I don't understand very well the quotient, so, I do not know if calculate is the correct word, but I need to simplify the expression $$ \mathbb{Z}[x] /\langle2x-1\rangle  $$
50951
Suppose it asks to show $\mathbb{Z}[x]/(2x-1) \cong \mathbb{Z}[\frac{1}{2}]$  Cand I do like this ? First of all elements of $\mathbb{Z}[x]/(2x-1)$ is of form $\frac{m}{2^n}+(2x-1)$ and also all elements of $\mathbb{Z}[\frac{1}{2}]$ are of form $\frac{m}{2^n}$. Now define a normal projection $\mathbb{Z}[\frac{1}{2}] \to \mathbb{Z}[x]/(2x-1)$ where $a \to a+(2x-1)$ for all $a\in \mathbb{Z}[\frac{1}{2}]$. So kernal is srt of all $a$ such that $2x-1|a \implies a=0$ so both are isomorphic.
28006
Problem: Show that the group given by the presentation $$\langle x,y,z  \mid xyx^{-1}y^{-2}\, , \, yzy^{-1}z^{-2}\, , \, zxz^{-1}x^{-2} \rangle $$ is equivalent to the trivial group.   I have tried all sorts of manners to try to show that the relations given by the presentation above imply that $x=y=z=e$. However, I am stuck and would appreciate any hints as to how I should move forward.
85857
I tried the following $$I = \langle X^2,X+1\rangle =\langle X^2,X+1,X^2+2(X+1)\rangle =\langle X^2,X+1,(X+1)^2+1 \rangle$$  Yet no matter how I arrange it, I cannot obtain $1$. Can someone help me out?
114710
Find the quotient and remainder when $6x^4-11x^3+5x^2-7x+9$ is divided by $(2x-3)$. I expressed the divisor $(2x-3)$ as $2(x-\frac{3}{2})$ and conducted synthetic division by $\frac{3}{2}$ and obtained the coefficients of the quotients as $6, -2, 2$ and $-4$ and the remainder as $3$. Then I divided the quotients by $2$. Why should I do this? and why should I not divide the remainder by $2$?
19753
I'm wondering whether there any nice identities (or relationships) that can simplify or possibly compact the following expressions:  $$\operatorname{Li}_2(\beta e^{\alpha x}) - \operatorname{Li}_2(\beta e^{-\alpha x}) $$  and  $$\operatorname{Li}_2(pe^{\alpha x}) + \operatorname{Li}_2(qe^{\alpha x}) $$
11328
Express $-1+i$ in exponential form.  My attempt so far  Let $z=-1+i$  $$r=|z|=\sqrt2$$  $$\theta=\tan^{-1}(-1)=-\frac{\pi}{4}$$  Now, this is where I go wrong (I don't know why it's wrong!):  So in exponential form: $-1+i=\sqrt2 e^{-i\pi/4}$  According to the solutions, $\theta=3\pi/4$.  Thanks in advance
71588
How to deal with combinatoric interpretation (or just solving it in algebraic way) when we have $(-1)^i$ factor in our sum?  Example task:  Simplify the sum: $$ \sum_{i=0}^{k}(-1)^i i \binom{n}{i} \binom{n}{k-i} \text{ for } 0\le k \le n $$  For task without $(-1)^i$  $$ \sum_{i=0}^{k} i \binom{n}{i} \binom{n}{k-i} = n \binom{2 n-1}{k-1} $$ I can write that interpretation:    I have $n$ rabbits and $k$ slots Each rabbit can be in both slot of first type and second type slots of first type + second type = $k$ Lets double rabbits I choose one rabbit as an king and it will be also a rabbit to slot of first type so I need to choose $2n-1$ rabbit for $k-1$ slots  But I don't know how to deal with $(-1)^i$
120382
I am trying to complete the square on $xy - xz + yz = 1$.  I have tried plugging in $u = x + y$. But I still end up with unsolvable cross terms.
78816
I need to express $z = 4e^{-i\pi/3}$ in the form of $x + yi$ and represent it on the Argand diagram.  I think that $4 = \sqrt{x^{2} + y^{2}}$ and that $\tan (\pi/3) = y/x$ but I haven't been able to do anything useful with this information...  Is this solvable via simultaneous equations?  Thanks!
36141
Show that $(\mathbb{Z}[x]/(x^{n+1}))^{\times}\cong \mathbb{Z}/2\mathbb{Z}\times\Pi_{i=1}^n\mathbb{Z}$.   Anyway how and what method is used to prove this. I still have no idea now. really thanks for your help
28394
Consider the quotient $\mathbb{Z}-\text{module}$ $$M=\frac{\mathbb{Z}^3}{\langle (3,3,1),(2,2,2)\rangle}$$ Prove that $M$ is isomorphic to $\mathbb{Z} \oplus \mathbb{Z}_4$.  I think the first step is to find a basis of $M$. Clearly $(1,0,0)$ and $(0,0,1)$ are in a basis of $M$. I don't know how to find the third element in that basis. Also the basis in $\mathbb{Z} \oplus \mathbb{Z}_4$ are $(1,0)$ and $(0,1)$.
4165
As stated in the title, I'm trying to simplify the following expression: $xy + xy'z + x'yz'$ I've only gotten as far as step 3:  $xy + xy'z + x'yz'$  $=x(y+y’z) + x’(yz’)$  $=x(y+y’z)+x(y’+z)$  But I don't know where to go from this step, I'm not sure if I'm allowe to rewrite y'z as y+z' (I'm not even sure if that would help)
95635
I want to simplify the following:  $\sum_{n=0}^x 1 - (1 - I[a \ne 0] * I[a \ge b]) * (1 - I[b \ne 0] * I[b \ge c]) * (1 - I[c \ne 0] * I[c \ge d]) * (1 - I[d = 0])$
34362
Can someone explain simply what a quotient group is?  I've read a lot of convoluted and unnecessarily tough descriptions on it but it seems like a really simple idea.  Question Show that $$8 \mathbb{Z} / 56 \mathbb{Z} \simeq \mathbb{Z}_7$$  I'm sure this is a really easy question if I can just realize what a quotient group is. If you got an easy way to describe isomorphism, that would be great too!
46658
Consider the quotient ring $R:=\mathbb{Z}[x]/(2x)$. Could somebody help me to identify $R$ with some well-known ring?    Thank You.
28582
Hi I don't know how to show that $\langle 2,x \rangle$ is not principal and the definition of a principal ideal is unclear to me. I need help on this, please.   The ring that I am talking about is $\mathbb{Z}[x]$ so $\langle 2,x \rangle$ refers to $2g(x) + xf(x)$ where $g(x)$, $f(x)$ belongs to $\mathbb{Z}[x]$.
70385
The reasoning my book uses is as follows:   Since $(3+i)(3-i) = 10$, $10 + \langle 3+i\rangle = 0 + \langle 3+i\rangle $. Also, $i + \langle 3+i\rangle = -3 + \langle 3+i\rangle = 7 + \langle 3+i\rangle$. So, $\mathbb{Z}[i]/\langle 3+i\rangle = \{k + \langle 3+i\rangle \mid k = 0,1,\ldots ,9\}$, since $1 + \langle3+i\rangle$ has additive order $10$.   I do not understand why the additive order of $1 + \langle 3+i\rangle$ is $10$, or why the fact that $(3+i)(3-i) = 10$, shows the additive order or $1 + \langle 3+i\rangle$  is $10$. If someone could please explain this I would greatly appreciate it.   Note: I apologize for not using MathJax to write this post. I intend to learn it over the coming weekend.