Difference between revisions of "Absolute value"

Revision as of 11:25, 4 November 2006

The absolute value of a real number $x$ , denoted $|x|$ , is its distance from 0. 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 $|z|=\sqrt{x^2+y^2}$ .

Introductory Concepts

Example Problems & How To Solve for Middle School Math Problems

Example Problem ________________

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.

Divide both sides by 3 and get:

|7x|=49

Now, using the above information, I think we can solve this problem. Try to finish this by yourself, then look at the solution.

Solution:

We first get rid of the absolute value by making two equations. Thus, we make:

7x=49 and 7x=-49 - - - - 7 7 7 7

x=7 or x=-7

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

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 solved it like you did above, you get 99 as the maximum and 43 as the minimum.

Obviously, that might be too easy for absolute value, but when the numbers get tricky, this format could definately help.

--Jhredsox 10:25, 4 November 2006 (EST)

Generalized absolute values

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

$|x|\ge 0$ for all $x$ , and $|x|=0$ if and only if $x=0$ .
$|x\times y|=|x|\times |y|$ .
$|x+y| \le |x|+|y|$ . (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 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 $K$ .

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Difference between revisions of "Absolute value"

Revision as of 11:25, 4 November 2006

Contents

Introductory Concepts

Example Problems & How To Solve for Middle School Math Problems

Generalized absolute values

See also

@@ Line 5: / Line 5: @@
 == Introductory Concepts ==
-=== Example Problems ===
+=== Example Problems & How To Solve for Middle School Math Problems===
+Example Problem
+________________
 * [[2000_AMC_12/Problem_5 | 2000 AMC 12 Problem 5]]
+Amc 12 2000, Problem 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.
++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:
+|7x|=147
+Keep on isolating the variable.
+Divide both sides by 3 and get:
+|7x|=49
+Now, using the above information, I think we can solve this problem. Try to finish this by yourself, then look at the solution.
+----------------------
+'''Solution:
+We first get rid of the absolute value by making two equations. Thus, we make:
+x=49 and 7x=-49
+-  -      -   -
+  7      7   7
+x=7 or x=-7'''
+--------------------
+Now, try some of these problems (if you really are having trouble, post something on this forum.
+-|x|=x-6
+|b|=21
++8|4x|=69
+-------------------
+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 solved it like you did above, you get 99 as the maximum and 43 as the minimum.
+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==