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Jan
16
comment Why are primes considered to be the “building blocks” of the integers?
@zod Edited, thanks!
Jan
16
revised Why are primes considered to be the “building blocks” of the integers?
edited body
Jan
13
awarded  Nice Answer
Jan
11
comment $X^2$irreducible but not prime
(Are $X$ and $x$ the same thing, and do you mean $X^i$ instead of $X_i$?) $X^2$ certainly looks prime here to me.
Jan
11
answered Reverse Product Rule ODE
Jan
11
reviewed Approve $2\times 2$ matrices forming the Klein $4$-group.
Jan
11
revised Why are primes considered to be the “building blocks” of the integers?
added 44 characters in body
Jan
11
answered Why are primes considered to be the “building blocks” of the integers?
Jan
11
comment Group Duality with respect to Generators and Relations
(2/2) On the other hand, there certainly are some notions of duality that arose from mathematical questions, which you may be interested in looking up. A simple one would be the concept of a dual vector space (vector spaces are, of course, continuous groups too), which arises by considering functions on that space (or matrix transposition). A harder one: Pontryagin duality comes from trying to do Fourier analysis on certain groups, and crops up very naturally in a bunch of places, including e.g. the character theory of finite abelian groups.
Jan
11
comment Group Duality with respect to Generators and Relations
(1/2) I think my real point is: in order for such a definition to be mathematically meaningful, its conception needs to be mathematically well-motivated. You can't just say "I want duals to exist; where are they?". Better questions: what do you want $G^D$ to do? Or how do you want it to look relative to $G$? I see no mathematical reason to swap the generators and relations of a group (and no guarantee that it could be made to work). My geometric proposal was just something slightly more well-motivated (but, if you work through the details, you'll see it's also not very interesting).
Jan
10
comment Group Duality with respect to Generators and Relations
Suppose $G = \langle x | x^2 \rangle$. What does your proposed object $G^D$ mean? (But, if you want your 'dual' group to be the group of symmetries of the respective 'dual' polyhedron - which seems like quite a reasonable request to me - then the right definition is probably simpler than you think.)
Jan
10
comment How to prove that a set spans a plane
@Pow: aha, okay - take an element (x,y,z) on your plane. Supposing $c \neq 0$, we can more helpfully write this as (x,y,$\frac{-ax-by}{c}$). Now it suffices for you to find r and s (in terms of a, b, c, x and y) such that r(b,-a,0) + s(0,c,-b) = (x,y,$\frac{-ax-by}{c}$) - shouldn't be too hard. (Then of course you have to deal with the case $c = 0$...) :)
Jan
10
answered How to prove that a set spans a plane
Sep
10
awarded  Yearling
Jan
25
comment What's the difference between $\mathbb{R}^2$ and the complex plane?
(You might also find it helpful to note that the "natural" multiplication on $\mathbb{R}^2$, namely $(u,v)\cdot (x,y) = (ux, vy)$, does not agree with the natural multiplication on $\mathbb{C}$ for any choice of $g\in \mathrm{GL}_2(\mathbb{R})$.)
Jan
25
comment What's the difference between $\mathbb{R}^2$ and the complex plane?
@laovultai: Let $f$ be the map $\mathbb{R}^2\to \mathbb{C}$, $(a,b) \mapsto a+ib$, and let $g$ be any element of $\mathrm{GL}_2(\mathbb{R})$. Then $f\circ g$ is an isomorphism $\mathbb{R}^2\to \mathbb{C}$ as $\mathbb{R}$-vector spaces. Now simply "pull back" the multiplication from $\mathbb{C}$ to $\mathbb{R}^2$ along the map $f\circ g$ (e.g. when $g$ is the identity map, the multiplication inherited is $(u,v)\cdot (x,y)=(ux−vy,uy+vx)$), and you get an isomorphism of rings (or, equivalently, $\mathbb{C}$-vector spaces). Does that answer your question?
Jan
17
comment How should I understand $R[x]/(f)$ for a ring $R$?
f should be irreducible, otherwise it is not true that R[alpha] = R[x]/f.
Jan
17
awarded  Custodian
Jan
17
reviewed Approve Optimization Word Problem
Jan
17
answered If $X$ and $Y$ are objects of $\mathrm{Set}$, is there any reason not to regard $\mathrm{Hom}(X,Y)$ as an object, too?