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Analysis problem

  1. Jul 15, 2016 #1
    1. The problem statement, all variables and given/known data
    show that if ##x## is rational and ##y## is irrational, then ## x+y## is irrational. Assume that ##x## is irrational and that ##y## is also irrational. Is ##xy## irrational?


    2. Relevant equations


    3. The attempt at a solution
    lol
    ##{1/2} +{√2}##
    is irrational.
    ##{{1/√2}}×{√2}=2 ## which is rational and therefore not irrational
     
    Last edited: Jul 15, 2016
  2. jcsd
  3. Jul 15, 2016 #2

    Math_QED

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    ##{{1/√2}}×{√2}=1## in fact and this is rational. Do you know what a rational number is?

    For the x*y part, with x and y irrational. They want you to proof or disproof that this product is irrational. Hint: you can disproof something by giving a counterexample.
     
  4. Jul 15, 2016 #3
    sorry typo error i have corrected i definetely know what a rational number is.
     
  5. Jul 15, 2016 #4

    BvU

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    Chwala has a problem with brackets. I don't think it has to do with a broken keyboard :smile: .

    In this thread I at first spied two exercises:
    1. Show that if x is rational and y is irrational, then x + y is irrational
    2. Assume that x is irrational and that y is also irrational. Is xy irrational?
    Or am I playing dumb again and should I read:
    Show that if x is rational and y is irrational, then x + y is irrational. (Hint: assume that x is irrational and that y is also irrational. Is xy irrational ? ).​

    Anyway it's clear that xy is not necessarly irrational if x and y are. from the counter-example.

    From post #3 I gather it is also clear that a rational number can be written as a ratio of two integers (hence the name rational...), right ?

    In order to forward our lol (?) attempt on part 1, I propose we rewrite it as: show that x(rational) + y(irrational) can NOT be written as M(integer) / N(integer) . Would that be a good strategy ?
     
  6. Jul 18, 2016 #5

    haruspex

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    Not sure I understand the attempt. Is that supposed to be a proof?
     
  7. Jul 18, 2016 #6

    Ray Vickson

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    What happens if ##x \neq 1/2## and/or ##y \neq \sqrt{2}##? Just showing the first result for two numerical examples of ##x## and ##y## does NOT constitute a proof.
     
  8. Jul 19, 2016 #7
    To show that something is irrational it's usually easier to assume it's rational and arrive at a contradiction.
    In general when doing proofs always go back to the definition.
    What does it mean that ##x## is rational exactly? What's the definition?

    A hint at solution steps:
    Assume ##x+y## is rational. What does this mean?
    What does this mean for ##y##?
     
  9. Aug 17, 2016 #8
    Ray how do we show this? in regards to your post number 6...
     
  10. Aug 17, 2016 #9

    BvU

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    Look at the hint in post #7 once more ...
     
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