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

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SUMMARY

The discussion centers on proving that for each positive integer n and each real number x, the equation f(nx) = nf(x) holds true, given that f(x) is an additive function defined by f(x+y) = f(x) + f(y). Participants suggest using mathematical induction as a method for proof. Key points include the necessity of correctly applying the definition of an additive function and clarifying the notation used for the function's iterations. The consensus is that while the induction approach is valid, the initial assumptions and definitions must be clearly established to avoid confusion.

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

Homework Equations


f(x) is an additive function (f(x+y)=f(x)+f(x))

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



Homework Equations


f(x) is an additive function (f(x+y)=f(x)+f(x))



The Attempt at a Solution


Well I'm thinking that I can just use mathematical induction to show that:
Yes, induction is the way to go.

1) f((1)x)=1f(x)
2)f(nx)=nf(x)=f_{}1(x)+f_{}2(x)+...+f_{}n(x) (assume this is true)
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).

3)f((n+1)x)=(n+1)f(x)=f_{}1(x)+f_{}2(x)+...+f_{}n+1(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?

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
 

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