Thread: 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

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 you can write .
Therefore, .

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

many thanks

Originally Posted by James0502

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

many thanks

Write, .
Therefore, .

Now you can prove it for positive rational numbers.