Bill Cook
Reputation
17,067
Next privilege 20,000 Rep.
Access 'trusted user' tools
26 49
Impact
~482k people reached

# 1,469 Actions

 1d answered Quotient Space Notation Nov 11 comment Comparing cardinality of sets Sure, no problem. Counting problems always leave me a little uneasy -- it's so easy to miss a case or double count something. It's nice to have a solid formula to hang your hat on. :) Nov 11 answered Comparing cardinality of sets Nov 4 comment Determine the Type and the General Solution of an ODE sosmath.com/diffeq/first/lineareq/lineareq.html Nov 3 comment Use group homomorphism to prove that for systems of linear equations, general=particular+homogeneous If y is a vector of functions as well as g and if A is a matrix of functions, this is essentially the definition of a (non homogeneous) linear system. Nov 3 comment Affine Weyl group as Coxeter group I would recommend checking a copy of "James E. Humphreys, Reflection Groups and Coxeter Groups, Cambridge studies in advanced mathematics, 29 (1990)" out of your library. Humphreys' text is quite readable and has the details you need. Nov 3 answered when can we make Function from its partial derivatives? Nov 3 answered Use group homomorphism to prove that for systems of linear equations, general=particular+homogeneous Nov 3 answered Polynomial with irrational coefficients Nov 3 comment Are $\mathbb{R}$ and $\mathbb{R}^{\infty}$ isomorphic as groups under addition? For infinite dimensional vector spaces over a field $\mathbb{K}$ (where $|X|$ denoted cardinality) we have $|V| = |\mathbb{K}| \mathrm{dim}(V)$. So for $\mathbb{K}=\mathbb{Q}$, we have $|V| = |\mathbb{Q}| \mathrm{dim}(V) = \mathrm{dim}(V)$ (since $\mathrm{dim}(V)$ is infinite). Thus the cardinality of these vector spaces and their dimensions are the same thing. This particular result now follows from: $(2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0 \cdot \aleph_0} = 2^{\aleph_0}$ Nov 3 comment Smart way to check the property $ab=0\implies a=0$ or $b=0$ to form an integral domain No problem. Glad to help. :) Nov 3 answered Smart way to check the property $ab=0\implies a=0$ or $b=0$ to form an integral domain Oct 28 comment Why does a group homomorphism preserve more structure than a monoid homomorphism while satisfying fewer equations ...of examples that you don't need extra equations. Groups add on this inversion map: $x \mapsto x^{-1}$. But because the axioms of a group are so restrictive, we don't need to force our homomorphisms to map inverses to inverses - we get it for free. This is just because groups are so much nicer than our previous class of objects (i.e. monoids). Monoids really aren't that much nicer than semigroups (or that's my opinion anyway). Oct 28 comment Why does a group homomorphism preserve more structure than a monoid homomorphism while satisfying fewer equations You add an equation because you've added structure that you want preserved. Just like going from an abelian group to a ring. Or a ring without identity to a ring with identity. As you add on structure, you usually want your new homomorphisms to preserve that structure. Another example: Every vector space is an abelian group. But abelian group homomorphisms don't preserve scalar multiplication (in general) so we add a condition to restrict our attention to nice homomorphisms (i.e. linear transformations). The point of all of this is sometime when adding structure you so limit your class... Oct 28 answered The order deduced from relations in $D_n$ Oct 28 awarded Nice Answer Oct 27 comment Why does a group homomorphism preserve more structure than a monoid homomorphism while satisfying fewer equations Every monoid morphism is a semigroup morphism. So the identity restriction is moving from a wider class of "morphisms" to a special subclass. From this viewpoint we see that in the context of groups there is no distinction between these "generalized" morphisms and regular morphisms. Oct 27 answered Evaluate the integral by changing to spherical coordinates. Oct 27 comment Evaluate the integral by changing to spherical coordinates. Notice that you are integrating above a cone: $z=\sqrt{x^2+y^2}$ and below a sphere: $z=\sqrt{72-x^2-y^2}$. This is an "ice cream cone" shaped region. The outer bounds yield a quarter circle. Your bounds for $\phi$ and $\theta$ aren't correct. The cone gives you an upper bound for $\phi$ and $\theta$ only goes a quarter turn. Oct 27 revised What is the geometric meaning of those vectors? added 1 character in body; edited title