Oliver Kayende
• Member for 2 years, 4 months
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## 135 Answers

Accepted answer
5 votes

This is equivalent to the problem of showing $$\lim_{n\rightarrow\infty}\int_0^1\frac {x^n}{1+x}\;dx = 0$$ Use the compound inequality $$0\leq\int_0^1\frac {x^n}{1+x}\;dx\leq\int_0^1 x^n\;dx =\frac{1}{... View answer Accepted answer 4 votes Yes. Note that (x+a)^p=\sum_{r=0}^p\binom{p}{r}x^ra^{p-r}=x^p+a^p because p divides \binom{p}{r} except when r=0,p. View answer Accepted answer 4 votes Let f(x)=-\cos(x)+x^3+x^2+4x. Note f(0)=-1<0 and f(\frac{\pi}{2})>0. Therefore, by the Intermediate Value Theorem f must admit a zero in [0,\frac{\pi}{2}]. If f had another zero in [... View answer Accepted answer 3 votes The OP claim is false. The singleton subsets \{1\},\{2\},\{3\},\dots are connected in \Bbb R and pairwise disjoint but \Bbb N:=\bigcup_{n=1}^\infty\;\{n\} is not a connected subspace of \Bbb R ... View answer 3 votes \;\;\;\;\; From the OP then I=\Bbb Z\times\{0\}. By the first homomorphism theorem then$$\frac{\Bbb Z\times\Bbb Z}{\Bbb Z\times\{0\}}\approx\Bbb Z$$as \Bbb Z\times\{0\} is the kernel of the ... View answer Accepted answer 3 votes B and \frac{\Bbb Z}{2\Bbb Z} act faithfully on A and \Bbb Z, respectively, via A automorphisms and \Bbb Z automorphisms. The map$$\Psi:\langle f_2\circ f_1\rangle\rtimes \langle f_2\...

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Accepted answer
3 votes

By the fundamental theorem of finite abelian groups we may choose a $G$ subgroup $G_n$ of size $n$. Lagrange's theorem gaurantees $G_n\leq\ker(\varphi_n)$ where $\varphi_n$ denotes the $G$ ...

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3 votes

Consider maybe the the polynomial ring $\Bbb Z[X_1,X_2,\dots]$ with infinitely many indeterminates. This ring is not finitely generated and the only units are $\pm 1$.

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3 votes

$X,Y$ are coprime in $\Bbb Q[X,Y]$. Therefore any generator of $I:=\;${$Xa+Yb | a,b\in\Bbb Q[X,Y]$} would be a common divisor of $X,Y$ and thus constant. Since $I$ is a proper ideal, i.e. $f(0,0) = 0$ ...

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3 votes

Under pointwise addition the family of functions $\Bbb F_2^{X}$ from any set $X$ to the finite field $\Bbb F_2$ of two elements is an abelian group all of whose non-trivial elements have finite order ...

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Accepted answer
2 votes

Define $u:[0,3]\to[0,1]$ by $u(x)=\sin(\frac{x}{2})$. Note $u''(x)^2=\frac{1}{16}\sin^2(\frac{x}{2})$ and thus $u(0)=u''(0)=0$. Finally, $$\Vert u\Vert^2=\int_0^3\sin^2(x)\;dx=16\cdot\int_0^3\frac{1}{... View answer Accepted answer 2 votes$$f(x,y)=y^2-x$$is irreducible in \Bbb F[x][y] because it is a monic quadratic polynomial in y with no roots in \Bbb F(x). Equivalently, f(x,y) is irreducible over \Bbb F[x,y] because$$g(x,...

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Accepted answer
2 votes

No. The size of the quotient need not be a prime power. Consider the field $\Bbb F_{31}:=\frac{\Bbb Z}{31\Bbb Z}$ and denote by $\mathbf x$ the residue class $x+31\Bbb Z$. As presecribed in the OP the ...

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2 votes

Suppose $3<n<p$ and $\Bbb F_p^+$ is the additive group of the finite field $\Bbb F_p:=\frac{\Bbb Z}{p\Bbb Z}=\{\mathbf 0,\mathbf 1\,\dots,\mathbf{p-1}\}$ of size $p$. Let $$x=(u,\mathbf 1)\;\;\;\... View answer 2 votes Let A=\bigcap_{i\in I}A_i. By definition, A\subseteq A_j for each j\in I and thus j\in I\implies A\in\mathcal P(A_j).$$\therefore A\in\bigcap_{i\in I}\mathcal P(A_i)$$View answer Accepted answer 2 votes Because \mathcal Z(G)\subset\mathcal C(a_i)\neq G we must have p | |\mathcal C(a_i)| | p^3. Therefore |\mathcal C(a_i)|=p^2.$$\therefore p^3=p+np\therefore n=p^2-1$$View answer 2 votes f(x)=1-2\cdot\lim_{t\to\infty}\frac{1}{|a+\sin(\pi x)|^t+1}= \begin{cases} -1 & \text{|a+\sin(\pi x)|<1} \\ 0 & \text{|a+\sin(\pi x)|=1}\\ 1 & \text{|a+\sin(\pi x)|>1} \end{... View answer Accepted answer 2 votes From the OP we have G=\Bbb Z_{60}^+ ; G is the additive group of the ring \frac{\Bbb Z}{60\Bbb Z} ; \phi:\bar a\mapsto\overline {3a}. Since G is cyclic, i.e. G=\langle\bar 1\rangle, then \... View answer 2 votes From Jose, xp(x) would have positive degree and thus x(p(x))-1 would be a non-constant polynomial with infinitely many roots which is impossible. View answer 2 votes Let H'=\varphi^{-1}[H] and note \ker(\varphi)=\varphi^{-1}[\{N\}]\subseteq H'. Now, \varphi|_{H'} is a group epimorphism from H' to H with \ker(\varphi|_{H'})=H'\cap\ker(\varphi)=\ker(\... View answer 2 votes \pi:g\mapsto \langle g\rangle defines a map from G-\text{ker}(\psi) onto the family \mathcal F of all cyclic G subgroups of size p^2 where G is an abelian group of exponent p^2 and \psi:... View answer Accepted answer 2 votes \;\;\;\;$$\Bbb F_p^*$is cyclic and of size$p-1$which is divisible by$4$and therefore we may fix$\alpha\in\Bbb F_p^*$of order$4$so that$\alpha^2$has order$2$; i.e.$\alpha^2=\mathbf{-1}$.... View answer 2 votes Except for the common identity element between them the cyclic subgroups$\langle x\rangle,\langle y\rangle,\langle z\rangle$are pairwise disjoint because their sizes are coprime.$\langle x\rangle$... View answer 2 votes The integrand$\sqrt{x^2+y^2}$is symmetric about the line$y=x$yielding $$\iint_{[0,1]\times [0,1]}\sqrt{x^2+y^2}\;dydx=2\int_0^1\int_0^x\sqrt{x^2+y^2}\;dydx=2\int_0^\frac{\pi}{4}\int_0^{\sec(\theta)... View answer Accepted answer 2 votes Firstly, the biconditional statement xax^{-1}\in H\iff a\in H is apparently meant to be read as$$\mathbf B(x,a):(\forall a\in G \;\; xax^{-1}\in H\iff a\in H) $$with \forall unnecessary as G ... View answer 2 votes Let P := <2 > + <x^2+x+1> be the \Bbb Z[x] ideal generated by the polynomial x^2+x+1 and the constant 2.$$\frac {\Bbb Z[x]}{P}\approx\frac{\frac {\Bbb Z[x]}{<2>}}{<x^2+x+... View answer Accepted answer 1 votes Given an$S$accumulation point$z$choose a convergent one-to-one$S\setminus\{z\}$sequence$(\frac{i^{A_n(z)}}{A_n(z)})_{n=1}^\infty$and a monotonic subsequence$(a_n(z))_{n=1}^\infty$of$(A_n(z))...

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1 votes

Proceed by contradiction and assume $(G,\cdot)$ is a doubly transitive permutation group on $\Omega$ but $2\nmid|G|$ where $|\Omega|>1$. By the double transitivity of $G$ we may fix distinct $a,b\... View answer 1 votes Given a finite abelian group$(G,\cdot)$with identity$\mathbf 1$then $$\psi(|G|)=\prod_{j=1}^m{\frak p}(e_j)$$ using the prime factorization$|G|=\prod_{j=1}^mp_j^{e_j}$where$\psi(n)$is the ... View answer 1 votes Denote by$e$the$G$identity and let$n=|\langle x\rangle|=\text{ord}(x)$. (a) becomes clear because $$\ker(\varphi)=\mathcal S:=\{k\in\Bbb Z:x^k=e\}$$ where$\varphi:k\mapsto x^k\$ defines the group ...

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