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

[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.

 

[jgens]

You're not understanding what you're proving then. If you prove part (c) and (d) correctly, there's no need to assume that any of the numbers must be rational. You may want to go back and re-examine or re-do the proofs of these facts if you haven't understood them already.

 

[SpringPhysics]

I think I've got it now. Thanks for putting up with my incompetency.