# Thread: proof by induction on rationals

1. ## 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

2. Originally Posted by James0502
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 $\displaystyle nx$ you can write $\displaystyle \underbrace{x+...+x}_{n \text{ times} }$.
Therefore, $\displaystyle f(nx) = f(x+...+x) = f(x)+...+f(x) = nf(x)$.

3. Yes, I understand that. I am trying to prove this result for rational numbers..

many thanks

4. Originally Posted by James0502
Yes, I understand that. I am trying to prove this result for rational numbers..

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

Now you can prove it for positive rational numbers.