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Math Help - qa is rational if and only if q = 0

  1. #1
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    qa is rational if and only if q = 0

    This is exactly how the book states it.

    Let q ∈ Q and a ∈ R\Q

    Does a mean real number? I don't even know what R\Q is..

    qa is rational if and only if q=0
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by glover_m View Post
    This is exactly how the book states it.

    Let q ∈ Q and a ∈ R\Q

    Does a mean real number? I don't even know what R\Q is..

    qa is rational if and only if q=0
    \mathbb{R} \backslash \mathbb{Q} is the reals with the rationals removed.

    So x \in \mathbb{R} \backslash \mathbb{Q} if and only if x \in \mathbb{R} and x \not\in \mathbb{Q}.

    With that the rest should be easy enough.

    CB
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    \mathbb{R} \backslash \mathbb{Q} is the reals with the rationals removed.

    So x \in \mathbb{R} \backslash \mathbb{Q} if and only if x \in \mathbb{R} and x \not\in \mathbb{Q}.

    With that the rest should be easy enough.

    CB
    yah not really, i figured it was the irrationals, but not sure about proving that heh

    the way my teacher does it is he proves both ways.

    so in -<<<
    Suppose q = 0 = x/y

    qa = 0a = 0 =x/y

    therefore qa is rational if q=0

    but going ->>>>
    suppose qa is rational

    q=
    .
    .
    .
    .
    .
    =0

    that's what i dont understand how to do, i think i understand the first part, but not the second direction.
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  4. #4
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    Suppose that q0.

    Since q is rational and not zero

    where x0.

    qa is rational, so


    Combining these two we get

    (We can divide by x because it is not zero)
    But this means that a is rational, contradiction.
    Therefore it must be q=0.
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  5. #5
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    Quote Originally Posted by Halmos Rules View Post
    Suppose that q0.

    Since q is rational and not zero

    where x0.

    qa is rational, so


    Combining these two we get

    (We can divide by x because it is not zero)
    But this means that a is rational, contradiction.
    Therefore it must be q=0.
    You don't really need to go back to the x/y definition of rationals (although you can of course). If q is not 0, then q has a multiplicative inverse and the rationals are closed under multiplication.
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