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visits member for 2 years, 2 months
seen Mar 17 at 8:09

Jul
2
awarded  Curious
Dec
5
comment Preimage of $x^2$
You are looking for all $x$ such that $-1 \leq x^2 \leq 1$. This is true for all $x \in [-1,1]$.
Nov
26
awarded  Yearling
Nov
24
awarded  Commentator
Nov
24
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.
Nov
24
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.
Nov
24
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.
Nov
18
revised Set of all affine maps as affine space
edited tags
Nov
18
asked Set of all affine maps as affine space
Nov
14
asked “Most important absolute property in mathematics” according to Osborne
May
25
awarded  Nice Question
Mar
5
answered What is the equation stands for in geometry(intuitively)?
Mar
4
revised Reference Request: Vector Spaces
Corrected spelling.
Mar
4
suggested suggested edit on Reference Request: Vector Spaces
Mar
4
answered prime divisor of $3n+2$ proof
Feb
17
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.
Jan
25
awarded  Revival
Nov
15
asked Pathologies in finite-dimensional linear algebra?
Sep
24
awarded  Scholar
Sep
24
accepted Can all convex polytopes be realized with vertices on surface of convex body?