If <math>|ax| = k</math>, for some real numbers <math>a</math>, <math>k</math>, then <math>ax = k</math> or <math>ax = -k</math>, and therefore <math>x = \frac{k}{a}</math> or <math>x = -\frac{k}{a}</math>.
  
Lets say you have a problem that goes like this:
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==Problems==
  
In Mrs. Barnett's class, the scores on a certain test varied 28 points from 71. What were the minumum and maximum scores on the test?
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# Find all real values of <math>x</math> if <math>-|x| = x-6</math>.
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# Find all real values of <math>x</math> if <math>5 + 8 \cdot |4x| = 69</math>.
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# ([[2000 AMC 12 Problems/Problem 5|AMC 12 2000]]) If <math>|x - 2| = p</math>, where <math>x < 2</math>, then find <math>x - p</math>.
  
You would have <math>|x-71|=28</math> as your equation, and if you solve it, you get 99 as the maximum and 43 as the minimum.
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==See Also==
 
 
=== Problems from Competitions ===
 
* [[2000_AMC_12/Problem_5 | 2000 AMC 12 Problem 5]]
 
 
 
== Generalized Absolute Values ==
 
The absolute value functions listed above have three very important properties:
 
 
 
*<math> |x|\ge 0</math> for all <math>x</math>, and <math>|x|=0</math> if and only if <math>x=0</math>. (Non-negative)
 
*<math> |x\times y|=|x|\times |y|</math>.  (Completely Multiplicative)
 
*<math> |x+y| \le |x|+|y|</math>. (The [[triangle inequality]])
 
 
 
We call ''any'' function satisfying these three properties ''an absolute value'', or a ''norm''.
 
 
 
Another example of an absolute value is the [[p-adic]] absolute value of <math>\mathbb{Q}</math>, the [[rational number]]s. Let <math>x=\prod_{i=1}^n p_i^{e_i}</math>, where the <math>p_{i}</math>'s are distinct [[prime number]]s, and the <math>e_i</math>'s are ([[positive]], [[negative]], or [[zero (constant) | zero]]) [[integer]]s. Define <math>|x|_{p_i}=p_i^{-e_i}</math>. This defines an absolute value on <math>\mathbb{Q}</math>. This absolute value satisfies a stronger triangle inequality, often known as the Ultrametric Inequality:
 
  
*<math> |x+y|\le\max(|x|,|y|)</math>.
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* [[Magnitude]]
 
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* [[Norm]]
An absolute value satisfying this strong triangle inequality is called [[nonarchimedian]]. If an absolute value does not satisfy the strong triangle inequality, then it is called [[archimedian]]. The ordinary absolute value on <math>\mathbb{R}</math> or <math>\mathbb{C}</math> is archimedian.
 
 
 
The theory of absolute values is important in [[algebraic number theory]]. Let <math>K/\mathbb{Q}</math> be a [[finite]] [[Galois extension]] with <math>[K:\mathbb{Q}]=n</math>, and let <math>\sigma_1,\ldots,\sigma_n</math> be the [[field automorphisms]] of <math>K</math> over <math>\mathbb{Q}</math>. Then the only absolute values are the archimedian ones given by <math>|x|_i=|\sigma_i(x)|</math> (the ordinary real or complex absolute values) and the nonarchimedian ones given by <math>|x|_{\mathfrak{p}}</math> for some prime <math>{\mathfrak{p}}</math> of <math>K</math>.
 
 
 
==See Also==
 
* [[Algebraic number theory]]
 
* [[Completion]]
 
 
* [[Valuation]]
 
* [[Valuation]]

Latest revision as of 09:37, 5 January 2009

The absolute value of a real number $x$, denoted $|x|$, is the unsigned portion of $x$. Geometrically, $|x|$ is the distance between $x$ and zero on the real number line.

The absolute value function exists among other contexts as well, including complex numbers.

Real numbers

When $x$ is real, $|x|$ is defined as \[|x| = \begin{cases} x & \text{for } x \ge 0,\\ -x & \text{for } x \le 0.\end{cases}\] For all real numbers $x$ and $y$, we have the following properties:

  • (Alternative definition) $|x| = \sqrt{x^2}$
  • (Non-negativity) $|x| \ge 0$
  • (Positive-definiteness) $|x| = 0 \iff x=0$
  • (Multiplicativeness) $|xy| = |x| |y|$
  • (Triangle Inequality) $|x+y| \le |x|+|y|$
  • (Symmetry) $|x| = |-x|$

Note that

\[|x| \le y \iff -y \le x \le y\]

and

\[|x| \ge y \iff x \ge y \text{ or } x \le -y.\]

Complex numbers

For complex numbers $z$, the absolute value is defined as $|z| = \sqrt{x^2+y^2}$, where $x$ and $y$ are the real and imaginary parts of $z$, respectively. It is equivalent to the distance between $z$ and the origin, and is usually called the complex modulus.

Note that $|z| = |\overline{z}| = \sqrt{z\overline{z}}$, where $\overline{z}$ is the complex conjugate of $z$.

Examples

  1. If $|x|=k$, for some real number $k$, then $x=k$ or $x=-k$.
  2. If $|ax| = k$, for some real numbers $a$, $k$, then $ax = k$ or $ax = -k$, and therefore $x = \frac{k}{a}$ or $x = -\frac{k}{a}$.

Problems

  1. Find all real values of $x$ if $-|x| = x-6$.
  2. Find all real values of $x$ if $5 + 8 \cdot |4x| = 69$.
  3. (AMC 12 2000) If $|x - 2| = p$, where $x < 2$, then find $x - p$.

See Also