Recent content by NoName3

I'm told that graph of the surface $g(x,y,z) = x^2+y^2+4z^2 = 1$ looks like: Is that correct? And if so, I've the following question. When considering the slices the graph of $g(x,y, z)$ is a circle in the $xy$ plane and an ellipse in the $xz$ plane and $zy$ plane. The circle is bigger than...

Hi, I like Serena. ;) Thanks for the reply. It was supposed to be $\displaystyle n_U=\displaystyle \frac{-x \vec{i}-y \vec{j}+\sqrt{x^2+y^2+1} \vec{k}}{\sqrt{x^2+y^2+2}}$ They probably misplaced the $2$ then. Why is the answer to $n_D$ a constant vector? EDIT: I see because $z =2$. I've...

Let the surface $S \subset \mathbb{R}^3$ be the graph of the function $f: \mathbb{R}^2 \to \mathbb{R}$, $f(x,y)= \sqrt{x^2+y^2+1}$. Let $U$ be the finite region of $S$ below the plane $z = 2$ and let $D$ be the finite region of the plane $z = 2$ for which $x^2 + y^2 \le 3$. Let $n_U (x, y, z)$...

Let $R$ be a commutative ring and let $\text{M}_2(R)$ denote the ring of $2 \times 2$ matrices with coefficients in $R$. (a) Show that the group of units in $\text{M}_2(R)$ is $\text{GL}_2(R) = \left\{A \in \text{M}_2(R): \text{det}(A) \in R^{\times} \right\}$; (b) show tha $\text{GL}_2(R)...

Thank you, very helpful!

Actually my proof above is not valid for $\epsilon \le \frac{1}{2}$. Let $x_n = \dfrac{(-1)^n(n-8)}{4n}.$ Clearly a lower bound for $x_n$ is $-\dfrac{3}{4}$. We need to show that $-\dfrac{3}{4}+\epsilon$ is not a lower bound for $\epsilon > 0$. If $\epsilon > 0$, there exists an even integer...

It appears that I was having an off-day here. The $\epsilon-\delta$ proof is far simpler than that. Let $x_n = \dfrac{(-1)^n(n-8)}{4n}.$ A lower bound for $x_n$ is $-\dfrac{3}{4}$. We want to show that $-\dfrac{3}{4}+\epsilon$ is not a lower bound for any $\epsilon > 0$. If $\epsilon > 0$, then...

I want to know whether the following the counts as a proof that infimum of the set $S = \left\{2(-1)^n+\frac{5}{n^2+2}: n \in \mathbb{N}^{+} \right\}$ is $\text{inf}(S) = -2$. Let $A \subseteq X$, where $X$ is some ordered field. Then $\text{inf}(A)$ is $m \in X$ such that for any $x \in A$...

Thanks, mathmari. Since the equality is never attained, we have $x_j < x_{j+1}$ thus $x_j$ is strictly increasing from $-3/4$ to $1/4$. Letting $i= 2k+1$ we have $x_i = \dfrac{8-i}{4i}$ and $x_{i+1} = \dfrac{7-i}{4(i+1)}$. Therefore $x_{i+1} > x_i$. Hence $x_i$ is decreasing from $7/4$ to...

Find $\text{glb}(A)$ if $A = \left\{(-1)^n \left(\frac{1}{4}-\frac{2}{n} \right): n \in \mathbb{N}\right\}$.$\displaystyle x_n = (-1)^n \left(\frac{1}{4}-\frac{2}{n} \right)$ then $ \displaystyle x_{2k} = \frac{1}{4}-\dfrac{1}{k} = \frac{k-4}{4k}$ and $ \displaystyle x_{2k+1} =...

So $\mathbb Z^{\times}_{20} \simeq \mathbb Z_{2} \times \mathbb Z^\times_{5}$? As for why, two groups of the same order are isomorphic if $\gcd(n, \phi(n)) = 1$. This satisfies that.

Hi, I like Serena, Thanks for the reply. Yes, I think they're isomorphic.

How do I show that $\mathbb{Z}^{\times} _{20} ≅ \mathbb{Z}_{2} \times \mathbb{Z}_{4}$? I read that the chinese remainder theorem is the way to go but there are many versions and I can't find the right one. Most versions that I have found are statements between multiplicative groups, not from...

Thank you. So let $x = h_1g$. Then $x = h_1hg' \in H$ because since $h_1, h \in H$ we have $h_1 h \in H$. Therefore $Hg \subseteq Hg'$. For the converse, suppose $h_2 \in H$ and let $y = h_2 g' = h_2h^{-1}g \in H$ because since $h_2 , h^{-1} \in H$ we have $h_2 h^{-1} \in H$. Therefore $Hg'...

Let $H$ be be a subgroup of a group $G$. Let $g'$ and $g$ elements of $G$. Prove that the following are equivalent: $(a)$ $g' \in Hg$, $(b)$ $g'g^{-1} \in H$, and $(c)$ $Hg = Hg'$. $g' \in Hg$ means $g' = hg$ for some $g \in G$ and $h \in H$. And $g' = hg \implies g'g^{-1} = hgg^{-1} = h$. But...