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Sep
4
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
Sep
4
comment Is this a valid proof technique regarding the divisibility of numbers?
@Herald Hanche-Olsen: Thanks!
Sep
4
comment Is this a valid proof technique regarding the divisibility of numbers?
Um. Where? Is it embarrassing or something?
Sep
4
asked Is this a valid proof technique regarding the divisibility of numbers?
Sep
4
revised Verify that $\sqrt{2}\left\| z \right\| \ge \left|\Re(z)\right| + \left|\Im(z)\right|$
edited body
Sep
4
answered Verify that $\sqrt{2}\left\| z \right\| \ge \left|\Re(z)\right| + \left|\Im(z)\right|$
Sep
1
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
Sep
1
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. :)
Sep
1
accepted Help with determining irrationality of a number?
Sep
1
comment Help with determining irrationality of a number?
@Winther: Thank you so much for your help! I will accept!
Sep
1
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?
Sep
1
comment Help with determining irrationality of a number?
@alex.jordan: Honestly, I thought it had to be real...
Sep
1
asked Help with determining irrationality of a number?
Sep
1
comment May I have a hint for this gcd problem?
Yep, that's what I meant. Thanks!
Sep
1
accepted May I have a hint for this gcd problem?
Sep
1
comment May I have a hint for this gcd problem?
Am I correct in my interpretation?
Sep
1
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 $\gcd(2^a - 1, 2^b-1) = 2^{(a,b)} - 1$, $\therefore \gcd(2^a - 1, 2^b-1) = 2^{(a,b)} - 1$?
Sep
1
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.
Sep
1
asked May I have a hint for this gcd problem?
Aug
30
accepted Question about proof of Cauchy-Schwarz inequality.