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 Jun 26 awarded Yearling Jun 26 awarded Yearling Jun 26 awarded Yearling Dec 14 comment Is the square for a number $n\in \mathbb{Z}$ even, then is also $n$ even - prove this via contraposition? $k$ may be even, but $n$ is odd. Dec 9 answered Is the set of measures of all measurable sets with finite measure closed? Dec 7 revised Evaluate $\lim_{n\to\infty} \int_0^{2 \pi} \frac{1}{x+\sin^n x+ \cos^n x} \ dx$ deleted 29 characters in body Dec 7 answered Evaluate $\lim_{n\to\infty} \int_0^{2 \pi} \frac{1}{x+\sin^n x+ \cos^n x} \ dx$ Dec 6 comment Is this a property of an integral domain that is not a field? That's not a domain. Dec 2 comment In ring $(R,+,*)$, if $-x\in R$, can we prove (or assume) $x\in R$? I can think of one reasonable interpretation of the question: if $R$ is a subring of some larger ring $S$ and we have some element $x\in S$ and we know $-x\in R$, do we then also know $x\in R$? (For an explicit example, one might consider $R=\mathbb{Q}$, $S=\mathbb{R}$.) Then the usual proof using the ring axioms works fine. Dec 1 comment A set of objects that satisfy $a^2 = \alpha x$ and commute When you say "multiples of x", do you just mean scalar multiples ($2x, pi\cdot x$, etc.)? Nov 26 answered Given $a + \sqrt{b}$ with positive integer $a,b$, find $a$ and $b$? Nov 25 answered Proof that the open Ball could not be written as a finite union of intervals Nov 25 comment Generating valid x and y that result in perfect squares Note that $(N+2)^2=N^2+4N+4$. Nov 10 comment How do I read this distribution function: $\min(X,Y)$? $min(X,Y)$ is defined by $min(X,Y)(\omega)=min(X(\omega),Y(\omega))$ for every $\omega\in\Omega$. Does that help? Nov 5 answered Let $x,y,z$ be independent uniform distribution on $(0,\pi)$. What's the probability that $z\leq \cos^2(x)\sin^2(y)$? Nov 5 comment Square root of a matrix Yes. You have such a factorization if and only if it is positive semi-definite. Oct 27 comment Why does $1234^{1234^{1234}\ \bmod\ 10^{10}}$ = $1234^{1234^{1234}\ \bmod \ \phi(10^{10})}$ The method requires $a$ and $10^{10}$ to be coprime, not just for $a$ to not be a multiple of 10. It is not true that $2^{\phi(10^{10})+1}\equiv 2$mod $10^{10}$; it's just really likely you'll get the right answer anyway since most equivalence classes mod $10^{10}$ are greater than $10$. Oct 27 comment Why does $1234^{1234^{1234}\ \bmod\ 10^{10}}$ = $1234^{1234^{1234}\ \bmod \ \phi(10^{10})}$ Well, uh, the basic method is already faulty (in that $10^{\phi(10^{10})+1}\equiv 0$ mod $10^{10}$ and not $10$ mod $10^{10}$. My answer isn't particularly more wrong. Oct 26 comment Why does $1234^{1234^{1234}\ \bmod\ 10^{10}}$ = $1234^{1234^{1234}\ \bmod \ \phi(10^{10})}$ You're right. I've worked it out fully now. Oct 26 revised Why does $1234^{1234^{1234}\ \bmod\ 10^{10}}$ = $1234^{1234^{1234}\ \bmod \ \phi(10^{10})}$ added 799 characters in body