Difference between revisions of "Absolute value"

(Example Problems And How To Solve Easy Problems)
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The '''absolute value''' of a [[real number]] <math>x</math>, denoted <math>|x|</math>, is its distance from 0. Therefore, 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''' of a [[real number]] <math>x</math>, denoted <math>|x|</math>, is its distance from 0 on a [[number line]]. Therefore, 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."
  
 
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>.
 
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>.
  
  
== Introductory Concepts ==
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== Example Problems ==
=== Example Problems & How To Solve for Middle School Math Problems===
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=== Simple Absolute Value Problems ===
Example Problem
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<math>|x|=5</math>
________________
 
* [[2000_AMC_12/Problem_5 | 2000 AMC 12 Problem 5]]
 
 
 
Amc 12 2000, Problem 5:
 
 
 
5. If |x − 2| = p, where x < 2, then x − p
 
(A) −2
 
(B) 2
 
(C) 2 − 2p
 
(D) 2p − 2
 
(E) |2p − 2|
 
 
 
 
 
How to solve simple Absolute Value Problems:
 
 
 
Example:
 
|x|=7
 
 
 
Solution: For this problem, you have to know that absolute value is always a positive number or 0. For example, if you put -7 in, you get 7 out. Another way to look at it, is it takes takes the value inside the absolute value symbols and instead puts the positive difference between itself and 0. If you have just an absolute value function on a graph, it will always appear in quadrant 1 and 2, hence those quadrants have a positive y value or 0. This is how you solve an absolute value problem: You make 2 equations out of this when you take away the absolute value signs. The 2 equations would be x=7 and x=-7. If you havn't learned this already you probably don't know why this works. This is because the other number (x as to -x - opposite) that works, is the number that would equal 0 if you add it to the original number, or in other words, it's opposite.
 
 
 
----------------------------------
 
 
 
Now, let's say that you have functions outside your absolute value.
 
 
 
4+3|7x|=151
 
 
 
Wait one second big guy.........you can't JUST use the information above!
 
 
 
Now, just a precaution. Treat numbers outside absolute value functions as it was in ().
 
 
 
So, you would first -4 from both sides in your effort to isolate the variable (If you don't know why, you should probably attempt to read an article in Algebra as to why you do this.........I'm not sure if this that is in there).
 
 
 
Your new equation is this:
 
 
 
3|7x|=147
 
  
Keep on isolating the variable.
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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>.
  
Divide both sides by 3 and get:
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----
  
|7x|=49
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Now, let's say that you have functions outside your absolute value: <math>4+3|7x|=151</math>.
  
Now, using the above information, I think we can solve this problem. Try to finish this by yourself, then look at the solution.
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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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Now, try to solve it by yourself.
  
----------------------
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Solution: We first get rid of the absolute value by making two equations: <math>7x=49</math> and <math>7x=-49</math>. Divide both sides of both equations by 7 to get the answer: <math>x=\{-7,\,7\}</math>.
'''Solution:
 
  
We first get rid of the absolute value by making two equations. Thus, we make:
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=== Practice Problems ===
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<math>-|x|=x-6</math>
  
7x=49 and 7x=-49
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<math>|7b|=21</math>
-  -      -  -
 
7  7      7  7
 
  
x=7 or x=-7'''
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<math>5+8|4x|=69</math>
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Now, try some of these problems (if you really are having trouble, post something on this forum.
 
 
 
-|x|=x-6
 
 
 
7|b|=21
 
 
 
5+8|4x|=69
 
-------------------
 
  
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=== Word Problems ===
 
Absolute Value Functions are also very useful for solving problems.
 
Absolute Value Functions are also very useful for solving problems.
  
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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?
 
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?
  
You would have
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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.
  
|x-71|=28 as your equation, and if you solved it like you did above, you get 99 as the maximum and 43 as the minimum.
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=== Problems from Competitions ===
 
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* [[2000_AMC_12/Problem_5 | 2000 AMC 12 Problem 5]]
Obviously, that might be too easy for absolute value, but when the numbers get tricky, this format could definately help.
 
 
 
--[[User:Jhredsox|Jhredsox]] 10:25, 4 November 2006 (EST)
 
 
 
==Generalized absolute values==
 
  
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== Generalized Absolute Values ==
 
The absolute value functions listed above have three very important properties:
 
The absolute value functions listed above have three very important properties:
  
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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>.
 
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==
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==See Also==
  
 
*[[Algebraic number theory]]
 
*[[Algebraic number theory]]
 
*[[Completion]]
 
*[[Completion]]
*[[p-adic number]]
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*[[P-adic number]]
 
*[[Valuation]]
 
*[[Valuation]]

Revision as of 12:42, 4 November 2006

The absolute value of a real number $x$, denoted $|x|$, is its distance from 0 on a number line. Therefore, if $x\ge 0$, then $|x|=x$, and if $x<0$, then $\displaystyle |x|=-x$. This is equivalent to "dropping the minus sign."

Similarly, the absolute value of a complex number $z=x+iy$, where $x,y\in\mathbb{R}$, is $|z|=\sqrt{x^2+y^2}$.


Example Problems

Simple Absolute Value Problems

$|x|=5$

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: $x=5$ and $x=-5$. This works because x can be both positive and negative, and will still give the same result. The answer is $x=\{-5,\,5\}$.


Now, let's say that you have functions outside your absolute value: $4+3|7x|=151$.

Just like in the other problem, you must isolate the variable. First, sutract 4 from both sides to get $3|7x|=147$. Then, divide by three to get $|7x|=49$.

Now, try to solve it by yourself.

Solution: We first get rid of the absolute value by making two equations: $7x=49$ and $7x=-49$. Divide both sides of both equations by 7 to get the answer: $x=\{-7,\,7\}$.

Practice Problems

$-|x|=x-6$

$|7b|=21$

$5+8|4x|=69$

Word Problems

Absolute Value Functions are also very useful for solving problems.

Lets say you have a problem that goes like this:

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?

You would have $|x-71|=28$ as your equation, and if you solve it, you get 99 as the maximum and 43 as the minimum.

Problems from Competitions

Generalized Absolute Values

The absolute value functions listed above have three very important properties:

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 on $\mathbb{Q}$, the rational numbers. Let $x=\prod_{i=1}^n p_i^{e_i}$, where the $p_{i}$'s are distinct prime numbers, and the $e_i$'s are (positive, negative, or zero) integers. Define $|x|_{p_i}=p_i^{-e_i}$. This defines an absolute value on $\mathbb{Q}$. This absolute value satisfies a stronger triangle inequality:

  • $|x+y|\le\max(|x|,|y|)$.

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 $\mathbb{R}$ or $\mathbb{C}$ is archimedian.

The theory of absolute values is important in algebraic number theory. Let $K/\mathbb{Q}$ be a finite Galois extension with $[K:\mathbb{Q}]=n$, and let $\sigma_1,\ldots,\sigma_n$ be the field automorphisms of $K$ over $\mathbb{Q}$. Then the only absolute values are the archimedian ones given by $|x|_i=|\sigma_i(x)|$ (the ordinary real or complex absolute values) and the nonarchimedian ones given by $|x|_{\mathfrak{p}}$ for some prime ${\mathfrak{p}}$ of $K$.

See Also