# user667648

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 Sep6 comment Is 0.9 repeating = 1 disproved by asymptotes? @MarcvanLeeuwen: Very true. Fortunately, I have yet to really meet any series that either interpretation really matters. Spivak also later mentions that it is a rather small technicality that is of little importance. Sep6 comment Is 0.9 repeating = 1 disproved by asymptotes? LOL! I just commented about this on Hardy's answer! Sep6 comment Is 0.9 repeating = 1 disproved by asymptotes? According to Spivak's Calculus: "This terminology is somewhat peculiar, because at best the symbol $\sum_{n=1}^{\infty} a_n$ denotes a number (so it can't "converge"), and it doesn't denote anything at all unless $\{a_n\}$ is summable. If I am interpreting Spivak correctly, you don't even need to worry about this whole "asymptotes" business, since when we put the = it is just defined that way. But, I could be very easily misunderstanding this. Sep4 comment Is this a valid proof technique regarding the divisibility of numbers? I was thinking induction, but my discrete math skills are bleh, so I just throw things out there and hope something works haha. What ever works, works. Sep4 comment Is this a valid proof technique regarding the divisibility of numbers? @Jlamprong: Now I am lost... Sep4 comment Is this a valid proof technique regarding the divisibility of numbers? @Jlamprong: It says the only numerical solution is 0... I'm just used to seeing True... Sep4 comment Is this a valid proof technique regarding the divisibility of numbers? @Jlamprong: No, you're right? Sep4 comment Is this a valid proof technique regarding the divisibility of numbers? @Jlamprong: Darn, I'm so used to looking at wolfram alpha, and checking to see if there is a "True", that I guess I didn't look close enough. wolframalpha.com/input/?i=2%5E%282*2%5Ek%29+%3D+4%5E%282%5Ek%29 Sep4 comment Is this a valid proof technique regarding the divisibility of numbers? @Herald Hanche-Olsen: Thanks! Sep4 comment Is this a valid proof technique regarding the divisibility of numbers? Um. Where? Is it embarrassing or something? Sep1 comment Help with determining irrationality of a number? Interesting! I have never heard of a Chebyshev polynomial! I will definitely look into this when I get further along in my mathematical career! :D Sep1 comment Help with determining irrationality of a number? @alex.jordan: Hahaha! I was taught that to say something is irrational you write $\alpha \in \mathbb{R} \setminus \mathbb{Q}$. XD Didn't even consider doing it your way. Silly me. :) Sep1 comment Help with determining irrationality of a number? @Winther: Thank you so much for your help! I will accept! Sep1 comment Help with determining irrationality of a number? @Winther: Oh! Oh! I think it get it! So is it since $(-1)^a$ where $a$ is an integer implies that the solution is either $-1$ or $1$, but since we can see that $(1/3 + i\frac{2\sqrt{2}}{3})^b$ is not an integer then this implies that $\alpha$ is not an integer? Is my thinking correct? Sep1 comment Help with determining irrationality of a number? @alex.jordan: Honestly, I thought it had to be real... Sep1 comment May I have a hint for this gcd problem? Yep, that's what I meant. Thanks! Sep1 comment May I have a hint for this gcd problem? Am I correct in my interpretation? Sep1 comment May I have a hint for this gcd problem? Ahh, I actually did try something like this, but forgot you could multiply $(2^b - 1)$ by $2^{a-b}$ too... This makes things clearer. So it's kind of like the identity property $\gcd(a,b) = \gcd(a,b)$, mixed with the fact that $\gcd(a,b)$ is (I believe) an isomorphism to $\gcd(2^a - 1, 2^b-1) = 2^{(a,b)} - 1$, $\therefore \gcd(2^a - 1, 2^b-1) = 2^{(a,b)} - 1$? Sep1 comment May I have a hint for this gcd problem? @ThomasAndrews: The set of notes I'm looking at here (cs.berkeley.edu/~oholtz/H90/integers.pdf) puts "Congruences" after this, so I'm trying to avoid modular arithmetic. (Unless I am mistaken and they are not the same thing, which could very well be the case). But thank you for the suggestion. Aug30 comment Question about proof of Cauchy-Schwarz inequality. I see, the usage of the unit vector is kind of like justification for normalization, ok, ok. Sorry, I didn't stare at it hard enough I guess.