Being in $U(R)$ is the natural analogy of being $1$. In $\mathbb{N}$, $1$ is the only unit. In $\mathbb{Z}$, there are two; $\pm 1$. In $\mathbb{Z}[i]$ there are four; $\pm 1$ and $\pm i$. In $F[t]... View answer Accepted answer 7 votes If the Pythagorean Triple$(a, b, c)$is not primitive, then$b+c$is trivially composite. Thus, let it be primitive. Then $$a=m^2-n^2$$ $$b=2mn$$ $$c=m^2+n^2$$ where$b$and$a$are interchangeable.... View answer 0 votes Let$L_n$be the left-hand side, and let$R_{m,n}$be the right-hand side. Note that$L_n$is divisible by all of the greatest prime powers$p^r \leq n$. In fact, $$L_n= \prod_{p \leq n} p^r$$ where ... View answer 17 votes The sum is essentially $$\sum_{a=1}^9 \sum_{b=1}^9 ab =\sum_{a=1}^9 a \sum_{b=1} ^9 b=\sum_{a=1}^9 a \frac{9\cdot 10}{2}=\left(\frac{9\cdot 10}{2}\right)^2$$ View answer 5 votes It doesn't get a new limit, it actually just didn't have a value before (because it was$0/0$), but it still had that same limit. The original function and the new function are actually different ... View answer Accepted answer 1 votes Before his edit Pretty sure it only works for functions$p(n)=o(1)$(using little-$o$notation). Otherwise,$p(n)=\Omega(1)$, so$1-(1-p(n))^n=\Omega(p(n)^n)$. Assuming$p \rightarrow \infty$, we ... View answer 2 votes I believe from context$\varphi_n(t)=\frac{1}{\sqrt{2\pi}}e^{int}=\frac{1}{\sqrt{2\pi}}(\cos(nt)+i\sin(nt))i^2=-1n \in \mathbb Z$with argument$t$. View answer 2 votes$0<\frac{a_n}{n}<\frac{a_1}{n} \rightarrow 0$, so Squeeze Theorem, and$\sum \frac{a_n}{n^2} < \sum \frac{a_1}{n^2}=a_1\pi^2/6$View answer 2 votes Note that$\sin(x)$is an integer. But then$\lfloor \sin(x) \rfloor=\sin(x)$, so your equation reduces to $$\lceil \cos(x) \rceil=2$$ which is not possible. View answer 1 votes Yes of course. Let$y=k\frac{z}{\gcd(x,z)}$and$x=x'\gcd(x,z)$. Then $$\frac{xy}{z}=\frac{x'\gcd(x,z)\cdot k\frac{z}{\gcd(x,z)}}{z}=x'k$$ Furthermore, if$xy \equiv 0 \mod z$, then $$\frac{xy}{z}... View answer 22 votes 3^{2m+1}-3=3(3^m-1)(3^m+1), and both of the factors 3^m \pm 1 are even, so their product is divisible by 4. View answer 0 votes \{x^3\}=0 and \lfloor x^4 \rfloor=1 implies$$x \in \{\sqrt{n} | n \in \mathbb{Z} \} \cap \left([1, \sqrt{2}) \cup (-\sqrt{2}, -1]\right)=\{\pm 1\}$$, so x=\pm 1 View answer Accepted answer 2 votes The reason is because the horizontal asymptote is identically the x-axis for your function. Also, an asymptote only describes end behavior, so it's perfectly fine for the function to cross it ... View answer 1 votes The solutions should just be 2e^{\frac{2\pi ki}{10}}, for odd integers 0 \leq k <10. They are the divisions of the circle of radius 2 centered at the origin, divided into 5 pieces, rotated ... View answer 5 votes Limits don't necessarily preserve strict inequalities. For example, 1-\frac{1}{n}<1+\frac{1}{n}, yet they have the same limit as n goes to \infty. View answer Accepted answer 0 votes By set inclusion: Consider x \in \cap_{i\in I}(B \setminus A_i). Then x \in B \setminus A_i for all i \in I \implies x \notin A_i for all i \in I \implies x \notin \cup_{i\in I} A_i \implies ... View answer 0 votes We know the minimal polynomial of \zeta_n is the nth cyclotomic polynomial \Phi_n (x), which is defined as$$\Phi_n (x)=\prod_{k \in \mathbb{Z}_n^{\times}}(x-\zeta_n^k)$$where \mathbb{Z}_n^{\... View answer 1 votes I assume the p_i are meant to be distinct. Let y_i satisfy$$-x_1p_1 \equiv y_1 \mod p_1-x_1p_1 \equiv y_2 \mod p_2-x_1p_1 \equiv y_3 \mod p_3 \\$$Clearly$y_1 \equiv 0 \mod p_1$... View answer 3 votes Stirling's approximation for$n!$says$n!$can be estimated by$\sqrt{2\pi n}(\frac{n}{e})^n$for large values. The terms of your series are therefore asymptotic to terms of the form$\frac{1}{\sqrt{...