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Math Help - [SOLVED] Absolute distance

  1. #1
    Member courteous's Avatar
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    Question [SOLVED] Absolute distance

    "Write it down as an inequality: number x is (on a number line, I guess) less distant from 1 (one), than it is from x^2."

    |x-1|<|x-x^2|=|x(x-1)|

    For x\geq1: x-1<x-x^2 \Rightarrow x^2<1 FALSE ( x is greater than 1)
    For 0\leq x<1: -(x-1)<-(x-x^2) \Rightarrow x^2>1 FALSE ( x is less than 1 and greater than 0)
    For x<0: -(x-1)<x-x^2 \rightarrow x^2-2x+1<0 \Rightarrow (x-1)^2<0 FALSE (always positive)

    \Rightarrow inequality has no solutions

    Is this absolutely(!) correct?
    Last edited by courteous; January 23rd 2009 at 08:36 AM. Reason: larger arrows
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  2. #2
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    Quote Originally Posted by courteous View Post
    "Write it down as an inequality: number x is (on a number line, I guess) less distant from 1 (one), than it is from x^2."

    |x-1|<|x-x^2|=|x(x-1)|

    For x\geq1: x-1<x-x^2 \rightarrow x^2<1 FALSE ( x is greater than 1)
    If x > 1 then x-x^2 < 0 for all x. Therefore you have to solve the inequality:

    x\geq1: x-1<x^2-x \rightarrow 0 < x^2 -2x +1 ~\implies~ 0< (x-1)^2

    This is true for all x > 1

    For 0\leq x<1: -(x-1)<-(x-x^2) \rightarrow x^2>1 FALSE ( x is less than 1 and greater than 0)
    For x<0: -(x-1)<x-x^2 \implies x^2-2x+1<0 \rightarrow (x-1)^2<0 FALSE (always positive)

    \Rightarrow inequality has no solutions

    Is this absolutely(!) correct?
    I'm not quite sure, but in my opinion the last line of your solution isn't correct either. (Unfortunately I haven't spotted the error yet)

    By first inspection and playing around with some values I assume that the solution could be:

    x\in \left((-\infty,-1) \cup (1,\infty)\right)
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  3. #3
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    Quote Originally Posted by courteous View Post
    "Write it down as an inequality: number x is less distant from 1 (one), than it is from x^2."
    If the problem is to find real numbers x such that the distance from x to 1 is less than the distance from x to x^2, then the correct expression is \left| {x - 1} \right| < \left| {x^2 - x} \right|.
    It is worth noting that \left| {x - 1} \right| = \left| {1 - x} \right|\;\& \,\left| {x^2 - x} \right| = \left| {x - x^2} \right|, that is distance is symmetric.
    As pointed out above the solution set is \left( { - \infty , - 1} \right) \cup \left( {1,\infty } \right).
    To see this note that \left| {x^2 - x} \right| = \left| x \right|\left| {x - 1} \right| and if x \not= 1 then
    \left| {x - 1} \right| < \left| x \right|\left| {x - 1} \right|\; \Rightarrow \;1 < \left| x \right|
    Last edited by Plato; January 23rd 2009 at 08:06 AM.
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  4. #4
    Member courteous's Avatar
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    Quote Originally Posted by Plato View Post
    If the problem is to find real numbers x such that the distance from x to 1 is less than the distance from x to x^2, then the correct expression is \left| {x - 1} \right| < \left| {x^2 - x} \right|.[/tex]
    So, even as distance is symmetric, then the correct initial expression is only \left| {x - 1} \right| < \left| {x^2 - x} \right|. Why (it makes sense, besides, the other way around yields "no solution" result)? What if you've had some hard-line initial conditions, not x^2?
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  5. #5
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    Quote Originally Posted by courteous View Post
    Why (it makes sense, besides, the other way around
    Quote Originally Posted by courteous View Post
    yields "no solution" result)? What if you've had some hard-line initial conditions, not x^2?
    \begin{array}{l}<br />
 \left| {x - 1} \right| < \left| {x - x^2 } \right| \\ <br />
 \left| {1 - x} \right| < \left| {x - x^2 } \right| \\ <br />
 \left| {x - 1} \right| < \left| {x^2  - x} \right| \\ <br />
 \left| {1 - x} \right| < \left| {x^2 - x } \right| \\ <br />
 \end{array}
    Each of the above has the solution ( - \infty , - 1) \cup (1,\infty ).
    The order makes no difference.
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  6. #6
    Member courteous's Avatar
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    Lightbulb

    Quote Originally Posted by courteous View Post
    |x-1|<|x-x^2|=|x(x-1)|
    Indeed! The mischievous |x-x^2|=|x(x-1)| has mislead me. All clear (now).
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