A necessary and sufficient condition for homomorphism of algebras
In Atiyah Macdonald, an A-algebra is defined as a ring B together with a ring homomorphism $f:A\to B$ that induces an A-module structure. Given two A-algebras $f:A\to B$ and $g:A\to C$, an A-algebra...
View ArticleAnswer by xaphi for expected value of the product of two Bernoulli variables
This statement isn't true. Consider the example where X and Y are independent Bernoulli variables with parameter $p=\frac{1}{2}$.$$\mathbb{E}[XY] = P(X=1,Y=1) = \frac{1}{4}$$On the other hand, $$P(X=Y)...
View ArticleAnswer by xaphi for show that a quotient space is a CW complex.
Hints:Construct an obvious family of maps $\{\Phi_{\alpha}\}$ consisting of homeomorphisms from the 0-disk to each vertex of $P$, from the 1-disk to each edge of $P$, and from the 2-disk to the face of...
View ArticleAnswer by xaphi for 15 players in groups of 3. In 7 rounds can they all play...
Yes. Moreover, this can be done in such a way that each person plays each other person exactly one time. This is equivalent to the existence of a (15,3,1)-design, which is answered in the affirmative...
View ArticleAnswer by xaphi for Inverting a relationship describing one matrix in terms...
The situation you've described is that there are $n^2$ elements ($B_{ij}$), each of which is expressed as a particular linear combination of $n^2$ other elements ($A_{ij}$).This is more suited to be...
View ArticleAnswer by xaphi for Question regarding mathematical notation or concept
$\{-1,1\}^n$ is the set of n-tuples $(x_1,x_2,...,x_n)$ such that $x_i \in \{-1,1\} \forall i$.$\#$ here is referring to the cardinality of the set - i.e. the number of components of the n-tuple $E$...
View ArticleAnswer by xaphi for I think I just proved that Pi + e is irrational, but I...
$-e^{ix} = e^{iy}$ implies that $x = y + (2n+1)\pi$ for some $n \in \mathbb{Z}.$So in your case, it simply shows that $e = Q + (2n+1)\pi$ for some $n \in \mathbb{Z},$ which we knew already from your...
View ArticleAnswer by xaphi for The set of all finite sequences in $\mathbb{Z}$ is...
Assuming that you're content relying on $\aleph_0 + \aleph_0 = \aleph_0$ and $\aleph_0 \cdot \aleph_0 = \aleph_0,$ then yes the proof is fine.Regarding finding an explicit bijection, you could try the...
View Article
