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 Oct15 comment Clarification for proof of $\mathbb{Q}$ being dense in $\mathbb{R}$ (Rudin's PMA) @ThomasAndrews And if it is the $nx$ obtained from $n(y-x)>1$, how does one know to focus on $nx$ and not $ny$? Oct14 comment Clarification for proof of $\mathbb{Q}$ being dense in $\mathbb{R}$ (Rudin's PMA) @ThomasAndrews When you first mention $nx$, are you referring to the $nx$ obtained from $n(y-x)>1$? Oct14 comment Baby Rudin Theorem 1.20 (b) Proof @BrianM.Scott Why doesn't the set begin with "-m_2"? Jun26 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. Jun17 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? Jun17 comment Open sets in $\Bbb{R}^2$ and $\Bbb{R}^3$ @Berci Thanks for that Jun17 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 $\Bbb{R}^2$? Jun17 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? Jun9 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? Jun9 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 $1/n$? Jun8 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? Jun7 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. Jun7 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 Jun3 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? Jun3 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? Jun3 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? Jun2 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? Jun2 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. Jun2 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 $p^1, \dots, p^n$. Jun2 comment Find all positive integers $x$ such that $13 \mid (x^2 + 1)$ Is there a general form of the first sentence you stated?