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Math: Absolute value (Modulus)

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Math: Absolute value (Modulus) [#permalink] New post 06 Nov 2009, 18:49
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ABSOLUTE VALUE
(Modulus)

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This post is a part of [GMAT MATH BOOK]

created by: walker
edited by: bb, Bunuel

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Definition

The absolute value (or modulus) |x| of a real number x is x's numerical value without regard to its sign.

For example, |3| = 3; |-12| = 12; |-1.3|=1.3



Graph:
Image


Important properties:

|x|\geq0

|x| = \sqrt{x^2}

|0|=0

|-x|=|x|

|x|+|y|\geq|x+y|


3-steps approach:

General approach to solving equalities and inequalities with absolute value:

1. Open modulus and set conditions.
To solve/open a modulus, you need to consider 2 situations to find all roots:
  • Positive (or rather non-negative)
  • Negative

For example, |x-1|=4
a) Positive: if (x-1)\geq0, we can rewrite the equation as: x-1=4
b) Negative: if (x-1)<0, we can rewrite the equation as: -(x-1)=4
We can also think about conditions like graphics. x=1 is a key point in which the expression under modulus equals zero. All points right are the first condition (x>1) and all points left are second condition (x<1).
Image

2. Solve new equations:
a) x-1=4 --> x=5
b) -x+1=4 --> x=-3

3. Check conditions for each solution:
a) x=5 has to satisfy initial condition x-1>=0. 5-1=4>0. It satisfies. Otherwise, we would have to reject x=5.
b) x=-3 has to satisfy initial condition x-1<0. -3-1=-4<0. It satisfies. Otherwise, we would have to reject x=-3.


3-steps approach for complex problems

Let’s consider following examples,

Example #1
Q.: |x+3| - |4-x| = |8+x|. How many solutions does the equation have?
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) x < -8. -(x+3) - (4-x) = -(8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not less than -8)

b) -8 \leq x < -3. -(x+3) - (4-x) = (8+x) --> x = -15. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)

c) -3 \leq x < 4. (x+3) - (4-x) = (8+x) --> x = 9. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)

d) x \geq 4. (x+3) + (4-x) = (8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not more than 4)

(Optional) The following illustration may help you understand how to open modulus at different conditions.
Image
Answer: 0

Example #2
Q.: |x^2-4| = 1. What is x?
Solution: There are 2 conditions:

a) (x^2-4)\geq0 --> x \leq -2 or x\geq2. x^2-4=1 --> x^2 = 5. x e {-\sqrt{5}, \sqrt{5}} and both solutions satisfy the condition.

b) (x^2-4)<0 --> -2 < x < 2. -(x^2-4) = 1 --> x^2 = 3. x e {-\sqrt{3}, \sqrt{3}} and both solutions satisfy the condition.

(Optional) The following illustration may help you understand how to open modulus at different conditions.
Image
Answer: -\sqrt{5}, -\sqrt{3}, \sqrt{3}, \sqrt{5}


Tip & Tricks

The 3-steps method works in almost all cases. At the same time, often there are shortcuts and tricks that allow you to solve absolute value problems in 10-20 sec.

I. Thinking of inequality with modulus as a segment at the number line.

For example,
Problem: 1<x<9. What inequality represents this condition?
Image
A. |x|<3
B. |x+5|<4
C. |x-1|<9
D. |-5+x|<4
E. |3+x|<5
Solution: 10sec. Traditional 3-steps method is too time-consume technique. First of all we find length (9-1)=8 and center (1+8/2=5) of the segment represented by 1<x<9. Now, let’s look at our options. Only B and D has 8/2=4 on the right side and D had left site 0 at x=5. Therefore, answer is D.

II. Converting inequalities with modulus into a range expression.
In many cases, especially in DS problems, it helps avoid silly mistakes.

For example,
|x|<5 is equal to x e (-5,5).
|x+3|>3 is equal to x e (-inf,-6)&(0,+inf)

III. Thinking about absolute values as the distance between points at the number line.

For example,
Problem: A<X<Y<B. Is |A-X| <|X-B|?
1) |Y-A|<|B-Y|
Solution:
Image
We can think of absolute values here as the distance between points. Statement 1 means than the distance between Y and A is less than that between Y and B. Because X is between A and Y, |X-A| < |Y-A| and at the same time the distance between X and B will be larger than that between Y and B (|B-Y|<|B-X|). Therefore, statement 1 is sufficient.


Pitfalls

The most typical pitfall is ignoring the third step in opening modulus - always check whether your solution satisfies conditions.


Official GMAC Books:

The Official Guide, 12th Edition: PS #22; PS #50; PS #130; DS #1; DS #153;
The Official Guide, Quantitative 2th Edition: PS #152; PS #156; DS #96; DS #120;
The Official Guide, 11th Edition: DT #9; PS #20; PS #130; DS #3; DS #105; DS #128;

Generated from [GMAT ToolKit]


Resources

Absolute value DS problems: [search]
Absolute value PS problems: [search]

Fig's post with absolute value problems: [Absolute Value Problems]



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Last edited by walker on 02 Jun 2013, 14:01, edited 27 times in total.
added a new example + illustrations
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Re: Math: Absolute value (Modulus) [#permalink] New post 06 Nov 2009, 20:02
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Great post!! Kudos!! :)

If some problems links with categorization (like 700 level, 600-700 level) are posted, it will be great. I was in search of range approach for solving modules problems, I got some stuff here. Thanks again!
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Re: Math: Absolute value (Modulus) [#permalink] New post 06 Nov 2009, 20:06
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Great post walker!!! + 1 to you.

PS. I have just made 2 corrections above.
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Re: Math: Absolute value (Modulus) [#permalink] New post 06 Nov 2009, 20:26
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hgp2k wrote:
PS. I have just made 2 corrections above.


+1
Thanks :)
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Re: Math: Absolute value (Modulus) [#permalink] New post 06 Nov 2009, 20:34
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Hussain15 wrote:
If some problems links with categorization (like 700 level, 600-700 level) are posted, it will be great. I was in search of range approach for solving modules problems, I got some stuff here. Thanks again!


Thanks! I added link to Fig's post with absolute value problems.
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Re: Math: Absolute value (Modulus) [#permalink] New post 07 Nov 2009, 16:06
kudos! :shock:
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Re: Math: Absolute value (Modulus) [#permalink] New post 13 Nov 2009, 22:07
Hi Walker, Thanks for posting this.

You've written a property that:
|X + Y| >= |X| + |Y|
Is the same true for negative?
|X - Y| <= |X| - |Y|
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Re: Math: Absolute value (Modulus) [#permalink] New post 14 Nov 2009, 03:53
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yangsta8 wrote:
Hi Walker, Thanks for posting this.

You've written a property that:
|X + Y| >= |X| + |Y|
Is the same true for negative?
|X - Y| <= |X| - |Y|


|X + Y| <= |X| + |Y|

|X - Y| >=|X| - |Y|
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Re: Math: Absolute value (Modulus) [#permalink] New post 13 Dec 2009, 04:40
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Not that it matters in any way, but can you correct the "witch" under 1b in the initial post. Other than that: awesome post! :)
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Re: Math: Absolute value (Modulus) [#permalink] New post 15 Dec 2009, 22:45
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Hi Walker!
I found a mistake in your expl.
In 3-step approach for more than one module: (d) is not -9, it's -1 :)
Correct me if I'm wrong.
Thank you.
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Re: Math: Absolute value (Modulus) [#permalink] New post 15 Dec 2009, 23:08
Expert's post
Igor010 wrote:
Hi Walker!
I found a mistake in your expl.
In 3-step approach for more than one module: (d) is not -9, it's -1 :)
Correct me if I'm wrong.
Thank you.


Thanks! You are right
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Re: Math: Absolute value (Modulus) [#permalink] New post 21 Dec 2009, 20:49
Thank you! Thank you! Thank you! This is the clearest explanation of absolute value that I've come across thus far.
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Re: Math: Absolute value (Modulus) [#permalink] New post 23 Dec 2009, 19:42
This is exactly what I was looking for past few days. Great Post.
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Re: Math: Absolute value (Modulus) [#permalink] New post 27 Dec 2009, 03:26
Sorry, but please clarify why is it not...
a) x < -8. -(x+3) + (4-x) = -(8+x)

The -ve sign is given in the question stem, once we take the solution -(4-x) with the -ve sign in the equation, then two -ves should be positive, shouldn't it?
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Re: Math: Absolute value (Modulus) [#permalink] New post 27 Dec 2009, 06:55
wonderful post walker :-D
kudos..
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Re: Math: Absolute value (Modulus) [#permalink] New post 27 Dec 2009, 07:05
arjunrampal wrote:
Sorry, but please clarify why is it not...
a) x < -8. -(x+3) + (4-x) = -(8+x)

The -ve sign is given in the question stem, once we take the solution -(4-x) with the -ve sign in the equation, then two -ves should be positive, shouldn't it?


hai arjunrampal..

when x< -8, (x+3) and (8+x) are both negative, while (4-x) is positive. (if you want to confirm this, u can plug in values for x and try)

when x<0 |x| = -x ( i.e. |x| = - (-ve x), ultimately modulus x is positive. i hope this point is clear)
when x>0 |x| = x

so for x< -8, |(4-x)| remains positive while modulus of the other two expressions become negative.

hope it is clear...
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Re: Math: Absolute value (Modulus) [#permalink] New post 27 Dec 2009, 09:22
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I've added a new example:


Example #2
Q.: |x^2-4| = 1. What is x?
Solution: There are 2 conditions:

a) (x^2-4)\geq0 --> x \leq -2 or x\geq2. x^2-4=1 --> x^2 = 5. x e {-\sqrt{5}, \sqrt{5}} and both solutions satisfy the condition.

b) (x^2-4)<0 --> -2 < x < 2. -(x^2-4) = 1 --> x^2 = 3. x e {-\sqrt{3}, \sqrt{3}} and both solutions satisfy the condition.

Answer: -\sqrt{5}, -\sqrt{3}, \sqrt{3}, \sqrt{5}
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Re: Math: Absolute value (Modulus) [#permalink] New post 27 Dec 2009, 10:32
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arjunrampal wrote:
Sorry, but please clarify why is it not...
a) x < -8. -(x+3) + (4-x) = -(8+x)

The -ve sign is given in the question stem, once we take the solution -(4-x) with the -ve sign in the equation, then two -ves should be positive, shouldn't it?


Thanks for the question. I've added illustration to the example as well as one more example and hope they help
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Re: Math: Absolute value (Modulus) [#permalink] New post 27 Dec 2009, 11:16
Many thanks for the illustration. This is now made clear. Kudos!
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Re: Math: Absolute value (Modulus) [#permalink] New post 26 Apr 2010, 04:56
Thank you walker. This post helped me understand the concepts of absolute value
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Re: Math: Absolute value (Modulus)   [#permalink] 26 Apr 2010, 04:56
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