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 Jul2 awarded Curious Dec5 comment Preimage of $x^2$ You are looking for all $x$ such that $-1 \leq x^2 \leq 1$. This is true for all $x \in [-1,1]$. Nov26 awarded Yearling Nov24 awarded Commentator Nov24 comment $\begin{bmatrix} -2 & 5 & 4 \\-1 & 0 & 0 \\0 & 4 & 3 \end{bmatrix}^{2013}$ =? @MarcvanLeeuwen: Or they maybe want them to get used to CA systems. Nov24 comment $\begin{bmatrix} -2 & 5 & 4 \\-1 & 0 & 0 \\0 & 4 & 3 \end{bmatrix}^{2013}$ =? @OriaGruber: Well, then compute the first ten powers by hand. Nov24 comment $\begin{bmatrix} -2 & 5 & 4 \\-1 & 0 & 0 \\0 & 4 & 3 \end{bmatrix}^{2013}$ =? You could use Sage to compute the first 10 powers. Then it shouldn't be difficult to spot a pattern. Or you can use Sage to just compute the answer immediately. Nov14 asked “Most important absolute property in mathematics” according to Osborne May25 awarded Nice Question Mar5 answered What is the equation stands for in geometry(intuitively)? Mar4 revised Reference Request: Vector Spaces Corrected spelling. Mar4 suggested approved edit on Reference Request: Vector Spaces Mar4 answered prime divisor of $3n+2$ proof Feb17 comment Separability of a field Extension. There are actually two slightly different definitions of 'separable polynomial'. The first one is: A polynomial over $K$ is separable if it has distinct roots in some algebraic closure of $K$. The second one says: A polynomial over $K$ is separable if each of its irreducible factors has no repeated roots. By the first definition $x^n - 1$ is not separable in general, by the second one it is. Jan25 awarded Revival Nov15 asked Pathologies in finite-dimensional linear algebra? Sep24 awarded Scholar Sep24 accepted Can all convex polytopes be realized with vertices on surface of convex body? Sep24 answered Can all convex polytopes be realized with vertices on surface of convex body? Sep24 awarded Nice Answer