Reputation
6,748
Next privilege 10,000 Rep.
Access moderator tools
Badges
1 11 35
Impact
~73k people reached

Feb
7
comment How can we prove a statement is provable?
I suspect the answer, in the vast majority of specific cases, is going to be, quite simply, "We don't." I've never heard of a non-independence result that doesn't itself discern whether the statement is true or false. I would be interested in finding out if such a thing exists - by the consistency theorem, you could start with two models, one of the statement and one of its negation, and try and derive a contradiction.
Feb
4
comment Why the canonical bundle of a complex manifold is a line bundle?
Well, the Wikipedia article on this topic does give the definitions and basis. Specifically, if $\{e_1, \ldots, e_n\}$ is a basis for $V$, then $\{e_{i_1} \land \cdots \land e_{i_k} \mid 1 \le i_1 < \cdots < i_k \le n\}$ forms a basis for $\Lambda^k V$. Since the basis is enumerated by ordered sets of $k$ distinct elements from the basis of $V$, the dimension formula follows.
Feb
4
comment Why the canonical bundle of a complex manifold is a line bundle?
Presumably you mean the $m^{\text{th}}$ exterior product of the tangent bundle of an $m$-dimensional complex manifold forms a line bundle. This is really a linear algebra problem on the level of tangent spaces: for a vector space $V$, you define the exterior powers $\Lambda^k V$. If $V$ has dimension $n$, then $\Lambda^k V$ can readily be seen to have dimension $\binom{n}{k}$.
Feb
3
comment Deducing the compactness theorem from the completeness theorem (in first order logic)
Every proof uses only finitely many statements from $\Sigma$.
Feb
2
comment If $G \cong \mathbb Z/3\mathbb Z$, then $\mathbb R[G] \cong \mathbb R \times \mathbb C$
Note that knowing what $\varphi(g)$ is actually determines the ring homomorphism (assuming $\mathbb R$-linearity), and $\varphi(g)$ must be carried to a primitive cube root of unity. There aren't very many choices in $\mathbb R \times \mathbb C$ for cube roots of unity.
Feb
1
comment How many non-negative integer solutions are there for the equation $x+y+z = 11$ when $x \geq 1$, $y \geq 2$, and $z \geq 3$?
... although I have never read the Wikipedia entry on that before, and frankly, the statement of the two theorems on there looks pretty horrendous...
Feb
1
comment How many non-negative integer solutions are there for the equation $x+y+z = 11$ when $x \geq 1$, $y \geq 2$, and $z \geq 3$?
This is an instance of the stars and bars problem.
Jan
31
comment Why can quotient groups only be defined for subgroups?
@downvoter Why... ?
Jan
30
revised Why can quotient groups only be defined for subgroups?
added converse
Jan
30
answered Why can quotient groups only be defined for subgroups?
Jan
27
comment Sharing a pepperoni pizza with your worst enemy
Oh I see. ${}{}$
Jan
27
comment Sharing a pepperoni pizza with your worst enemy
"Number the slices $1,2,1,2,1,2,1,2,$ such that any two consecutive slices have different numbers." Why are you able to assume this?
Jan
26
comment Prove that identity matrix is the only idempotent $n x n$ matrix that is invertible.
That's perfect, and I suspect that is as good a proof as you're going to get.
Jan
24
reviewed Approve proof of a convergent subrow in every row in B using diagonal argument
Jan
24
reviewed Approve Proving associative property, floor function
Jan
23
comment Prove that $\{ (x,y) \in \mathbb{R}^2 \mid -2 < x - y^2 < 8 \}$ is open.
I mean it's probably right - I haven't checked the details, because, as you say, it's messy. It's just a highly nonstandard use of the symbols $\epsilon$ and $\delta$. Usually they refer to the definition of continuity, which isn't how you're using them here. But you probably shouldn't listen to what I say; I'm an algebraist, not an analyst.
Jan
22
comment Prove that $\{ (x,y) \in \mathbb{R}^2 \mid -2 < x - y^2 < 8 \}$ is open.
That's not how you use $\epsilon$ and $\delta$.
Jan
21
comment Every matrix $A\in M_2(\mathbb{C})$ is similar to one of two forms
Why do you have $au-T(u)=-v$? It seems you've assumed $a_2=1$ to get that. Note that it is entirely possible to have $a_2 \neq 1$ - you will need to find a possibly new basis to get that coordinate to be $1$.
Jan
21
comment Every matrix $A\in M_2(\mathbb{C})$ is similar to one of two forms
Note that the diagonal elements of a lower triangular matrix give the roots of the characteristic polynomial, so you know $a_1 = a$ right away. So $a_2$ is the only part that takes any real work.
Jan
18
comment Short exact sequence of $\mathbb{R}[X]$-modules that does not split
@Bob1993 your example works as well. It's just not following egreg's hint.