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Math Help - Inequality problem

  1. #1
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    Inequality problem

    Hi thanks in advance.
    I canīt grasp what the inequality is asking for?
    For what positive number (s) is the following statement true?
    Ix-3I<s=>Ix-4I<2 (those strange I:s represent abs: )

    I solved Ix-4I<2
    That gives me (-2)<x-4<2 / add +4
    2<x<6
    And now I donīt understand what I am suppose to do.
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  2. #2
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    Dear Henryt999,

    We have to find S\in{Z^+} such that, \left|x-3\right|< S\Rightarrow\left|x-4\right|<2

    As you have mentioned, \left|x-4\right|<2\Rightarrow2<x<6-------------A

    Similarly, \left|x-3\right|< S\Rightarrow-S<x-3<S

    Therefore, 3-S<x<3+S--------------B

    From A and B, 3-S\leq{2} and S+3\geq{6}

    Therefore, S\geq{1} and S\geq{3}

    So we would get, S\geq{3} as the answer.

    Hope this helps.
    Last edited by Sudharaka; January 26th 2010 at 04:40 PM.
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  3. #3
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    Hello, Henryt999!

    For what positive number s is the following statements true?

    . . \begin{array}{cccc}|x-3| &<& s & [1]\\ \\[-3mm] |x-4| &<& 2 & [2] \end{array}
    Note that: .  s \,\geq\,0


    From [1], we have: . -s \:<\:x-3 \:<\:s \quad\Rightarrow\quad -s+3 \:<\:x\:<\:s + 3 .[3]

    From [2], we have: . -2 \:<\:x-4\:<\:2 \quad\Rightarrow\quad 2 \:<\:x\:<\:6 .[4]



    [3] tells us that x is between 3-s and 3+s
    Code:
    
        - - - o = = = = = = o - - -
             3-s           3+s


    [4] tells us that x is between 2 and 6.
    Code:
    
        - - o = = = = = = = = o - -
            2                 6



    For both statements to be true,
    . . the first interval must be "inside" the second interval.
    Code:
    
        - - - o = = = = = = o - - -
             3-s           3+s
    
        - - o = = = = = = = = o - -
            2                 6

    So we have: . \begin{array}{ccccccc}3-s \:\geq\:2 & \Rightarrow & -s \:\geq \:-1 & \Rightarrow & s \:\leq\:1 \\<br /> <br />
3+s \:\leq\:6 & \Rightarrow & s \:\leq\:3 \end{array}

    . . which are both true for: .  s \,\leq\,1


    Therefore, the inequalities are true for: . 0 \,\leq s\,\leq\,1

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  4. #4
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    Dear Sorban,

    You had considered the situation where the 3-s<x<3+x is inside 2<x<6. But I have considered the situation where 2<x<6 is inside 3-s<x<3+x. I think both situations must be considered since in both situations, the two inequalities hold.

    Therefore the answer could be either, 0 \,\leq S\,\leq\,1<br />
or S\geq{3}\Rightarrow{S}\in[0,1]\cup[3,\infty)
    Last edited by Sudharaka; January 26th 2010 at 05:00 PM.
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  5. #5
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    Thanks too both of you guys but..


    Therefore, the inequalities are true for: . 0 \,\leq s\,\leq\,1

    Thanks for the help, one part I dont understand.
    We get that s<1
    how did this get too 0<s<1?
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  6. #6
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    [QUOTE=Henryt999;445369]For what positive number (s) is the following statement true? |x-3|<s=>|x-4|<2 [/QUOTE

    Quote Originally Posted by Sudharaka View Post
    Therefore the answer could be either, 0 \,\leq S\,\leq\,1 or S\geq{3}\Rightarrow{S}\in[0,1]\cup[3,\infty)
    Quote Originally Posted by Henryt999 View Post
    Therefore, the inequalities are true for: . 0 \,\leq s\,\leq\,1
    Sudharaka, the solution Sorban gave is correct and is the only solution.
    I think that you have misread the question or you failed to see the implication.
    |x-3|<s is the set of numbers (3-s,3+s).
    |x-4|<2 is the set of numbers (2,6).

    The implication in the question, |x-3|<s~{\color{blue}\Rightarrow} |x-4|<2, means that x \in \left( {3 - s,3 + s} \right) \Rightarrow ~ x \in \left( {2,6} \right).
    We know that 7\in [3,\infty) but we cannot have s=7.
    For then we would have 0\in (-4,10) but 0 \notin (2,6) .

    You want to find all values of s for which  \left( {3 - s,3 + s} \right) \subseteq \left( {2,6} \right) is true.
    Last edited by Plato; January 27th 2010 at 09:32 AM.
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  7. #7
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    Dear Plato,

    What you have said is correct. Now I see the mistake I had made and thank you so much for telling me that.
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