1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Analysis: Prove f(u+v)=f(u)+f(v) implies f(x)=f(1)x

  1. Oct 3, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose that a function f R->R has the property that f(u+v) = f(u)+f(v). Prove that f(x)=f(1)x for all rational x. Then, show that if f(x) is continuous that f(x)=f(1)x for all real x.

    3. The attempt at a solution
    I've proved that f(x)=f(1)x for all natural x by breaking up x into f(1)+f(x-1) and using induction, but I can't figure out how to do it for all rationals.
  2. jcsd
  3. Oct 3, 2011 #2
    What is


    ??? (if you take n sums)
  4. Oct 3, 2011 #3


    User Avatar
    Science Advisor
    Homework Helper

    f(1)=f(1/2+1/2)=f(1/2)+f(1/2), right? That means f(1/2)=f(1)*(1/2). Can you show f(1)=f(1/3+1/3+1/3)=f(1/3)+f(1/3)+f(1/3)? How about using induction to show f(1/n)=f(1)*(1/n)??
  5. Oct 4, 2011 #4
    Ok, I see that I can show it for f(1/n). How can I combine the two though, since I don't have a rule to say f(m/n)=f(1)*(m/n)? Also, how can I include negative numbers?

  6. Oct 4, 2011 #5


    User Avatar
    Science Advisor

    I would be inclined to look at the basic "sets of numbers" and use their defining properties. For example, the "natural numbers" or "counting numbers" are defined by "induction"- 1 is a natural number and, if n is a natural number, then n+ 1 is also. Obviously, for any number, x, f(x)= f(x)(1). Use induction to show that for any positive integer, n, then f(n)= f(x)n and then take x= 1. (You will use the general "x" below.)

    The "whole numbers" are just the positive integers together with 0. And 0 is the "additive identity", for any positive integer, n, n+ 0= n. Look at f(n+ 0)= f(n)+ f(0).

    The "integers" includes the negative integers: we now have additive inverse. Any negative integer is of the form -n for some positive integer n. f(n+ (-n))= f(n)+ f(-n) and, of course, f(n+(-n))= f(0).

    Finally, look at numbers of the form 1/n, n non-zero. Obviously, n(1/n)= 1 so f(n(1/n))= f(1). But you have already shown that f(nx)= f(x)n. So take x= 1/n.
  7. Oct 5, 2011 #6

    So helpful, got it from there. Thanks!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook