# Proof of f(x) = x

1. Sep 26, 2009

### SpringPhysics

1. The problem statement, all variables and given/known data
Prove that f(x) = x (for x is rational) if f(x + y) = f(x) + f(y) and f(xy) = f(x)f(y).

2. Relevant equations

3. The attempt at a solution
I substituted y = x to get f(2x) = f(2x), which means that you can pull constants out of the function. Therefore, for all x and y, f(xy)=xy and so f(x) = x and f(y) = y.

However, I am not sure that this solution works because it seems to simple and illogical in a way. I also didn't prove that x is rational.

2. Sep 26, 2009

### gabbagabbahey

Is this a typo? Don't you mean f(2x)=2f(x)?

3. Sep 26, 2009

### SpringPhysics

Sorry, I meant f(2x) = 2f(x). Thanks.

4. Sep 26, 2009

### HallsofIvy

Staff Emeritus
I would be inclined to work with the various number systems. For example, you should be able to use induction to show that f(nx)= nf(x)= f(n)f(x) for any positive integer n and thus that f(n)= n.
Then use the fact that f(n+0)= f(n)+ f(0) to show that f(0)= 0.
Use the fact that f(-n)= f((-1)(n))= nf(-1) while f(1- 1)= f(1)+ f(-1)= f(0)= 0 to get f(-n)= -n.

Finally, use the fact that f(1)= f(m(1/m))= mf(1/m)= 1 to get rational numbers.

By the way, if you add the requirement that f is continuous for all x, then f(x)= x for all real numbers. Without that added requirement, that is not true.

5. Sep 26, 2009

### D H

Staff Emeritus
As gabbagabbahey noted, surely you meant f(2x)=2f(x). All that means you can take a factor of 2 out of the argument. It does not necessarily mean you can "pull constants out of the function". BTW, if this were a valid step, why take the extra step of going to f(xy)=xy, etc.? You have already proved the assertion! (Invalidly, of course).

6. Sep 26, 2009

### SpringPhysics

Thanks for the reply. So in general this would prove that f(x) = x, since I start with the base case (f(0) = 0), then prove it for positive integers using the induction method, then for negative integers, then rational numbers. Aren't the proofs a little redundant? I don't quite see the point in proving negative integers for rational numbers - is it to ascertain that the function holds for negative rational numbers as well? There's also a hint in there to use the fact that rational numbers exist between any number (in order to prove f(x) = x for all x). So wouldn't this have already proven that?

7. Sep 26, 2009

### SpringPhysics

I had wanted to prove that f(x) = x in general by using that specific example, but I guess I would have to work through all the number systems for that.

8. Sep 26, 2009

### SpringPhysics

Would it be possible to prove that for irrational numbers? (The exact same thing, except defining n as an irrational number?)

9. Sep 27, 2009

### nietzsche

are you by any chance in regina rotman's class at u of t?

10. Sep 27, 2009

### Office_Shredder

Staff Emeritus
You can prove that f(x)=x for all algebraic numbers (essentially, f(a polynomial of x) = the polynomial evaluated at f(x)) but for transcendental numbers it seems you can define f(x) to be something different. There was another thread about the same function a couple days ago that covered the details, I'll see if I can find the link

EDIT:

Here it is

11. Sep 27, 2009

### SpringPhysics

Thanks, I'll try that.

12. Sep 27, 2009

### Hurkyl

Staff Emeritus
13. Sep 27, 2009

### nietzsche

This question is part of my assigned homework too, and I can't figure out how to prove that f(x) = x for irrational numbers. The book has a hint that says "use the fact that between any two numbers there is a rational number".

In previous parts of the question, I've already proved that:
f(1) = 1,
f(x) = x if x is rational,
f(x)>0 if x>0, and
f(x)>f(y) if x>y.

So I'm assuming that there is an easy way to prove this for the irrational numbers, using only these four facts, the two properties of the function [ f(xy) = f(x)f(y) and f(x+y) = f(x)+f(y) ] and the hint that between any two numbers, there exists a rational number.

Any suggestions?

14. Sep 27, 2009

### Hurkyl

Staff Emeritus
Can f(pi) be bigger than 4? 3.2?

15. Sep 27, 2009

### nietzsche

Well...

Assume that f(pi) > 4

\begin{align*} f(\pi) &> 4\\ f(\pi) &> f(4)\\ f(\pi-\pi) &> f(4-\pi)\\ f(0) &> f(4-\pi)\\ 0 &> f(4-\pi) \end{align*}

but this contradicts the statement that if x > 0, f(x) > 0. So f(pi) cannot be greater than 4.

But I'm not sure how that helps me. Or if what I just did there is valid.

16. Sep 27, 2009

### nietzsche

I guess we can sort of generalize it that f(pi) will never be larger than f(x) when x is a rational number larger than pi?

17. Sep 27, 2009

### Hurkyl

Staff Emeritus
That sounds correct. Can you find a less convoluted way of saying it?

18. Sep 27, 2009

### jgens

Your proof can be simplified considerably: since $\pi < 4$ we know that $f(\pi) < f(4) = 4$. It's really the same argument, it just uses another fact that you've already proven.

To help you complete the proof, suppose that $f(\alpha) \neq \alpha$ for some $\alpha$ irrational. Then we have that $f(\alpha) = \alpha + \epsilon$ for some $\epsilon \neq 0$. Can you see how this produces a contradtion?

19. Sep 27, 2009

### nietzsche

is the epsilon considered to be a rational number?

20. Sep 27, 2009

### nietzsche

i'm sorry, i'm still confused. i can't see how it produces a contradiction.