Recent content by pzzldstudent

Cool, thanks a lot! I think I got it now. If not, I'll post again here. Thanks for all your help thus far!

Where's the radical? I'm still not quite "seeing" it. Where does the radical 'nth root' go? I understand the comparison of 2^n+3^n<=3^n+3^n+3^n=3*3^n=3^(n+1), but I'm still struggling with how to relate it to the problem that has (2^{n} + 3^{n})^{1/n}. Where are you putting the (1/n)...

I've forgotten how to take logs and limits of logs. Is it log 2^(1/n) can be rewritten as (1/n) = log_{2} x? How do I find the limit from there? Take ln?

Will this work? 3 \leq\sqrt[n]{2^{n}+3^{n}} \leq 3\sqrt[n]{2} How do I know the limit of 3\sqrt[n]{2} is 3? So is the limit of 3^{n} = 3 = limit of 3^{n} + 3^{n}. But I am taking \sqrt[n]{2^{n}+3^{n}} and not just {2^{n}+3^{n}}

Thanks for looking at this thread. We haven't gotten to the Binomial Theorem yet, and I don't think our professor will be emphasizing it that much, so I think that if I use that theorem, it would appear unrealistic for me to do that since we haven't covered it yet. Thanks for the input...

We were given the hint to try using the Squeeze Theorem in order to find the limit as n → ∞ of the sequence {(2^n + 3^n)^(1/n)}. I understand the concept of the squeeze theorem that I need to find functions A and B such that A ≤ {(2^n + 3^n)^(1/n)} ≤ B, and A and B limit to the same...

by george, i think i got it! :D With more help I think I got it: X~ Y implies that there is a function f: X → Y that is 1-1 and onto. Y ~ Z implies that there is a function g: Y → Z that is 1-1 and onto. Consider the function h: X → Z where h = f [SIZE="1"]o g. To show 1-1: Take a...

Okay, here's what I got so far: X~ Y implies that there is a function f: X → Y that is 1-1 and onto. Y ~ Z implies that there is a function g: Y → Z that is 1-1 and onto. I need to prove X ~ Z which implies there is a function h: X → Z that is 1-1 and onto. To show 1-1: Take a and b in...

Thanks for the feedback. I could also go the other direction, right? Since A and B are countable, that also means N ~ A and N ~ B. Which means for functions f and g where f: N → A and g: N → B, the functions are both 1-1 and onto. There are bijections from N to A and N to B. So I can show there...

Suppose X, Y, Z are sets. If X ~ Y and Y ~ Z, prove that X ~ Z. My work on the proof so far is: Suppose X, Y, Z are sets. Let X ~ Y and Y ~ Z. By equivalence, there are functions f and g such that f: X → Y where f is 1-1 and onto, and g: Y → Z where g is 1-1 and onto. So now I have to...

Awesome, thanks again! :cool:

Thanks for the encouraging note :) My professor did give a list of tips that to prove something uncountable we'd need a contradiction and so we would first assume it countable. Thanks to everyone's helps/tips, I'm slowly getting the hang of this. This forum rocks and is a tremendous lifesaver...

*light goes on in head* I think I see it now: Okay, I think I got it (thanks to your help!): My proof would be: Let A be a countable subset of an uncountable set X. Assume X \ A is countable. A union X \ A = X which is countable because the union of two countable sets is countable. BUT...

Oh, I'm afraid I don't know what that is since my professor did not cover that. All that was mentioned was some diagonalizing, but we're not going into that this semester. Thanks for your time though.

Statement to prove: If A is a countable subset of an uncountable set X, prove that X \ A (or "X remove A" or "X - A") is uncountable. My work so far: Let A be a countable subset of an uncountable set X. (N denotes the set of all naturals) So A is equivalent to N (or "A ~ N") by the...