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Aug
15
comment Show that $\mathbb Z[x]$ and $\mathbb Q_{>0}$ are isomorphic
Perhaps this could be of help: Think of elements of $(\mathbb{Z}(x), +)$ as infinite tuples which are eventually zero. Explicitly, you can identify $a_{0}+a_{1}x+a_{2}x^2+\cdots + a_{n} x^{n}$ with the infinite tuple $(a_{0}, a_{1}, …, a_{n}, 0, 0, …)$ where $a_{i}\in\mathbb{Z}$. Can you think of a map that would send this element to $(\mathbb{Q}_{>}, \cdot)$? (Another hint is to consider uniqueness of prime factorization).
Aug
15
reviewed Approve suggested edit on Show that $\mathbb Z[x]$ and $\mathbb Q_{>0}$ are isomorphic
Aug
14
answered Understanding what the Sylow theorems say about $p$-groups
Aug
14
comment Deduction of usual Cayley-Hamilton Theorem from “Determinant Trick”
@SammyBlack: Hmmm, why does it make sense only for those endomorphisms acting on the right?
Aug
14
reviewed Close topology products, unions and bijections
Aug
14
awarded  Custodian
Aug
14
asked Deduction of usual Cayley-Hamilton Theorem from “Determinant Trick”
Aug
13
comment Normalizer and centralizer of abelian subgroups of a group are equal
@user1729: No worries! :)
Aug
13
comment Normalizer and centralizer of abelian subgroups of a group are equal
@kammy: Yes. Using $H=Z(G)$ only gives $G=G$.
Aug
13
comment Normalizer and centralizer of abelian subgroups of a group are equal
@user1729: I don't understand. Do you want to use hypothesis (which asserts $C_{G}(H)=N_{G}(H)$ for abelian subgroups $H$ of $G$) to conclude $C_{G}(G)=N_{G}(G)$? That would only work if $G$ is abelian to begin with. And that would be circular.
Aug
13
answered Normalizer and centralizer of abelian subgroups of a group are equal
Aug
13
comment Prove that $\frac{a^3}{x} + \frac{b^3}{y} + \frac{c^3}{z} \ge \frac{(a+b+c)^3}{3(x+y+z)}$ a,b,c,x,y,z are positive real numbers.
@user169248: Technically yes… There is a way to prove it from the scratch. I could go ahead and re-prove the Cauchy-Schwarz (without actually saying it is C-S, but proving the special case) and re-proving AM-GM inequality in the special case of 3 terms. But there would be no point. I think it is more beneficial to master the well-known inequalities first (like Cauchy-Schwarz and AM-GM), and then apply them whenever you need to. You can find proofs of these famous inequalities by a simple Google search.
Aug
13
comment Minkowski inequality for $0<p<1$
@Ignacio: If there are mistakes, you can press "edit" button to modify the post :)
Aug
13
answered Existence of a metric space M with no continuous map from M to any other metric space
Aug
13
comment Is this proof that there are no perfect, odd, integer square numbers legitimate?
Nice proof (+1). Now try proving that there are no odd perfect numbers that are perfect powers ;)
Aug
13
comment Elementary theorems with several proofs?
I really like the Third Proof: Bounding with the Maximal Order. It also seems to work for any multiplicative group (group of units) of a field.
Aug
12
comment The topology on $\mathbb A^2$ is not the product topology
@MorganO: Brilliant! :)
Aug
12
comment The topology on $\mathbb A^2$ is not the product topology
I don't have a proof, but I am guessing that in general Zariski topology on $\mathbb{A}^{m+n}$ is not the product topology on $\mathbb{A}^{m}\times\mathbb{A}^{n}$ for any $m, n\in\mathbb{N}$. Could someone verify this?
Aug
12
answered Prove that $\frac{a^3}{x} + \frac{b^3}{y} + \frac{c^3}{z} \ge \frac{(a+b+c)^3}{3(x+y+z)}$ a,b,c,x,y,z are positive real numbers.
Aug
12
comment Finitely many embeddings of a finite extension in an algebraic closure
Re your edit: Every finite extension is algebraic.