Proof f(x)=x for x Rational w/ f(x+y)=f(x)+f(y), f(xy)=f(x)f(y)

[SpringPhysics]

Homework Statement

Prove that f(x) = x (for x is rational) if f(x + y) = f(x) + f(y) and f(xy) = f(x)f(y).

Homework Equations

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.

 

[gabbagabbahey]

I substituted y = x to get f(2x) = f(2x), which means that you can pull constants out of the function.

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

 

[SpringPhysics]

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

 

[HallsofIvy]

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.

 

[D H]

I substituted y = x to get f(2x) = f(2x), which means that you can pull constants out of the function.

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).

 

[SpringPhysics]

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.

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?

 

[SpringPhysics]

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).

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.

 

[SpringPhysics]

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

 

[nietzsche]

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

 

[Office_Shredder]

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

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:
https://www.physicsforums.com/showthread.php?t=339490

Here it is

 

[SpringPhysics]

Thanks, I'll try that.

 

[Hurkyl]

for transcendental numbers it seems you can define f(x) to be something different.
...
https://www.physicsforums.com/showthread.php?t=339490

Actually, it's a surprising fact that what you suggest doesn't actually work.

 

[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?

 

[Hurkyl]

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

 

[nietzsche]

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

Well...

Assume that f(pi) > 4

<br /> \begin{align*}<br /> f(\pi) &amp;&gt; 4\\<br /> f(\pi) &amp;&gt; f(4)\\<br /> f(\pi-\pi) &amp;&gt; f(4-\pi)\\<br /> f(0) &amp;&gt; f(4-\pi)\\<br /> 0 &amp;&gt; f(4-\pi)<br /> \end{align*}<br />

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.

 

[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?

 

[Hurkyl]

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?

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

 

[jgens]

Your proof can be simplified considerably: since \pi &lt; 4 we know that f(\pi) &lt; 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?

 

[nietzsche]

is the epsilon considered to be a rational number?

 

[nietzsche]

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

 

[jgens]

is the epsilon considered to be a rational number?

No, it doesn't need to be. Sorry for the confusion, I should have been more clear. Let's try again . . .

For definiteness, suppose that f(\alpha) &gt; \alpha. Now since there is a rational number between any real numbers, we can find a rational number a such that f(\alpha) &gt; a &gt; \alpha. Now, can you use this to produce a contradiction?

 

[nietzsche]

No, it doesn't need to be. Sorry for the confusion, I should have been more clear. Let's try again . . .

For definiteness, suppose that f(\alpha) &gt; \alpha. Now since there is a rational number between any real numbers, we can find a rational number a such that f(\alpha) &gt; a &gt; \alpha. Now, can you use this to produce a contradiction?

Ah yes, I do see now. Much appreciated, thank you!

 

[nietzsche]

How can this be used as a contradiction if infinity is not a number?

Also, since we're doing the same question, to prove that f(x) > 0 if x > 0 and that f(x) > f(y) if x > y, would you just prove them by contradiction? (Assuming that x < 0, but since f(x) = x, then x must be greater than 0?)

To prove that f(x) > 0 if x > 0, I said that if x > 0, then it can be written as a square of some number a so that x = a^2. As long as that assumption is true, you should be able to figure it out using the properties of the function.

The problem relies heavily on the information that you gather from the previous parts.

 

[nietzsche]

No, it doesn't need to be. Sorry for the confusion, I should have been more clear. Let's try again . . .

For definiteness, suppose that f(\alpha) &gt; \alpha. Now since there is a rational number between any real numbers, we can find a rational number a such that f(\alpha) &gt; a &gt; \alpha. Now, can you use this to produce a contradiction?

Now, I'm able to prove that f(a) cannot be less than a, nor can f(a) be greater than a. Does this imply that f(a) = a? Assuming that f(a) and a exist, then they must be equal, I think. But is it valid to say that?

 

[nietzsche]

Or how about this?

Since f(a) is not less than a, and f(a) is not greater than a, then there is no rational number q between a and f(a). Therefore, a = f(a). That sounds a bit more valid...

 

[SpringPhysics]

To prove that f(x) > 0 if x > 0, I said that if x > 0, then it can be written as a square of some number a so that x = a^2. As long as that assumption is true, you should be able to figure it out using the properties of the function.

The problem relies heavily on the information that you gather from the previous parts.

I suck at proofs and deducing information. I tried previously to prove that f(x^2) = [f(x)]^2, and then saying that since x^2 is always > 0 and [f(x)]^2 is always > 0. Square-rooting both sides would then give the inequality. I don't know how good of a proof that is though.

 

[nietzsche]

I suck at proofs and deducing information. I tried previously to prove that f(x^2) = [f(x)]^2, and then saying that since x^2 is always > 0 and [f(x)]^2 is always > 0. Square-rooting both sides would then give the inequality. I don't know how good of a proof that is though.

Yup, that's essentially what I did. Don't square root both sides though, because it's not necessarily true because there might be a negative in there. Instead, say that a = x^2, so f(a) = [f(x)]^2. So f(a) must be positive.

 

[SpringPhysics]

Yup, that's essentially what I did. Don't square root both sides though, because it's not necessarily true because there might be a negative in there. Instead, say that a = x^2, so f(a) = [f(x)]^2. So f(a) must be positive.

All that does is assert that x^2 is positive. But x could be negative.

And I'm also confused at the f(delta) > a > delta inequality and contradicting it.

Could you also divide both sides by x and f(x) to get them > 0?

 

[nietzsche]

All that does is assert that x^2 is positive. But x could be negative.

And I'm also confused at the f(delta) > a > delta inequality and contradicting it.

Don't square x. The question says that x > 0. If there is some number a such that a^2 = x, then f(a^2) = f(x). For all positive numbers, x will have a number such that a^2 = x (x has a square root). It's true that a could be negative. But when you show that f(a^2) = [f(a)]^2, then does it still matter that a is negative?

 

[SpringPhysics]

I see then. That elucidates the problem, thank you very much.

Could you also explain the f(alpha) > a > alpha with some hints please? (I would like to try this myself.)

Sorry for all the trouble.

 

[nietzsche]

I see then. That elucidates the problem, thank you very much.

Could you also explain the f(alpha) > a > alpha with some hints please? (I would like to try this myself.)

No problem. Are you working out of the Spivak textbook, by the way? And do you go to University of Toronto? Just curious, because this is part of a problem set, and I already met one person on Physics Forums who's in my class...

Anyway, part (e) of the question says to use the fact that between any numbers there is a rational number. So you have to prove that f(x) = x for all x. In earlier parts of the question, you proved that f(x) = x if x is rational, so now all you have to do is prove that it holds for when x is irrational.

It's a proof by contradiction. You have to assume that if a is an irrational number and q is a rational number, then

f(a) =/= a

This means that you are assuming that f(a) is either > a or < a.

Now this is where you use the fact that between f(a) and a, there will be a rational number.

 

[jgens]

Now, I'm able to prove that f(a) cannot be less than a, nor can f(a) be greater than a. Does this imply that f(a) = a? Assuming that f(a) and a exist, then they must be equal, I think. But is it valid to say that?

Yes. Since f(\alpha) is a real number, we know that one of the following must hold: f(\alpha) &gt; \alpha, f(\alpha) &lt; \alpha, or f(\alpha) = \alpha. Once you've proven that the first two cases cannot hold, the last one follows immediately.

 

[jgens]

It's a proof by contradiction.

I think it's also technically a proof by exhaustion. You prove that every other possible case cannot hold so the remaining case must follow.

 

[SpringPhysics]

No problem. Are you working out of the Spivak textbook, by the way? And do you go to University of Toronto? Just curious, because this is part of a problem set, and I already met one person on Physics Forums who's in my class...

Anyway, part (e) of the question says to use the fact that between any numbers there is a rational number. So you have to prove that f(x) = x for all x. In earlier parts of the question, you proved that f(x) = x if x is rational, so now all you have to do is prove that it holds for when x is irrational.

It's a proof by contradiction. You have to assume that if a is an irrational number and q is a rational number, then

f(a) =/= a

This means that you are assuming that f(a) is either > a or < a.

Now this is where you use the fact that between f(a) and a, there will be a rational number.

Yes, I do. I feel so ashamed to be in that class - everyone else is so smart. >_<

By the way, what is the "q" for?

Yes. Since f(\alpha) is a real number, we know that one of the following must hold: f(\alpha) &gt; \alpha, f(\alpha) &lt; \alpha, or f(\alpha) = \alpha. Once you've proven that the first two cases cannot hold, the last one follows immediately.

Okay, I will try that. Thanks so much to the both of you for your help!

 

[nietzsche]

Yes. Since f(\alpha) is a real number, we know that one of the following must hold: f(\alpha) &gt; \alpha, f(\alpha) &lt; \alpha, or f(\alpha) = \alpha. Once you've proven that the first two cases cannot hold, the last one follows immediately.

Thanks very much for your help.

 

[nietzsche]

Yes, I do. I feel so ashamed to be in that class - everyone else is so smart. >_<

By the way, what is the "q" for?



Okay, I will try that. Thanks so much to the both of you for your help!

q is just any rational number, I could have called it a or b or something else. But I think q is the normal letter they use for rationals, (q for quotient).

 

[SpringPhysics]

I'm really stuck now.

Let f(a) - a = p^2 > 0
Then f[f(a) - a] = f(p^2)
f[f(a)] - f(a) = [f(p)]^2
f[f(a)] = [f(p)][f(p)] + f[a]

And then I'm stuck.

 

[nietzsche]

I'm really stuck now.

Let f(a) - a = p^2 > 0
Then f[f(a) - a] = f(p^2)
f[f(a)] - f(a) = [f(p)]^2
f[f(a)] = [f(p)][f(p)] + f[a]

And then I'm stuck.

what part of the question is this?

 

[SpringPhysics]

I attempted part e), and tried to represent the difference between f(a) and a as another variable that is greater than zero.

 

[nietzsche]

no, you shouldn't need any extra f of f's...like f(f(x)).

you have to assume that f(x) =/= x for the irrational numbers, and look at each case separately. there will be f(x) > x and f(x) < x. that's where the thing about the rational numbers between any two numbers comes in.

 

[SpringPhysics]

no, you shouldn't need any extra f of f's...like f(f(x)).

you have to assume that f(x) =/= x for the irrational numbers, and look at each case separately. there will be f(x) > x and f(x) < x. that's where the thing about the rational numbers between any two numbers comes in.

Can we assume that the difference between f(x) and the rational number a is = the difference between a and the irrational number x?

 

[nietzsche]

you're on the right track, but the difference between them won't matter. try writing out the inequality, and see what you can figure out from that.

 

[jgens]

Some additional help. Let's suppose that f(\alpha) \neq \alpha for some \alpha irrational. Then either f(\alpha) &gt; \alpha or f(\alpha) &lt; \alpha. We'll treat the two cases separately.

Case 1: Let f(\alpha) &gt; \alpha. Since there is a rational number between every two real numbers, we can find a rational number a such that f(\alpha) &gt; a &gt;\alpha. Now, we've already managed to prove that f(x) = x for rational x so we have that f(\alpha) &gt; a = f(a) &gt; \alpha. This means that, for some \alpha and some a we have that a &gt; \alpha implies that f(\alpha) &gt; f(a). Can you see why this is a contradiction? Hint: Look at part (d).

I'll leave the second case to you.

 

[SpringPhysics]

What I got was the following:

if f(x) > f(a)
then f(x) - f(a) = p^2
then f(x-a) = f(p^2)
then x - a = p^2 > 0

which is not possible because a > x

But does that not assume indirectly that f(x) = x?

 

[jgens]

You assume that the difference between x and a is rational which isn't necessarily a valid conclusion. Follow the hint that I gave in the last post.

 

[SpringPhysics]

You assume that the difference between x and a is rational which isn't necessarily a valid conclusion. Follow the hint that I gave in the last post.

I treated part d) as I did part c), so I assumed then that x and y were still rational numbers.

 

[jgens]

Well, if x and a are both rational numbers then there difference is necessarily rational and you haven't shown anything of interest then (especially since you should have already proven that f(x) = x for rational x). Now, if you're trying to do part (e), completing the proof, then follow my hint that I posted above. If you're trying to prove part (d), then the proof is best done using a trick. You know that f(x) > 0 for all x > 0. Now let x = U - V > 0, then f(x) = f(U - V) = f(U) - f(V) > 0.

 

[SpringPhysics]

Well, if x and a are both rational numbers then there difference is necessarily rational and you haven't shown anything of interest then (especially since you should have already proven that f(x) = x for rational x). Now, if you're trying to do part (e), completing the proof, then follow my hint that I posted above. If you're trying to prove part (d), then the proof is best done using a trick. You know that f(x) > 0 for all x > 0. Now let x = U - V > 0, then f(x) = f(U - V) = f(U) - f(V) > 0.

I'm sorry but I really don't follow. So I tried to substitute values in:

f(x) > p/q > x,

In order to show that this is not always the case, I brought the q from p/q to another side of the inequality.

 

[jgens]

Okay, so part (e) . . .

If you read through my hint, I've managed to show that if f(\alpha) &gt; \alpha then for some a &gt; \alpha we have that f(\alpha) &gt; f(a). This is a contradiction since, if x &gt; y then f(x) &gt; f(y) (note, this is just part (d)). It's a really simple proof. Now try the second case.

 

[SpringPhysics]

Okay, so part (e) . . .

If you read through my hint, I've managed to show that if f(\alpha) &gt; \alpha then for some a &gt; \alpha we have that f(\alpha) &gt; f(a). This is a contradiction since, if x &gt; y then f(x) &gt; f(y) (note, this is just part (d)). It's a really simple proof. Now try the second case.

But that is a contradiction in itself because you're already assuming from d) that the formula works for irrational numbers.