anonymous

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visits member for 2 years, 2 months
seen Dec 15 '12 at 10:51

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
Oct
26
answered Why does $1234^{1234^{1234}\ \bmod\ 10^{10}}$ = $1234^{1234^{1234}\ \bmod \ \phi(10^{10})}$