# ALL functions such that:

1. ### iceblits

113
1. The problem statement, all variables and given/known data

With proof, Find all functions f: R->R such that f(x+y)=f(x)+f(y)+1 and f(1)=0

2. Relevant equations

3. The attempt at a solution

First I cried...then...

Since f(1)=0, f(2)=1, f(3)=2.....= x-1
I'm having a really hard time conceptualizing this problem and I'm not sure what the answer is suppose to look like. Is it a finite or infinite set of functions? I'm not sure. The proof part is kinda throwing me off too. Any help would be great..

Last edited: Sep 20, 2011
2. ### flyingpig

f(1) = 0 and f(x + y) = ?

3. ### Dick

25,853
You can find the values of more than just the integers. E.g. what's f(1/2)? Can you find the value of f(1/n) where n is an integer? Can you find the value of f(m/n) where m and n are integers? Are you given that f is continuous or anything like that?

4. ### iceblits

113
yes f is continuous, and yes i can find f(1/2), but what i dont understand is why this problem has more than one solution (y=x-1)

5. ### Dick

25,853
Why do you think it has more than one solution?

6. ### LCKurtz

8,371
And if you assume it is differentiable you can think about its derivative...

7. ### iceblits

113
@flyingpig: f(x+y)=f(x)+f(y)+1..it isn't given as anything else

8. ### flyingpig

Looking at all the post, this question is probably out of my league. So I will just erase my answer.

9. ### Dick

25,853
You don't need to assume it's differentiable. I think continuous is enough, isn't it?

10. ### LCKurtz

8,371
It might be, I don't know. I haven't worked through it. But if it is differentiable that gives an easy way to find a family of functions that work, which must be included in the "all" functions.

11. ### Dick

25,853
I did work through it. To use differentiability, which I DON'T think is given, you need to find the value of something like f(1/n). Then it's straightforward. But once you've found f(1/n) why not find f(m/n) (m and n integers) and use continuity and skip the differentiability assumption? That's what I'm trying to push iceblits to do.

12. ### iceblits

113
oh I see...I'll try. I'm just having a hard time conceptualizing this because I feel like y=x-1 is the only functions that can satisfy all the points. Thank-you for all the responses though..I will keep trying this problem until I solve it

13. ### Dick

25,853
y=x-1 IS the only function that works. You just have to prove it.

14. ### iceblits

113
oh! i see!!!!!!

15. ### iceblits

113
ok now i have some direction lol...I was thinking there was an infinite set of functions or something (like some wierd sine wave going diagonally up)..or like the line y=x-1 except if you imagine it like a string, push the string together to get various functions..or something like that

16. ### LCKurtz

8,371
No, it isn't the only one.

Woops, ignore that. I agree. I forgot f(1) = 0.

17. ### Dick

25,853
That makes me feel better.

18. ### iceblits

113
gahhh..ok so if I let x and y both equal n, then, f(2n)=2f(n)+1. if I let x and y equal 1/n I get f(2/n)=2f(1/n)+1 and if I let x and y equal m/n, I get f(2m/n)=2f(m/n)+1....I dont think that shows anything (sorry, I haven't taken a proof class or anything yet :( ). Should I be trying to manipulate f(x+y)=f(x)+f(y)+1 to achieve y=x-1 or should I assume y=x-1 is the only solution and then show that any if any other function exists to satisfy the condition, it would have to be y=x-1? I guess I'm still lost haha..I thought I would be able to prove that y=x-1 is the only solution really easily but i guess i cant...

19. ### Dick

25,853
You found f(1/2), right? How would you find f(1/3)? Does that help you to find a general way to find f(1/n)? Here's a hint. Can you figure out a formula for f(nx) where n is an integer and x is any real? Formally, it's an induction proof, but if you haven't done induction you should still be able to figure out the formula just by thinking about it.

20. ### iceblits

113
ye f(1/2+1/2)=f(1/2)+f(1/2)+1...f(1)=2f(1/2)+1, f(1/2)=-1/2..wouldn't the formula just be y=x-1?