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Inequalities and Absolue Values: Problem Solving Approach

  1. Feb 13, 2012 #1
    Hello all! :smile: 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?
     
  2. jcsd
  3. Feb 13, 2012 #2

    mathman

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    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.
     
    Last edited: Feb 13, 2012
  4. Feb 13, 2012 #3
    Hi mathman :smile: 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.
     
  5. Feb 13, 2012 #4

    micromass

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    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.
     
  6. Feb 13, 2012 #5

    LCKurtz

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  7. Feb 13, 2012 #6

    micromass

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    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.
     
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