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Contrapositive proof of irrational relations

  1. Feb 7, 2012 #1
    I'm confused with a question and wondered if anyone could help explain where I need to go...

    let x ε R. Prove that x is irrational thenI'm confused with a question and wondered if anyone could help explain where I need to go...

    let x ε R. Prove that x is irrational then ((5*x^(1/3))-2)/7) is irrational.

    I've learnt that this is a proof by contrapositive so set ((5*x^(1/3))-2)/7) to be rational and hence a/b with a,b ε Z and b ≠ 0, and there for x is rational.

    But I'm not sure where to go from here. Do I rearrange to find for x? x = ((7a - 10b)/5b)^3

    But this is still raational isn't it? Where is the contradiction?

    Thanks
     
  2. jcsd
  3. Feb 7, 2012 #2
    Call that big expression y, and solve for x. If y is rational, what does that tell you about x?
     
  4. Feb 7, 2012 #3
    I thought I did solve for x. If y is rational then x is rational but where is the contradiction? I don't see
     
  5. Feb 7, 2012 #4
    Ah, I missed that step in your initial post.

    Remember that contradiction is different from contraposition. For contraposition, you're actually already done.

    Contraposition takes advantage of the fact that a->b is equivalent to (not b) -> (not a).

    You showed that y is rational implies that x is rational. Therefore, you've shown that x is irrational implies y is irrational.
     
  6. Feb 9, 2012 #5

    lavinia

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    the rational numbers are closed under multiplication,division, and addition.
     
  7. Feb 9, 2012 #6

    Deveno

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    by assumption a, and b are INTEGERS, with b non-zero.

    well if a and b are integers, then surely k = 7a - 10b is also an integer.

    thus k3 and 125b3 are also integers, and moreover, 125b3 ≠ 0.

    thus x = k3/(125b3) is rational.

    but x was assumed irrational, contradiction.
     
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