Inequalities and Absolue Values: Problem Solving Approach


by Saladsamurai
Tags: absolue, inequalities, solving, values
Saladsamurai
Saladsamurai is offline
#1
Feb13-12, 02:11 PM
Saladsamurai's Avatar
P: 3,012
Hello all! In my quest to re-teach myself the basics of mathematics in a more rigorous fashion, I have found out that inequalities and absolute values are a weak point if mine. So I am working to address that. I am getting much better at it (with help from PF), but I have recently encountered seemingly simple problem that turned out to be a little trickier than I thought. Though I can arrive at the correct answer, I am not sure that my procedure is sound. Hopefully you can offer some insight. Take the following problem from chapter 1 of Spivak's Calculus (Problem 11 (iv)):

Find all ##x## for which ##|x-1|+|x-2| > 1 \qquad(1)##.

My approach to these has been to use the fact that the definition of absolute value is

[tex]
\begin{align}
|x| =
\begin{cases}
x, & \text{if }x\ge0 \\
-x, & \text{if }x\le0
\end{cases}
\end{align}
[/tex]

so then for each quantity enclosed by absolute value signs, there are 2 cases that needed to be evaluated. Applying this to (1) we have


Case 1: ##(x-1)>0 \wedge (x-2)>0## then

##(x-1) + (x-2) > 1 \implies x > 2.##


Case 2: ##(x-1)<0 \wedge (x-2)<0## then

## (1-x) + (2-x) > 1 \implies x<2.##


Case 3: ##(x-1)>0 \wedge (x-2)<0## then

## (x-1)+(1-x) > 1 \implies 0 >1. ##


Case 4: ##(x-1)<0 \wedge (x-2)>0## then

## (1-x) + (x+2) > 1 \implies 3 > 1. ##


Let's just look at Case 1 for a moment.

Assuming that ##(x-1)>0 \wedge (x-2)>0## is the same as assuming ## x > 1 \wedge x>2.## This is clearly only true for ##x>2##, so there is really no need to specify that ##x>1.## But when it comes time to solve the actual problem, I need to use the expression ##(x-1)## under the assumption that it is a positive quantity, which is the same as specifying that ##x>1##. The answer I got is ##x>2## and is valid, but I feel like I might miss something in future problems if I do not pay attention to this detail.

So my question is this: Do I simply solve the inequality as I have done and then restrict the solution to ##x>2## if I were to get something less than 2?

Does my question make sense?
Phys.Org News Partner Mathematics news on Phys.org
Researchers help Boston Marathon organizers plan for 2014 race
'Math detective' analyzes odds for suspicious lottery wins
Pseudo-mathematics and financial charlatanism
mathman
mathman is offline
#2
Feb13-12, 03:31 PM
Sci Advisor
P: 5,935
It is simpler just to look at the domains for x.
There are 3 cases, x > 2, 2 > x > 1, 1 > x. These correspond to your first 3 cases (not in the same order).
Your case 4 is non-existent, x > 2 and x < 1 is impossible.
Saladsamurai
Saladsamurai is offline
#3
Feb13-12, 03:42 PM
Saladsamurai's Avatar
P: 3,012
Quote Quote by mathman View Post
It is simpler just to look at the domains for x.
There are 3 cases, x > 2, 2 > x > 1, 1 > x. These correspond to your first 3 cases (not in the same order).
Your case 4 is non-existent, x > 2 and x < 1 is impossible.
Hi mathman Yes, we have not gotten to the other 3 cases yet. I am focusing on the simple one: Case 1. I am afraid people might not be understanding question, but I am not sure how else to word it.

micromass
micromass is online now
#4
Feb13-12, 04:13 PM
Mentor
micromass's Avatar
P: 16,528

Inequalities and Absolue Values: Problem Solving Approach


The case [itex]x>1~\wedge~x>2[/itex] is indeed equivalent to x>2.

If you solve the equation under the premisse that x>2, then every solution must satisfy that. If you find that the equation is true for all x>-2, then only the x>2 will count.

For example, if you solve [itex]|x-1|>-5[/itex] (I know you can easily see that all x will be a solution, but I'm just setting an example).

You can split up

1) [itex]x\geq 1[/itex], in that case |x-1|=x-1. So the equation is x-1>-5. This is true for x>-4. However, you originally set [itex]x>1[/itex], so in this case the only solutions are all [itex]x>1[/itex] (and not all x>-4).

2) x<1, in that case |x-1|=1-x. The equation becomes 1-x>5, or x<6. In this case the solutions are all x<1 (and not x<6).

Adding (1) and (2) yields that all real numbers are a solution.
LCKurtz
LCKurtz is offline
#5
Feb13-12, 04:15 PM
HW Helper
Thanks
PF Gold
LCKurtz's Avatar
P: 7,176
This thread is a duplicate of:
http://www.physicsforums.com/showthread.php?t=577008
micromass
micromass is online now
#6
Feb13-12, 09:23 PM
Mentor
micromass's Avatar
P: 16,528
Salad convinved me that a seperate thread does have some merit. This thread is not the same as the HW question, but is rather more general. So I'll allow this thread to stay open.


Register to reply

Related Discussions
Proofs: Absolute Values and Inequalities Calculus & Beyond Homework 5
Inequalities involving division of two absolute values Precalculus Mathematics Homework 2
Solving Inequalities Precalculus Mathematics Homework 3
Solving quadratic inequalities and absolute values Calculus & Beyond Homework 8
help solving problem involving inequalities General Math 7