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Inequalities and Absolue Values: Problem Solving Approach 
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#1
Feb1312, 02:11 PM

P: 3,015

Hello all! In my quest to reteach 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 ##x1+x2 > 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: ##(x1)>0 \wedge (x2)>0## then ##(x1) + (x2) > 1 \implies x > 2.## Case 2: ##(x1)<0 \wedge (x2)<0## then ## (1x) + (2x) > 1 \implies x<2.## Case 3: ##(x1)>0 \wedge (x2)<0## then ## (x1)+(1x) > 1 \implies 0 >1. ## Case 4: ##(x1)<0 \wedge (x2)>0## then ## (1x) + (x+2) > 1 \implies 3 > 1. ## Let's just look at Case 1 for a moment. Assuming that ##(x1)>0 \wedge (x2)>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 ##(x1)## 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? 


#2
Feb1312, 03:31 PM

Sci Advisor
P: 6,039

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 nonexistent, x > 2 and x < 1 is impossible. 


#3
Feb1312, 03:42 PM

P: 3,015




#4
Feb1312, 04:13 PM

Mentor
P: 18,040

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]x1>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 x1=x1. So the equation is x1>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 x1=1x. The equation becomes 1x>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. 


#5
Feb1312, 04:15 PM

HW Helper
Thanks
PF Gold
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This thread is a duplicate of:
http://www.physicsforums.com/showthread.php?t=577008 


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