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Proof Exercise II

  1. Jul 31, 2013 #1

    reenmachine

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    1. The problem statement, all variables and given/known data

    Suppose ##x## and ##y## are positive real numbers.If ##x < y## , then ##x^2 < y^2##.

    2. Relevant equations

    Assume ##x<y## with ##x,y \in ℝ^+##.This implies that ##\exists z \in ℝ^+## such that ##x+z=y##.We have ##y^2 = (x+z)^2 = x^2 + 2xz + z^2## , and this proves that if ##x < y## with ##x,y \in ℝ^+## , then ##x^2 < y^2##.

    any thoughts on that one? Thank you!
     
    Last edited: Jul 31, 2013
  2. jcsd
  3. Jul 31, 2013 #2

    tiny-tim

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    hi reenmachine! :wink:

    yes, that looks fine :smile:

    (alternatively, you could factor y2 - x2)
     
  4. Jul 31, 2013 #3

    hilbert2

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    You only need to know that if ##x>0##, ##y>0## and ##x<y## then ##x+y>0## and ##x-y<0##.

    Also, if ##a>0## and ##b<0## then ##ab<0##.
     
  5. Jul 31, 2013 #4

    pasmith

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    You need to explain why [itex]2xz + z^2[/itex] is strictly positive. It is obvious, but then so is the result you're trying to prove.

    An alternative proof is that if [itex]0 < x < y[/itex] then, since multiplying by a positive number preserves inequalities, we have [itex]x^2 < xy[/itex] on multiplying by [itex]x[/itex] and [itex]xy < y^2[/itex] on multiplying by [itex]y[/itex]. Putting these together we have [itex]x^2 < xy < y^2[/itex].
     
  6. Jul 31, 2013 #5

    reenmachine

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    Hi! :)

    thank you for taking the time to respond!
     
  7. Jul 31, 2013 #6

    reenmachine

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    I thought the fact that [itex]2xz + z^2[/itex] is positive was obvious from the fact x and z are in ##R^+##...

    I like your way of doing it , pretty good road to the proof.

    thank you!
     
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