Show an additive homomorphism f:Q->Q is Q linear

[johnqwertyful]

Homework Statement

Let f(x+y)=f(x)+f(y) where x, y, f(x) are all rational numbers. Show that f(qx)=qf(x) for all q, x rational.

Homework Equations

The Attempt at a Solution

It's easy to get to f(nx)=nf(x). Because f(x+...+x)=f(x)+...+f(x)=nf(x). But going from there to rational is hard. I think I've done it before, but I can't get it. I've been stumped.

 

[Dick]

Homework Statement

Let f(x+y)=f(x)+f(y) where x, y, f(x) are all rational numbers. Show that f(qx)=qf(x) for all q, x rational.

Homework Equations

The Attempt at a Solution

It's easy to get to f(nx)=nf(x). Because f(x+...+x)=f(x)+...+f(x)=nf(x). But going from there to rational is hard. I think I've done it before, but I can't get it. I've been stumped.

Start with a simple example. f((1/2)x)+f((1/2)x)=f((1/2)x+f((1/2)x)=f(x). So how is f((1/2)x) related to f(x)? Now generalize. If you've done it before this should click pretty fast.

 

[johnqwertyful]

Start with a simple example. f((1/2)x)+f((1/2)x)=f((1/2)x+(1/2)x)=f(x). So how is f((1/2)x) related to f(x). Now generalize.

...

It was that easy, wasn't it? Thank you.

 

[johnqwertyful]

I absolutely remember doing this before. I feel so silly

f(x)=f(nx/n)=nf(x/n)
so f(x)/n=f(x/n)

I appreciate it, that's been stumping me. I should have known it

 

[Dick]

I absolutely remember doing this before. I feel so silly

f(x)=f(nx/n)=nf(x/n)
so f(x)/n=f(x/n)

I appreciate it, that's been stumping me. I should have known it

You're welome and great, it did click fast.