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seen Dec 16 at 23:45

a real pythagorean


Dec
15
comment Double angle sine
Because $\alpha$ will play the role of $2x$ and $x$ is restricted to $0<x<\pi$. --- Oops, I corrected the constant.. :)
Dec
15
answered Double angle sine
Dec
12
answered Contravariant functor properties
Dec
12
answered Subcategory of sets with surjective mapping
Dec
12
answered Mutually orthogonal set of vectors
Dec
8
awarded  Nice Answer
Dec
7
comment game theory atomic selfish routing
at least a link should have been added to the text anyway, I guess
Dec
7
answered Product of two arbitrary elements in $O(2)$
Dec
7
comment What does $S^z$ mean for each $z\in\mathbb{C}$?
Related: math.stackexchange.com/questions/198514/…
Dec
7
revised Adjoint to forgetful functor from $\textbf{Met}$ to $\textbf{Set}$
added 3 characters in body
Dec
6
comment Let $H$ be a Hilbert space, $A$ is unitary and $S=\{Ax:x\in H\}$. Does $S^{\perp}=\operatorname{Null}(A)$?
In the title you write '$A$ is unitary'. Are you sure about this condition? Wouldn't it mean that $A$ is invertible (with $A^{-1}=A^*$)? But then $S=H$ and $S^\perp=0$.
Dec
6
answered Category defined by a finite commutative diagram
Dec
6
comment Subspace of differentiable real-valued functions over an interval
Are you sure it is '$f^{-1}$' written there? It just makes no sense. Most probably it wants to be either $f(-1)$ or $f'(-1)$, I guess...
Dec
6
answered Prove that H is a normal subgroup if and only if, $\forall a , b \in G, ab \in$ H implies $ba\in H.$
Dec
5
comment Group generated by two polynomials
I guess so..$\,$
Dec
4
comment Meaning of $f=me$ Factorization in Abelian Categories
or by its kernel, and these two ways are isomorphic!
Dec
4
answered Adjoint to forgetful functor from $\textbf{Met}$ to $\textbf{Set}$
Dec
1
comment Is $\lnot\forall x\;\lnot\forall y\;A$ the same as $\forall x\;\forall y\;A$?
No, it is not the same: 'For not all $x$ and not all $y$ holds $A$' $\ne$ 'For all $x$ and all $y$ holds $A$'. I suggest to start the proof over, on a clear page.
Dec
1
answered Z / 6Z being a set of well dedfined equivalence classes, and a congruent to b(mod 6)
Nov
30
answered Define f: Z /6Z by g(5[a]) = [5a]