azimut
• Member for 8 years, 11 months
• Last seen this week

As a child, the Fibonacci numbers $$1,\; 1,\; 2,\; 3,\; 5,\; 8,\; 13,\; 21,\; 34,\; 55,\;\ldots$$ were very fascinating to me. They are named after the the Italian mathematician Fibonacci, who ...
The localization $A_\mathfrak{p}$ is given by all fractions $\frac{a}{b}$ with $a\in R$ and $b\in R\setminus\mathfrak{p}$. So $\mathfrak{p}A_\mathfrak{p}$ consists of all fractions $\frac{a}{b}$ with $... View answer 41 votes Step 1: The characteristic of$A$is$2$(Credit for this observation goes to Jyrki Lahtonen) The mapping$x\mapsto -x$is an involution on$A^\times$. Since$\lvert A^\times\rvert = 2^n - 1$is odd,... View answer 33 votes I'm assuming that you are looking for a factorization of the polynomial $$f = x^4 + 12x^3 + 44x^2 + 72x + 36$$ in$\mathbb Q[x]$. By the Lemma of Gauss and the fact that$f$is monic, this is the ... View answer Accepted answer 31 votes This is a really nice question! Counterexample for$n=6$The sets $$\{0,1,4,5,11,13\}\\\{0,1,2,6,10,13\}$$ are affinely inequivalent, but the multiset of differences is in both cases $$1^2\cdot 2\... View answer 31 votes Informally, the historical definition of an Archimedean solid requires that all its faces are regular polygons and that each vertex of the solid locally looks the same. Since the ancient Greeks it was ... View answer Accepted answer 28 votes Let G be a group of order 30 and let n_p be the number of its p-Sylow subgroups. By Sylow, n_3\in\{1,10\} and n_5\in\{1,6\}. It is not possible that n_3 = 10 and in the same time n_5 = 6,... View answer 28 votes Hint: The smallest case where your "theorem" is obviously wrong is the set \{1,2\}. Now check every step of the "proof" with this example. What goes wrong? View answer 26 votes The theorem that$$\text{PRIMES is in P}$$was proven only in 2002 (via the AKS primality test). The algorithm certainly isn't rocket science, making it quite surprising that this wasn't found way ... View answer 23 votes By looking at the possible cycle types, we see that A_4 consists of the identity element (order 1), 3 double transpositions (order 2) and 8 3-cycles (order 3). Assume that A_4 has a ... View answer Accepted answer 21 votes Hint: By induction, show that for n\geq 2$$\frac{4^n}{n+1} < \binom{2n}{n} < 4^n.$$View answer 20 votes I try to give you an argument why this notation makes sense. Consider$$\displaystyle\bigvee_x A(x)$$as an infinite version of \vee. For example, if x comes from a countable set \{x_1,x_2,x_3,\... View answer 19 votes To completely understand the +1 answer of Martin Brandenburg, I had to add a few details for myself. I decided to document the result in this answer: Let a finite generating system of P be given by ... View answer Accepted answer 18 votes Hint Show that none of the 9 consecutive numbers can be a multiple of 13. Denote the product of the 9 consecutive numbers by a. Show that a must be square. Look at the remainder of a modulo ... View answer 18 votes For such abstract questions, it is important that you stick to the definitions of the involved notions. By definition, A is a subset of B if every element contained in A is also contained in B... View answer Accepted answer 17 votes There is exactly one such function, which is the empty set. Explanation: By definition, functions f : A \to B are certain subsets of the cartesian product A\times B (for the defining property, ... View answer 15 votes The following solution only needs basic group theory. Let G be an transitive abelian subgroup of S_n. By transitivity, for each i\in\{1,\ldots,n\} there is a \sigma\in G such that \sigma(1) =... View answer 15 votes First of all, as far as I know there is no precise definition of a sparse matrix. The word sparse is used for a series (A_n)_{n\in\mathbb{N}} of n\times n matrices whose fraction of non-zero ... View answer Accepted answer 14 votes There is no hard mathematical difference between the comma (,) and the semicolon(;). The semicolon is used sometimes to optically separate some variable group. So the semicolon is not more than a ... View answer Accepted answer 14 votes I think the given proof is a bit confusing. Try this one: An edge in L(G) is a 2-set \{e_1,e_2\} of edges in G with which are adjacent to a common vertex v. This vertex v is uniquely ... View answer 13 votes For the particular base of 16, there is this remarkable formula:$$\pi=\sum_{n=0}^\infty \left(\frac{4}{8n+1}-\frac{2}{8n+4}-\frac{1}{8n+5}-\frac{1}{8n+6}\right)\frac{1}{16^n}$$It allows the ... View answer 13 votes The smallest non-abelian group is the symmetric group S_3 of order 6. So all its proper subgroups are abelian (the trivial subgroup, three subgroups of order 2 and one subgroup of order 3). View answer 12 votes The key is the repeated application of the geometric summation formula$$ \sum_{i=0}^{z-1} q^i = \frac{q^z - 1}{q-1}. $$Plugging in z=n, q=x and assuming n = kl, we get$$p(x) = x^{n-1} + x^{n-2}... View answer Accepted answer 12 votes a) Algebraic extensions of$\mathbb{R}$have degree$1$or$2$. Since not all zeros of$x^{15} - 1$are in$\mathbb{R}$, we get$[L : \mathbb{R}] = 2$. b) Let$\zeta$be a primitive$15$th root of ... View answer 12 votes No, it isn't. For all$a,b\in\mathbb{Z}$it holds that$\langle (a,b)\rangle = \{(ka,kb) \mid k\in\mathbb{Z}\} \neq \mathbb{Z}^2$. So$\mathbb{Z}^2$is not generated by a single generator and hence ... View answer 12 votes$f : (\mathbb{Z},+) \to (\mathbb{Z}, +), \quad x\mapsto 2x$View answer 11 votes Let$a_d$denote the number of elements of order$d$and$u_d$the number of cyclic subgroups of order$d$. Then$a_d = \varphi(d) u_d$, where$\varphi$is the Euler phi function. Reason: A cyclic ... View answer Accepted answer 11 votes The best strategy is to reroll any rolled 1, of course. So using that strategy, what is the chance that you do not end up at a 6 with a single dice? The probability for an instant 6 is$1/6$, and for ... View answer 11 votes Two important applications which are involved at every phone call on your mobile phone: Coding theory (adding redundancy to the information, such that occuring errors can be compensated) and ... View answer Accepted answer 11 votes This is based on the comment of @Derek Holt: A counterexample is given by the subgroup $$V = \{(g,h) \in S_3\times S_3 \mid \operatorname{sgn}(gh) = +1\}$$ of$S_3\times S_3$. Reason: Since$V\$ is ...