# Probably easier than I think f(xn)=nf(x)?

1. Mar 15, 2009

### geoman

1. show that for each positive integer n and each real number x, f(nx)=nf(x).

2. Relevant equations
f(x) is an additive function (f(x+y)=f(x)+f(x))

3. The attempt at a solution
Well I'm thinking that I can just use mathematical induction to show that:
1) f((1)x)=1f(x)
2)f(nx)=nf(x)=f$$_{}1$$(x)+f$$_{}2$$(x)+....+f$$_{}n$$(x) (assume this is true)
3)f((n+1)x)=(n+1)f(x)=f$$_{}1$$(x)+f$$_{}2$$(x)+....+f$$_{}n+1$$(x)
from here it would just be algebra. is this right?
note: the numbered functions above are supposed to be subscripted but i messed it up somehow. They are only supposed to number the functions as I add the function n times and then n+1 times

Last edited: Mar 15, 2009
2. Mar 15, 2009

### Staff: Mentor

How you have presented this problem is confusing.
The statement you presented in section 1 is not true in general.
The statement you presented as a relevant equation is more than just relevant; it is part of the problem statement, I'm pretty sure. Also, you have a typo, and this equation should be f(x + y) = f(x) + f(y).

In your attempt at a solution, yes you should assume that f(nx) = nf(x), but how in the world do you go from nf(x) to a sum of powers of f(x)?

Look at what happens with f(x + x). What is that equal to by your original assumptions? You have a way to go before you will be able to tackle the induction step, IMO, so let's get up to speed with some simple calculations first.

3. Mar 16, 2009

### HallsofIvy

Staff Emeritus
Yes, induction is the way to go.

What??? Where in the world did that horribly complicated right end come from? You haven't even defined "f1", "f2", etc.
Just use f(nx)= nf(x).

You can't use "f((n+1)x)= (n+1)f(x)", that's what you want to prove. You can use the fact that f((n+1)x)= f(nx+x)= what?