smackcrane
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# 159 Actions

 Jan28 awarded Notable Question Jan8 awarded Popular Question Jul2 awarded Curious Jun12 awarded Yearling May15 awarded Popular Question May4 awarded Popular Question Apr15 awarded Nice Answer Mar10 awarded Popular Question Nov14 awarded Popular Question Jun12 awarded Yearling Dec24 answered prove $\lceil{x}\rceil=-\lfloor-x\rfloor$ Dec15 answered Different equivalence relations of the set $\{a,b\}$ Dec14 answered Expectation of a uniform point chosen out of a triangle with vertices $(0,0), (1,0), (0,2)$ Dec12 answered There exist an infinite subset $S\subseteq\mathbb{R}^3$ such that any three vectors in $S$ are linearly independent. Dec10 answered Calculating the dimension of a vector space in 2 different ways Dec2 accepted Show that $\mathbb{Z}[\theta]$ (where $\theta = (1 + \sqrt{19}i)/2$) is a principal ideal domain. Nov30 answered Show that $\mathbb{Z}[\theta]$ (where $\theta = (1 + \sqrt{19}i)/2$) is a principal ideal domain. Nov30 revised Show that $\mathbb{Z}[\theta]$ (where $\theta = (1 + \sqrt{19}i)/2$) is a principal ideal domain. problem partly solved Nov29 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. Nov29 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.