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Help with gsce irrational number question

  1. Jun 4, 2006 #1
    please please help me quick!!!!!

    hi i was practisin a gcse maths paper and need some help with last question;

    x and y are two positive irrational numbers. x + y is rational and so it x times y.
    a) by writing the 1/x + 1/y as a single fraction explain why 1/x + 1/y is always rational.

    b) explain why (x-y) is always rational

    c) write down two possible positive irrational values for x and y

    i can do part A easily enough its just the second 2.
     
  2. jcsd
  3. Jun 4, 2006 #2

    HallsofIvy

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    c) write down two possible positive irrational values for x and y

    Best way to do b) is to do c) so that you have an example to look at:
    My first thought was [itex]\sqrt{2}[/itex] and [itex]-\sqrt{2}[/itex] but those are not "both positive". Okay, sinc [itex]\sqrt{2}< 2[/itex], what about [itex]x= 2+ \sqrt{2}[/itex] and [itex]y= 2- \sqrt{2}[/itex]. They are both positive x+ y= 4 while xy= 4- 2= 2. Notice that I used (a- b)(a+ b)= a2- b2 there to avoid square roots? Perhaps you could show that any irrational x, y such that both x+ y and xy are rational must be of the form [itex]x= a+ b\sqrt{c}[/itex] and [itex]y= a- b\sqrt{c}[/itex] for some rational a, b, c.
     
  4. Jun 4, 2006 #3

    arildno

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    As it stands, b is FALSE.
    Counterexample:
    [itex]x=2+\sqrt{2}, y=2-\sqrt{2}, x-y=2\sqrt{2}[/tex]
    The difference is IR-rational, but the number satisfy the given properties.

    I'm pretty sure you were to show that [itex](x-y)^{2}[/itex] must always be rational.
     
  5. Jun 4, 2006 #4

    Hurkyl

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    You can't do (b) because you have it wrong.

    Anyways, it's harder for a number to be rational than to be irrational... so instead of picking the irrational numbers, and seeing if the right things are rational, why not pick the rational numbers, and see if the right things are irrational?
     
  6. Jun 4, 2006 #5
    That is true.
     
  7. Jun 4, 2006 #6
    k, thanks thats has really helped!!!!
    yeah...im pretty sure it was meant to be (x-y)2!!!!!
     
  8. Jun 4, 2006 #7

    HallsofIvy

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    My first thought was that b) couldn't be true but then I realized that my example did not restrict x and y to be positive. My first thought for an example for c) was, for example, [itex]\sqrt{2}[/itex] and [itex]-\sqrt{2}[/itex] but that's wrong: [itex]-\sqrt{2}[/itex] is not positive. Okay, but [itex]\sqrt{2}< 2[/itex] so what about [itex]x= 2+\sqrt{2}[/itex] and [itex]y= 2- \sqrt{2}[/itex]? x+ y= 4 which is rational and xy= 4- 2= 2 which is rational. Notice that this uses (a+ b)(a- b)= a2- b2 to get rid of the square roots.
    Unfortunately, [itex]x- y= (2+ \sqrt{2})- (2-\sqrt{2})= 2\sqrt{2}[/itex] which is irrational! Is it possible that your problem (b) is to show that, under these conditions, x- y is irrational?
     
  9. Jun 4, 2006 #8

    arildno

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    Note that this is a double post; the other is in General math, I believe.
     
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