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Math Help - proof by induction on rationals

  1. #1
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    proof by induction on rationals

    ok.. I have a function f such that f(x + y) = f(x) + f(y)

    I have proves f(nx) = nf(x) for all x and every natural number n

    I now need to show that this is true for f(rx) where f is a rational number n/m

    I've tried fixing m and doing induction on n but I cant do it by fixing n and doing induction on m, is this the right way of going about it?

    many thanks
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  2. #2
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    Quote Originally Posted by James0502 View Post
    ok.. I have a function f such that f(x + y) = f(x) + f(y)

    I have proves f(nx) = nf(x) for all x and every natural number n

    I now need to show that this is true for f(rx) where f is a rational number n/m

    I've tried fixing m and doing induction on n but I cant do it by fixing n and doing induction on m, is this the right way of going about it?

    many thanks
    When you have nx you can write \underbrace{x+...+x}_{n \text{ times} }.
    Therefore, f(nx) = f(x+...+x) = f(x)+...+f(x) = nf(x).
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  3. #3
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    Yes, I understand that. I am trying to prove this result for rational numbers..

    many thanks
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  4. #4
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    Quote Originally Posted by James0502 View Post
    Yes, I understand that. I am trying to prove this result for rational numbers..

    many thanks
    Write, 1 = \tfrac{1}{n}+...+\tfrac{1}{n}.
    Therefore, f(1) = f(\tfrac{1}{n}+...+\tfrac{1}{n}) = nf(\tfrac{1}{n}) \implies f(\tfrac{1}{n}) = \tfrac{1}{n}f(1).

    Now you can prove it for positive rational numbers.
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