SMM
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Look at the picture: From $\triangle ABE$ we have $(2+r)^2= 2^2+(2-r)^2$ so $r=1/2$. From $\square ECGF$ we have $CG^2=(1/2+s)^2-(1/2-s)^2= 2s$. From $\square ADGF$ we have $GD^2= (2+s)^2-(2-s)^2= 8s$...

I believe that this hint helps you to establish that $\mathbb Z[X]$ is countable. For a polynomial $f(X)=a_nX^n+\cdots +a_1X+a_0\in\mathbb Z[X]$, $a_n\neq 0$, define the number $h(f)$ by: $$h(f)= n+|... View answer 8 votes Consider a regular 36-gon A_1A_2\ldots A_{36} inscribed in a circle of radius R. Inscribed angle over any side is 5^\circ. We can see our configuration as it is shown on the picture. It ... View answer 8 votes Note that \varepsilon_8^i\sqrt[4]2, for i=1,3,5,7, are roots of this polynomial, where \varepsilon_8=\frac{\sqrt 2}{2}+i\frac{\sqrt 2}{2} is the eighth primitive root of unity. Hence splitting ... View answer Accepted answer 7 votes Consider the picture: Let A and B be the two points, and AC and BD be the desired chords; let AC+BD=a. Assume first that the situation is like on the picture, i.e. C and D are on the ... View answer Accepted answer 7 votes Note that by the definition g*\phi is in F(X,\mathbb C) defined by (g*\phi)(x):= \phi(g^{-1}x), so the calculation is the following:$$\begin{align} (h*(g*\phi))(x)&= (g*\phi)(h^{-1}x)\\ &...

Let $\bar{\mathbb{Q}}$ be the algebraic closure of $\mathbb Q$. Consider a maximal subfield $F$ of $\bar{\mathbb{Q}}$ not containing $\sqrt{2}$; this is straight-forward by Zorn's lemma. Clearly, $F$ ...

If $n_5=6$ and $P$ is a Sylow 5-subgroup, you have $|N_G(P)|=60$. Set $H=N_G(P)$, and we find a subgroup of order $10$ in $H$. Your arguments work. In $H$, $n_5$ is either $1$ or $6$. If $n_5=1$, ...

In a language of groups $\mathcal L=\{\cdot, ^{-1},e\}$ take $\phi$ to be the conjunction of: $\cdot$ is associative; $e$ is neutral for $\cdot$; $x^{-1}$ is inverse of $x$; $\forall x(x\cdot x\cdot ... View answer Accepted answer 6 votes If$c_i$is$i$th column of your second determinant, do$c_n= c_n-c_{n-1}$,$c_{n-1}=c_{n-1}-c_{n-2}$, ...,$c_2=c_2-c_1$to get: $$\left|\begin{array}{ccccccc} 1-x & x & 0 & 0 & \... View answer Accepted answer 5 votes Denote the points like on this picture: Consider the following composition of rotations: I= R_{C',60^\circ}\circ R_{A',60^\circ}\circ R_{B',60^\circ}. The classification of isometries says that I ... View answer Accepted answer 5 votes I am not sure if this answers the question. We can note that the squares of even numbers are on the diagonal of the second quadrant, so if we set:$$\hat n=\max\{2k\mid (2k)^2\leqslant n\},$$or in ... View answer Accepted answer 5 votes If K is cyclic then K\cong\mathbb Z/N\mathbb Z which you can write as \mathbb Z/1\mathbb Z\times\mathbb Z/N\mathbb Z. Assume that K is not cyclic. By the fundamental theorem of finite abelian ... View answer Accepted answer 4 votes First note that if B\subsetneq\mathbb Z_n is not contained in a proper subgroup and contains 0, then B\subsetneq B\oplus B: B\subseteq B\oplus B follows since 0\in B, and B\neq B\oplus B ... View answer 4 votes If N is a random graph (countable) then it is a fact that for any finite partition of N at least one of the members of the partition is a random graph (with induced graph structure). Since there ... View answer Accepted answer 4 votes Note that for real \alpha and positive integer q, \alpha\choose q is defined by:$${\alpha\choose q}=\frac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-(q-1))}{q!}.$$So:$${-p\choose q}= \frac{(-p)(-... View answer Accepted answer 4 votes If$e\neq 0$is a left zero divisor in$A$, then the set of zero divisors contains the left ideal$Ae$. Since this ideal is non-trivial (it contains$e$) and it is not equal to$A$(it doesn't contain ... View answer Accepted answer 4 votes No. You can take inclusion map$\{1,-1\}\longrightarrow \{1,-1,i,-i\}$, considered as subgroups of$\mathbb C^\times$.$S=\{i\}$is a generating set with empty preimage. View answer Accepted answer 3 votes Let$E$be the midpoint of$AC$. Since$EN$is a middle line of$\triangle CAD$, we have$EN\parallel AD$and$EN=AD/2$. Similarly,$EM\parallel CB$and$EM=CB/2$. In particular,$EN:EM=AD:CB$. Since$...

We will prove that this is not possible. Toward a contradiction, assume that $A,B,C,D,E$ are points on the sphere satisfying the condition, and let $O$ be the center of the sphere. Any isometry which ...

Denote by $X-X=\{x-y\mid x,y\in X\}$. Then: $X$ is disjoint from some of its shift iff $X-X\subsetneq\mathbb Z/n\mathbb Z$. Moreover: $X\cap (a+X)=\emptyset$ iff $a\notin X-X$. For $(\Rightarrow)$...

Note that, by the symmetry of the picture, $\angle OQP= \angle OPQ= \angle OPX_4$. Consider $\triangle OQP$ and the circumcircle. We have that the inscribed angle over $OP$, $\angle OQP$, is equal to ...

Maybe you can use the following. Consider a regular 9-gon $I_1I_2\ldots I_9$ with center $O$, and find your picture by setting $A=O$, $B=I_3$ and $C=I_5$. Then note that $M$ is the intersection of ...

If $A',B',C'$ are feet of the altitudes, from $ABA'\sim AHC'$ we have $AH\cdot h_a= AB\cdot AC'$. Similarly, from $BAB'\sim BHC'$, $BH\cdot h_b= AB\cdot BC'$ follows. By adding we have $AH\cdot h_a+... View answer Accepted answer 3 votes As you noted, if$|G:H|=k<p$, by simplicity we can embed$G$into$\mathbb S_k$. But$G$has a Sylow$p$-subgroup of order$p^m$for some$m\geqslant 1$, and$\mathbb S_k$doesn't have subgroups of ... View answer Accepted answer 3 votes Let$f$satisfy the equation. By setting$(x,y)$to be$(t,1)$,$(t+1,1)$and$(t,2)we get: \begin{align} (t-1)f(t+1)&=(t+1)(f(t)-f(1))\\ tf(t+2)&=(t+2)(f(t+1)-f(1))\\ (t-2)f(t+2)&=(t+2)(... View answer Accepted answer 3 votes It seems thatF\times\mathbb Z_4$is a counterexample for any field$F$. Namely,$I=F\times\{0\}$is a maximal regular ideal generated by the idempotent$(1,0)$, but$(0,2)^2=(0,0)\in I$and$(0,2)\...

Assume that $\{g_1,\ldots,g_n\}$ is a generating set for $G$ of the minimal size. If $Q$ is a quotient group $G/H$ for $H\triangleleft G$, note that $\{g_1H,\ldots,g_nH\}$ is a generating set for $Q$. ...
We will prove that such extension doesn't exist. Lemma 1. Equation $X^2=3Y^2+3Z^2$ doesn't have integer solution $(a,b,c)$ such that $a\neq 0$. (In fact, $(0,0,0)$ is the only solution.) Proof. If $(... View answer Accepted answer 3 votes Write: $$0= (A-E)^k= \sum_{i=0}^k{k\choose i}(-1)^iA^{k-i}= \sum_{i=0}^{k-1}{k\choose i}(-1)^iA^{k-i}+ (-1)^kE=$$ $$= A\sum_{i=0}^{k-1}{k\choose i}(-1)^iA^{k-i-1}+ (-1)^kE.$$ It follows that$A^{-1}= (...