60,185 reputation
476161
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location Buenos Aires, Argentina
age 21
visits member for 2 years, 7 months
seen 1 hour ago

(My avatar is a piece by artist Pollock named "Number 8".)

Some interesting questions, with great answers:

  1. The so-called Axiom of Choice

  2. Real numbers and sets

  3. How discontinuous can a derivative be?

  4. Why is summation by parts important? This is one example.

  5. Amazing work


1h
comment If each uncountable set $T$ has a countable subset, can we form $T$ by a union of countable subsets?
Cardinality. ${}{}{}$
1h
comment If each uncountable set $T$ has a countable subset, can we form $T$ by a union of countable subsets?
Since $T$ is uncountable, this means by definition that $\# T>\aleph_0$. But since the union is disjoint, $\# \bigcup_{i=1}^n C_i =\sum_{i=1}^n \# C_i=n\aleph_0=\aleph_0$. But of course $\aleph_0>\aleph_0$ is contradictory. One can prove that the finite union of countable sets is countable with no AOC, but probably someone will comment that countable union of sets countable or that every infinite set has a countable subset are AOC-dependent.
3h
comment $(a, b) = (b, c) = (a, c) = 1$ implies $(c^2, ab) = (ab, a^n - b^n) = (c^2, a^n - b^n) = 1$?
It is true that $(a,bc)=1$ follows from $(a,b),(a,c)=1$. It is also true that $(a,a^n-b^n)=(a,-b^n)=(a,b^n)=(a,b)=1$ since $a^n-b^n=-b^n\mod a$, the same with $(bc,a^n-b^n)$. In general, if $b=b'\mod a$, $(a,b)=(a,b')$.
1d
revised The Sum of ${11^{th}}$ power of the roots of the equation ${x^5+5x+1=0}$
deleted 27 characters in body; edited title
1d
comment The Sum of ${11^{th}}$ power of the roots of the equation ${x^5+5x+1=0}$
What's with the boldface...?
1d
comment No. of homomorphisms from $\mathbb Z_n$ to $\mathbb Q$
(I'm not sure if the OP is thinking $\Bbb Q$ as an additive of multiplicative group. In the former case the only torsion elt is $0$, in the latter, there are two torsion elts.)
1d
answered No. of homomorphisms from $\mathbb Z_n$ to $\mathbb Q$
2d
comment What's the name of $\sum_{k = 0}^{n} (-1)^k {n \choose k} (n-k)^w$?
Yes, and the formula is proven using inclusion-exclusion by counting surjections.
2d
comment What's the name of $\sum_{k = 0}^{n} (-1)^k {n \choose k} (n-k)^w$?
You might be thinking about the inclusion-exclusion principle.
Aug
17
comment Show that the kernel of the map $SL(n, \mathbb{Z}) \to SL(n, \mathbb{Z}/3\mathbb{Z})$ has no torsion.
(Given this argument fails for $2$ instead of $3$, I'm tempted to find a counterexample in the case of $\Bbb Z_2$, or an improved argument).
Aug
16
comment If $R$ is nonunital, is it true that $R/A$ is a field iff $A$ is a maximal ideal?
@JonasMeyer True, I missed that. =)
Aug
16
answered Is $\{f \in End_{\mathbb R}(\mathbb R^n) : d(f(x),f(y))=(x,y) \space \forall x,y \in \mathbb R^n\}$ a group?
Aug
16
comment If $R$ is nonunital, is it true that $R/A$ is a field iff $A$ is a maximal ideal?
A commutative ring is a field iff its only ideals are $(0)$ and $(1)$. If $R$ is any ring and $\mathfrak a$ an ideal, there is a correspondence between the ideals of $R/\mathfrak a$ and the ideals lying over $\mathfrak a$. A maximal ideal $\mathfrak m$ is such that the only ideals lying over $\mathfrak m$ are $\mathfrak m$ and $(1)$.
Aug
13
revised Use Sylow's theorem to show that $G = HN_G(P)$
added 421 characters in body
Aug
13
answered Use Sylow's theorem to show that $G = HN_G(P)$
Aug
13
revised Is the series $\sum _{n=1}^{\infty } (-1)^n / {n^2}$ convergent or absolutely convergent?
deleted 2 characters in body
Aug
12
awarded  Nice Answer
Aug
12
answered Prove that the Möbius function is multiplicative
Aug
12
comment Prove that a function $f(n)$ counting the number of odd divisors multiplicative
Well, you stated what is to be proven... maybe a hint could help? =)
Aug
12
reviewed Approve suggested edit on Prove that a function $f(n)$ counting the number of odd divisors multiplicative