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 Sep24 awarded Autobiographer 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 accepted Is this a valid proof technique regarding the divisibility of numbers? 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? Sep4 asked Is this a valid proof technique regarding the divisibility of numbers? Sep4 revised Verify that $\sqrt{2}\left\| z \right\| \ge \left|\Re(z)\right| + \left|\Im(z)\right|$ edited body Sep4 answered Verify that $\sqrt{2}\left\| z \right\| \ge \left|\Re(z)\right| + \left|\Im(z)\right|$ 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 accepted Help with determining irrationality of a number? 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?