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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
26
comment Showing path connected matrices of a group $G$ is a normal subgroup
In the first part, to allow for the inverse of $B$, do I need $\tau^{-1}$ to be continuous too or is it enough to say $\tau$ is continuous? Is this equivalent to saying if $B\in H$, we want to show $B^{-1}$ is in $H$? And you don't need to edit the post just to fix $I$, it's clear now.
Jun
17
comment Open sets in $\Bbb{R}^2$ and $\Bbb{R}^3$
So the unit intveral on the x-axis wouldn't be open in $\Bbb{R}^2$ right?
Jun
17
comment Open sets in $\Bbb{R}^2$ and $\Bbb{R}^3$
@Berci Thanks for that
Jun
17
comment Open sets in $\Bbb{R}^2$ and $\Bbb{R}^3$
Yes, but it is still a union of open balls. So in some sense, are open balls the basis of all open sets in $\Bbb{R}^2$?
Jun
17
comment Topologist's sine curve is connected
Doesn't disconnected mean that there are two disjoint open sets $A$ and $B$ such that the entire set $S$ is equal to $A\cup B$? But when you take your definitions of $A$ and $B$, the union of them is more than what is needed, no?
Jun
9
comment Can a group of order $55$ have exactly $20$ elements of order $11$?
How do you get that there would be 34 elements remaining with order 5?
Jun
9
comment Showing a compact metric space has a countable dense subset
I'm struggling with why $d(x,y)$ has to be less than $1/n$. If the definition of a dense set just requires that another point of $D$ lie in some open neighbourhood of $x \in X$, couldn't it be possible that $y$ is still in $B(x,\epsilon)$ but that the distance between the two points is not less than $1/n$?
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
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
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
comment Showing existence of an element with order $p$
Do you mind explaining further? I'm still not sure what to do.
May
27
comment Proving $\sum_{m=0}^M \binom{m+k}{k} = \binom{k+M+1}{k+1}$
but $r$ is indexed
May
27
comment Proving $\sum_{m=0}^M \binom{m+k}{k} = \binom{k+M+1}{k+1}$
In my text, I have an identity $\sum_{r\geq 0} \binom{r + n}{r} x^r = 1/(1-x)^{n+1}$ This may be the cause of my confusion, but is this identity correct and is it equivalent to the one you used?