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Spot the mistakes

  1. Oct 2, 2008 #1
    1. The problem statement, all variables and given/known data
    Spot the mistakes of a student


    2. Relevant equations
    1/x < x < 1
    therefore 1<x^2
    therefore 1<x
    but x<1 therefore there are no solution


    3. The attempt at a solution
    the questions requires me to spot the mistake made by a student
    so first of all
    in the 2nd line , it reads 1<x^2 . this statement is wrong since the student assumes that x is always >0

    i'm not too sure about about the second one , namely 1<x . Since it's related to the first one . What should i put ?
    is there any other mistakes i have not spotted ?
     
  2. jcsd
  3. Oct 2, 2008 #2

    danago

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    Gold Member

    If you are given the statement x2 > 1, it doesnt necessarily imply that x>1. Consider the case where x=-5. x2=25 which is certainly greater than 1, but x>1 is not true. Again, the student has assumed that x is positive.
     
  4. Oct 2, 2008 #3
    so does it mean there's only one mistake ?
     
  5. Oct 2, 2008 #4

    statdad

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    Homework Helper

    If [tex] x > 0 [/tex] you can't have

    [tex]
    0 < \frac 1 x < x < 1
    [/tex]

    because this is equivalent to

    [tex]
    0 < 1 < x^2 < x
    [/tex]
     
  6. Oct 2, 2008 #5
    If 1/x < x < 1, it does not necessarily follow that 1 < x^2.

    For example, for x=-1/2, we have -2 < -1/2 < 1, but 1 < 1/4 is false.

    In fact, the implication is false for all -1 < x < 0.
     
  7. Oct 2, 2008 #6

    statdad

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    Homework Helper

    If the previous post was directed at mine, you missed one of my points.
    I said
    If [tex] x > 0 [/tex] you can't have

    [tex]
    0 < \frac 1 x < x < 1
    [/tex]

    because (if you multiply through the inequality by [tex] x [/tex]) then you would have

    [tex]
    0 < 1 < x^2 < x
    [/tex]

    My initial comment ruled out negative values from consideration. logarithmic, if I misunderstood you post by assuming it was meant at me, I apologize.
     
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