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Math Help - Proove x,y must be rational when m is rational

  1. #1
    r45
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    Proove x,y must be rational when m is rational

    Let m= \frac{y}{x+1}. Prove that if m is rational, then x and y must also be rational.
    I am a bit stuck on this problem. Here is how I've approached it:

    Let m= \frac{c}{d}. Then \frac{c}{d} = \frac{y}{x+1}

    Then from here I don't know what to do. We can't express x or y solely in terms of c and d. And expressing x in terms of c, d and y (or y in terms of c, d and x) is inconclusive because we cannot assume that y and x respectively are rational, as that is what we are trying to show.

    Any help, please? Many thanks!
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    This is false. For example, let
    x = \sqrt{2} and
    y = \sqrt{2} + 1.

    Then
    \frac{y}{x+1} = 1
    which is rational, but x and y are irrational.
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  3. #3
    r45
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    Hmmmmmmmm. I shall post the original problem then. (The problem I posted was a smaller part of a larger problem).

    Let (a,b) be a point in the first quadrant which is on the unit circle x^2 + y^2 = 1 and let m denote the slope of the line joining (-1,0) to (a,b). Prove that m is rational if and only if a and b are both rational.
    Well m in terms of a and b is just m=\frac{b}{a+1}, right? Since the gradient of a line is given by change in vertical distance divided by change in horizontal distance?

    And so the question evolves into:

    Let m=\frac{b}{a+1}. Prove that m is rational if and only if a and b are both rational.
    I did it one way; i.e. assuming a and b are rational, and showing that this leads to a rational m. But it's the reverse way I can't show.

    And by way of counterexample, you've shown that m rational doesn't necessarily imply a,b rational.

    So have I simplified the problem incorrectly? Or is the original problem just incorrect?
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    you have to prove the biconditional (both forward and backwards). I suggest using proof by contrapositive and contradiction
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    Finding all the Pythagorean triples, eh?
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    Quote Originally Posted by r45 View Post
    Hmmmmmmmm. I shall post the original problem then. (The problem I posted was a smaller part of a larger problem).



    Well m in terms of a and b is just m=\frac{b}{a+1}, right? Since the gradient of a line is given by change in vertical distance divided by change in horizontal distance?

    And so the question evolves into:



    I did it one way; i.e. assuming a and b are rational, and showing that this leads to a rational m. But it's the reverse way I can't show.

    And by way of counterexample, you've shown that m rational doesn't necessarily imply a,b rational.

    So have I simplified the problem incorrectly? Or is the original problem just incorrect?
    You haven't used the fact that a^2+ b^2= 1
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