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

  • #1
395
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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.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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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.
 
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  • #3
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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. :redface:
 
  • #4
395
14
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
 
  • #5
Dick
Science Advisor
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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.
 

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