"The notation T for the circle group stems from the fact that, with the standard topology (see below), the circle group is a 1-torus. More generally Tn (the direct product of T with itself n times) is geometrically an n-torus."
Does it make sense to call points in projective space $RP^n$ projectively independent if the corresponding linear subspaces are transverse, i.e. their sum is $R^{n-1}$?
@AkivaWeinberger Thanks. In a paper I read the statement "the $m$ points in projective space $RP^{m-1}$ span an $(m-1)$ simplex in $RP^{m-1}$". So I guess I can translate this as "the corresponding linear subspaces in $R^m$ are transverse".
which makes no sense as the workings will mean G/N is always abelian (since I can always summond a conjugate in the workings to cancel out a left coset and convert it to a right coset)
A student is asked to show taht if $H$ is a normal subgroup of an abelian group $G$, then $G/H$ is abelian. The student's proof starts as follows: We must show that $G/H$ is abelian. Let $a$ and $b$ be two elements of $G/H$. a. Why does the instructor reading this proof expect to find nonsense from here on in the student's paper?
Please help me understand the bold part of this question: Find the value of a for which $(ax^2+3x-4)/(a+3x-4x^2)$ takes all values for all real values of x. I don't want the solution.
@AkivaWeinberger Yes, it does. Leaky confused me with strange wordings. I understood this part exactly as you stated then I became unsure after Leaky's explanation.
So $\dfrac{ax^2+3x-4}{a+3x-4x^2}=y$ always has a solution. This isn't quite a quadratic, so I can't take the discriminant of it yet, but I think we could manipulate it into a quadratic or something?
Let $x_1+x_2+\cdots+x_n=m$ then minimize the function
$f(x_1,x_2,\cdots,x_n)=\sum_{i=1}^n (x_i)^{\alpha}$
where $x_i,m,n$ are positive integers and $\alpha>1$.
My attempt: I applied the Lagrange's multiplier and found that minimum is obtained when all $x_i's$ are equal, but I am unable to pro...
An interesting question to ask in general is that: For a countable, linearly ordered set $S$ and some countable subset $X \subset S$, what property $X$ must have such that $S-X < \aleph_0$
Let $A,B$ be infinite sets with $A \subset B$. Then $|B-A|<\aleph_0$ if $A$ is cofinite, $|B-A|=\aleph_0$ if $A$ is cocountable and $|B-A|=\aleph_{\alpha}$ if error, circular definition
anyhow. I just finished an exercise that took me 2 seconds, while this previous one took me 2 hours:l how am I supposed to make a schedule, when the time it takes to do an exercise can be so arbitrary.
Guys. Let $G_1,G_2$ be groups. I have to show that $(G_1\times G_2)/H_1\cong G_2$, where $H_1=G_1\times\{e_2\}$. The exercise explicitly says that I need to use the following theorem to show this: Let $G$ be group, and $N\subset G$ a normal subgroup, $H\subset G$ a subgroup. Then $H/(H\cap N)\cong HN/N$.
My attempt: Let $N=G_1\times\{e_2\}$ and $H=G_1\times G_2$. This would give us $(G_1\times G_2)/(G_1\times\{e_2\})\cong HN/G_1\times\{e_2\}$. Now it seems to me that $HN=G_1\times G_2$, so we get a trivial result. I don't see how I can use this theorem to show it (I've already shown it wit…
Let $G$ be a group, and let $N$ be a normal subgroup of $G$. Let $G/N$ and $N$ be abelian. Show that for each subgroup $H$ of $G$, there exists a normal subgroup $N'$ of $H$ such that $H/N'$ and $N'$ are abelian. Well, obviously, $N'$ is going to be $H\cap N$. So $H\cap N$ is abelian.
All I need to show is that $H/(H\cap N)$ is abelian. One thing I thought I might use is that $G/N$ is abelian implies that $[G,G]\subset N$. So I'm guessing I need to show that $[H,H]\subset H\cap N$. Maybe I could show that $H\cap[G,G]=[H,H]$? I'm not sure.
Let $x_1+x_2+\cdots+x_n=m$ then minimize the function
$f(x_1,x_2,\cdots,x_n)=\sum_{i=1}^n (x_i)^{\alpha}$
where $x_i,m,n$ are positive integers and $\alpha>1$.
My attempt: I applied the Lagrange's multiplier and found that minimum is obtained when all $x_i's$ are equal, but I am unable to pro...
