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 16h comment Isomorphism between $\mathbb R^3$ and the the Heisenberg group. What do you mean by 'as a group'? There's no group isomorphism between $\mathbb{R}^3$ and the Heisenberg group because the former (under the usual group operation, at least) is abelian while the latter isn't. If you mean a three-dimensional parametrization of the Heisenberg group, there are several, with the easiest coming just by writing $z\in\mathbb{C}$ as $a+bi : a,b\in\mathbb{R}$... 1d comment Prove that for $k>1$ $a_n$ is a perfect square Have you tried explicitly computing $P(3)$, $P(4)$, etc. to see if you can find a pattern? 1d comment $n \equiv 7 \pmod{8}$, prove $\sigma(n) \equiv 0 \pmod{8}$ This is a much better answer than the hint I had in comments - really nice! 1d comment $n \equiv 7 \pmod{8}$, prove $\sigma(n) \equiv 0 \pmod{8}$ A broad hint: $\sigma(n)$ is a multiplicative function, and therefore so is $\sigma(n)\bmod 8$ (why?). If $n\equiv 7\pmod 8$, what are the possible factorizations of $n$? 2d awarded Enlightened 2d awarded Nice Answer Apr 24 comment How do I prove the following equation involving $\int_{0}^{+\infty}\left(\frac{sin(x)}{x}\right)^ndx$ It's easy to make a case that the integrals on the left are manifestly positive: since $\sin x\geq 0$ for $0\leq x\leq \pi$ and $\sin x\leq 0$ for $\pi\leq x\leq 2\pi$ and $\sin(x+\pi)=-\sin x$, then $\left|\frac{\sin(x+\pi)}{x+\pi}\right|\lt \left|\frac{\sin x}{x}\right|$ and so the 'negative' piece of each $2\pi$ period is smaller than the positive piece just before it - i.e., break the integral into integrals from $0$ to $\pi$, $\pi$ to $2\pi$, etc. and make the case that each integral is smaller in absolute value than the one before and they're alternating in sign with the first positive. Apr 19 comment If $G/Z(G)$ is cyclic, why is $G$ only abelian and not also cyclic? I'll go a step further - start with the simplest case, $C_2\times C_2$, and see what goes wrong. Apr 18 comment Is there an explicit irrational number which is not known to be either algebraic or transcendental? @lulu I believe that number is known to be transcendental. It's the sum of a rational number ($\frac19$) and a theta-value at a rational argument that I'm pretty sure is known not to be algebraic. Apr 15 comment A sum of irrational numbers is an algebraic integer If you need to find a suitable polynomial 'by hand' then you can peel off: $x-\sqrt{2}+\sqrt{17}(\frac12(7-\sqrt{13}))=\sqrt[3]{5}$, so you can cube both sides of this, etc. This gets painful in a hurry, though. Instead, you might consider what operations algebraic integers are closed under... Apr 15 comment Bijection between $[\mathbb N\to \{0,1\}]$ and $[\mathcal P(\mathbb N) \to \{0,1\}]$ @user3286435 If you work in a system where all relations etc. must be computable, then you're right that $\mathcal{P}_c(\mathbb{N})$ is countable in the 'base model' - that is externally - but there's no computable bijection between $\mathcal{P}_c(\mathbb{N})$ and $\mathbb{N}$ - that is, $\mathcal{P}_c(\mathbb{N})$ is 'internally' uncountable. Apr 15 answered Bijection between $[\mathbb N\to \{0,1\}]$ and $[\mathcal P(\mathbb N) \to \{0,1\}]$ Apr 13 comment Can you comb the hair on a 4-dimensional coconut? Spoiler from that Wikipedia article: the short answer is that it alternates, so that $S^3$ has a tangent field, $S^4$ doesn't, $S^5$ does, etc. Apr 12 comment How to integrate $x^2\sin(x^2)$? Whoops - yes, it is. Fixed, though of course that doesn't materially affect the result here. Apr 12 revised How to integrate $x^2\sin(x^2)$? Fixed a matho Apr 12 answered How to integrate $x^2\sin(x^2)$? Apr 12 comment What is the Order (Big O) of this polynomial? Two pieces of relevant nitpickery: (1) I presume you mean as $n\to\infty$, since $n$ is usually used for integer variables, but note that $p(n)$ is $\in O(1)$ as $n\to 0$, for instance - knowing what your asymptote is matters! (2) You (or your instructor) probably pedantically mean $\Theta()$ - it's also completely true to say that $p(n)\in O(n^{14})$ as $n\to\infty$, because $O()$ only talks about upper bounds, not precise orders. (Also, (3) strictly speaking $p(n)$ isn't a polynomial, but that's another matter entirely.) Apr 5 comment Rewriting ∃! using predicate logic expressions ( “=” excluded) It's a little strange for = to not be a valid symbol - in the language of set theory there are some reasonable definitions for equality, but most of those require at least $\in$. You might have a look at en.wikipedia.org/wiki/First-order_logic#Equality_and_its_axioms , but I would speak to your professor first. Apr 5 comment Rewriting ∃! using predicate logic expressions ( “=” excluded) @P.Lance No - in fact, what you've written is exactly equivalent to $\exists x A(x)$; can you see why? Apr 5 comment Finding the limit $\lim_{n\to\infty}{\sum_{i=0}^n \frac{(-1)^i}{i!}f^{(i)}(1)}$. What have you tried so far? Do you know any formulas that involve sums over the derivatives of a function, suitably scaled?