The Proof is Trivial!

It feels like you're using the Axiom of Choice here to select your pairwise disjoint open intervals. I'd much rather you didn't :P

A nice way of doing 485, which doesn't involve the axiom of choice, is setting up an equivalence relation.

You used choice in the bolded bit I think.

I don't think axiom of choice is needed. The rationals are unique from the equivalence relation and then the rationals can be well-ordered without the axiom of choice.

I don't think axiom of choice is needed. The rationals are unique from the equivalence relation and then the rationals can be well-ordered without the axiom of choice.

486, just let ln(9-x)=a*ln(3+x)

That was Putnam 1987/B1 if you're interested.

That was a Putnam problem? I'm surprised; I thought they tended to be a fair bit harder. The first couple of questions of the 2014 Putnam paper are quite nice ones if you are interested.

It does seem to be unusually easy. Most Putnam questions are beyond my reach at the moment but I'll see

Also, after unsuccessfully trying to compute jjpneed1's integral, I thought I'd search online and found this solution, which makes me feel better about my failed attempt.

That looks like a more complicated version of mine. How could I forget it equals $\pi /2$ though? What a nice result

Did you do it without complex analysis?

Split the original integral into a sum of two integrals (with same integrand): one from 0 to 1, other from 1 to infinity. Use x->1/x on the second and recombine the integrals. I think I then used integration by parts. From there you have to evaluate some relatively simpler integrals. It took me a while

Problem 487***

$f:\;(0,1)\to (0,1)$ satisfies $f(x)<x$. Let $\hat{f}(x) =\displaystyle \lim_{n\to\infty}\underbrace{f(f(\cdots f}_n(x)\cdots);$ can $\hat{f}$ take uncountably many values?

This seems too easy. yes since there are uncountably infinite real numbers between 0 and 1. Seems so easy I feel like I've missed something, or at least f(x) can take uncountably many values.