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

[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 $f(\alpha) > \alpha$ or $f(\alpha) < \alpha$. We'll treat the two cases separately.

Case 1: Let $f(\alpha) > \alpha$. Since there is a rational number between every two real numbers, we can find a rational number $a$ such that $f(\alpha) > a >\alpha$. Now, we've already managed to prove that $f(x) = x$ for rational $x$ so we have that $f(\alpha) > a = f(a) > \alpha$. This means that, for some $\alpha$ and some $a$ we have that $a > \alpha$ implies that $f(\alpha) > f(a)$. 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 $f(\alpha) > \alpha$ then for some $a > \alpha$ we have that $f(\alpha) > f(a)$. This is a contradiction since, if $x > y$ then $f(x) > f(y)$ (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 $f(\alpha) > \alpha$ then for some $a > \alpha$ we have that $f(\alpha) > f(a)$. This is a contradiction since, if $x > y$ then $f(x) > f(y)$ (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.