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visits member for 3 years, 1 month
seen Jul 11 at 16:49

Jul
2
awarded  Curious
Jun
12
awarded  Yearling
May
15
awarded  Popular Question
May
4
awarded  Popular Question
Apr
15
awarded  Nice Answer
Mar
10
awarded  Popular Question
Nov
14
awarded  Popular Question
Jun
12
awarded  Yearling
Dec
24
answered prove $\lceil{x}\rceil=-\lfloor-x\rfloor$
Dec
15
answered Different equivalence relations of the set $\{a,b\}$
Dec
14
answered Expectation of a uniform point chosen out of a triangle with vertices $(0,0), (1,0), (0,2)$
Dec
12
answered There exist an infinite subset $S\subseteq\mathbb{R}^3$ such that any three vectors in $S$ are linearly independent.
Dec
10
answered Calculating the dimension of a vector space in 2 different ways
Dec
2
accepted Show that $\mathbb{Z}[\theta]$ (where $\theta = (1 + \sqrt{19}i)/2$) is a principal ideal domain.
Nov
30
answered Show that $\mathbb{Z}[\theta]$ (where $\theta = (1 + \sqrt{19}i)/2$) is a principal ideal domain.
Nov
30
revised Show that $\mathbb{Z}[\theta]$ (where $\theta = (1 + \sqrt{19}i)/2$) is a principal ideal domain.
problem partly solved
Nov
29
comment Calculating the Odds of Victory in Risk
I agree! I wrote a Risk simulation a few years ago, and I believe this was the method I used to calculate winning probabilities in battles.
Nov
29
comment Show that $\mathbb{Z}[\theta]$ (where $\theta = (1 + \sqrt{19}i)/2$) is a principal ideal domain.
$\pm \theta$ is certainly good enough: as you probably saw, the point of saying $N(2a/b - \theta - m) < 1$ is simply that $N(b)N(2a/b - \theta - m) = N(2a - \theta b - mb) < N(b)$ contradicts $b$'s minimality.
Nov
29
asked Show that $\mathbb{Z}[\theta]$ (where $\theta = (1 + \sqrt{19}i)/2$) is a principal ideal domain.
Nov
29
accepted if $S$ is a ring (possibly without identity) with no proper left ideals, then either $S^2=0$ or $S$ is a division ring.