1,099 reputation
625
bio website math.stackexchange.com/users/…
location Antarctica
age 78
visits member for 1 year, 9 months
seen Jan 31 at 5:07

Self-learning mathematics. Many thanks to those who make time to help others on the site.


Jun
8
comment What can we say about the order of a group given the order of two elements?
To generalize, do I just take all divisors of $n$? If every nontrivial element has the same order, then it must be a $p$-group because the only divisors of $p$ are $1$ and $p$ itself (this means the group $G$ in my problem is certainly not a $p$-group). Is this correct?
Jun
8
asked Showing a compact metric space has a countable dense subset
Jun
7
comment What can we say about the order of a group given the order of two elements?
could we say it has an element of order 2, 3, and 5? $x^6 = (x^2)^3 = (x^3)^2$, and similarly for $x^{10}$. I know that doesn't say much about $G$, but I'm just trying to figure out what more can be said.
Jun
7
comment What can we say about the order of a group given the order of two elements?
@MartinArgerami Aren't they saying the same thing? Sorry, I was trying to be more formal about it. Thanks for pointing it out
Jun
7
revised What can we say about the order of a group given the order of two elements?
added 26 characters in body
Jun
7
asked What can we say about the order of a group given the order of two elements?
Jun
5
revised Central Limit Theorem Problem
changing title, adding tag
Jun
5
suggested suggested edit on Central Limit Theorem Problem
Jun
3
comment $f(x) = \frac {e^{2x-1}} {(1+e^{2x-1})}. $ What is the value of $ f(1/2009) + f(2/2009) + … + f(2008/2009) $?
@CalvinLin If you don't mind me asking, what is the difference between brilliant.org and sites such as art of problem solving?
Jun
3
comment solve system of linear congruences mod 13
Ah, okay. Thanks. Btw, when you multiply $-5$ to equation $I$, did you mean to write $-5$ on the RHS?
Jun
3
comment solve system of linear congruences mod 13
In my textbook, multiplication by a scalar m on $a \equiv b \pmod{c}$ gives $am \equiv bm \pmod{mc}$. Why is it that you don't multiply the modulus number $13 $in the beginning of your solution?
Jun
2
comment Showing existence of an element with order $p$
It's because the order of an element of a group has to divide the order of the group. The only possible divisors of $p^n$ are $1,p^2,\cdots, p^n$. The only element that has order 1 is the identity $e$. So we just pick an element not equal to $e$. Is this correct?
Jun
2
comment Has anyone studied this operator?
@KeyIdeas I had read one of your posts about Gauss's Algorithms I was wondering if it was called something else officially. I'd simply like to read up on it, but I can't find its wikipedia page.
Jun
2
comment Showing existence of an element with order $p$
Sorry, I was referring to how I need to pick that $a$ such that the order is one of $p^1, \dots, p^n$.
Jun
2
comment Find all positive integers $x$ such that $13 \mid (x^2 + 1)$
Is there a general form of the first sentence you stated?
Jun
2
accepted Find all positive integers $x$ such that $13 \mid (x^2 + 1)$
Jun
2
revised Find all positive integers $x$ such that $13 \mid (x^2 + 1)$
added 39 characters in body
Jun
2
asked Find all positive integers $x$ such that $13 \mid (x^2 + 1)$
Jun
2
comment Showing existence of an element with order $p$
Do you mind explaining further? I'm still not sure what to do.
Jun
2
accepted Showing existence of an element with order $p$