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...
Re: First step of this simple "limit" problem
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.
Re: Group.
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.
Re: What are my Grad school prospects?!?
That's a great metaphor. Also, Tobias Funke might be the best forum name I've ever seen. "The Man Inside Me" haha.
Re: What are my Grad school prospects?!?
perhaps as important as grammar and spelling is the ability to know your audience. the internet audience is different from a scholastic audience
Re: What are my Grad school prospects?!?
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.
when they say there is a reflection along the main diagonal they mean that entry i,j = j,i, (i.e. i + j = j + i, the definition of commutivity. it's not just magic!) which you can see is happening here.
For one the property a+(b+c) = (a+b)+c is the associative property not the commutative property. The commutative property is ab = ba. I can't see your picture yet, but the table should look like
01234
12340
23401
34012
40123
Where the entry in the position i,j is i + j (mod 5).
This table is...
consider the family of open sets of the form (1,1/n) where n = 1,2,...
note the definition of compact is that EVERY open cover contains a finite subcover.
edit: oops that should be (1/n,1) of course!
I always wondered what would happen if you substituted values other than integers (and replacing factorials with gamma functions) in cauchy's differentiation formula. Would this give the fractional derivative in the sense you guys are talking about?
Without Sylow's theorems!!
This was a problem at the end of a chapter on Lagrange's theorem. I know that every subgroup of order 77 is cyclic. But I don't know how to prove this using only Lagrange. Any suggestions?
you know that (p-2)! = 1 (mod p). So (p-3)!*(p-2) = 1 (mod p). In this situation, 34!*35 = 1 (mod 37). Call 34! 'x' and then solve 35x = 1 mod 37, which has a unique solution since gcd(35,37) = 1.
4n+/-1
6n+/-1 is composite iff there are nonzero integers a and b such that n = 6ab + a + b.
for instance 6(4) + 1 is composite since 4 = 6(-1)(-1) + (-1) + (- 1)
to take the laplace transform of the heaviside function, you want it to be of the form f(t)H(t). Since 2tH(t-1) isn't of this form, the author uses some simple algebra to get something that is.