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Thread: 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?
    $\displaystyle Ix-3I<s=>Ix-4I<2$ (those strange I:s represent abs: )

    I solved $\displaystyle Ix-4I<2$
    That gives me $\displaystyle (-2)<x-4<2$ / add +4
    $\displaystyle 2<x<6$
    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 $\displaystyle S\in{Z^+}$ such that, $\displaystyle \left|x-3\right|< S\Rightarrow\left|x-4\right|<2$

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

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

    Therefore, $\displaystyle 3-S<x<3+S$--------------B

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

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

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

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

    For what positive number $\displaystyle s$ is the following statements true?

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


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

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



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


    [4] tells us that $\displaystyle x$ 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: .$\displaystyle \begin{array}{ccccccc}3-s \:\geq\:2 & \Rightarrow & -s \:\geq \:-1 & \Rightarrow & s \:\leq\:1 \\

    3+s \:\leq\:6 & \Rightarrow & s \:\leq\:3 \end{array}$

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


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

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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, $\displaystyle 0 \,\leq S\,\leq\,1
    $ or $\displaystyle S\geq{3}\Rightarrow{S}\in[0,1]\cup[3,\infty)$
    Last edited by Sudharaka; Jan 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: .$\displaystyle 0 \,\leq s\,\leq\,1$

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

    Quote Originally Posted by Sudharaka View Post
    Therefore the answer could be either, $\displaystyle 0 \,\leq S\,\leq\,1 $ or $\displaystyle S\geq{3}\Rightarrow{S}\in[0,1]\cup[3,\infty)$
    Quote Originally Posted by Henryt999 View Post
    Therefore, the inequalities are true for: .$\displaystyle 0 \,\leq s\,\leq\,1$
    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.
    $\displaystyle |x-3|<s$ is the set of numbers $\displaystyle (3-s,3+s)$.
    $\displaystyle |x-4|<2$ is the set of numbers $\displaystyle (2,6)$.

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

    You want to find all values of $\displaystyle s$ for which $\displaystyle \left( {3 - s,3 + s} \right) \subseteq \left( {2,6} \right) $ is true.
    Last edited by Plato; Jan 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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