Recent content by matticus

are the entries positive integers?

in your example you have completely ignored prime factorization. suppose we write a number as a list of exponents n=(e1,e2,e3,...) where only finitely many ei's are non-zero, and so that e1 is the exponent of 2, e2 is the exponent of 3, and en is the exponent of the nth prime. in your example...

l'hospital's rule applies nicely. if you haven't learned that, then just factor the numerator and denominator.

say you go to the casino every week with $127 and play a game that pays 1:1. You start by betting a dollar, double your bet with every loss, and start back at a dollar with every win. So, you would need to win 127 times without hitting a streak of seven losses. What would your probability of...

The world series would be pretty boring if there was only one team there :) Also, there is a 0% chance there's two teams from LA because the yankees are going.

Suppose there is another prime q that divides the order of the group and show there must be an element of order q.

a number cannot be closed. sets are closed if they contain all their limit points. the set you list is not closed, since 12 is a limit point and 12 is not in the set.

That's a great metaphor. Also, Tobias Funke might be the best forum name I've ever seen. "The Man Inside Me" haha.

perhaps as important as grammar and spelling is the ability to know your audience. the internet audience is different from a scholastic audience

vanadium is right. while there is a vocabulary portion on the GRE test, the admissions board doesn't actually look at it. rather, they go back and check all your old forum posts and make sure you capitalized your 'i's.

can you be more clear on how you got this? the inductive step as far as a i can see would be: suppose (xy)^{3k} = x^{3k}y^{3k}. then (xy)^{3(k+1)}= x^{3k}y^{3k}x^{3}y^{3}. but then what... edit: wait i think i got it x^{3k}y^{3k}x^{3}y^{3} =...

if G is a group such that (xy)^{3} = x^{3}y^{3} for all x,y in G, and if 3 does not divide the order of G, then G is abelian. I proved an earlier result that said if there exists an n such that (xy)^{n} = x^{n}y^{n} (xy)^{n+1} = x^{n+1}y^{n+1} (xy)^{n+2} = x^{n+2}y^{n+2} for all x,y in G...

ug...nothing is worse than when I'm reading an algebra book and they say "oh yeah I took algebra in high school." even people that know I've taken a lot of math will say it. do they think I've been studying math at school for years and am still at the same place they were in high school? i...

I am just looking at this again after a long while, does this work? Let G be a group of order 77. If G is cyclic we're done. If not, then either the hypothesis holds or all elements have order 7 or they all have order 11 (excluding the identity). Supposing all non-identity elements have order...

lol. suppose no one was at the party. then all of the hypothesis hold vacuously. i don't claim that this is a unique solution, and most likely not the intended one.