Reputation
2,658
Top tag
Next privilege 3,000 Rep.
Cast close & reopen votes
Badges
3 22
Newest
 Tumbleweed
Impact
~26k people reached

Jun
22
revised Diagonals of $2n$-gon bisecting area implies?
edited tags
Jun
21
asked Diagonals of $2n$-gon bisecting area implies?
Jun
18
awarded  Tumbleweed
Jun
17
comment Fundamental polygon square $abab$
He identified all of the vertices here, though, so it should only have one vertex. Also I lied about the homology; it has the same homology as that of the klein bottle
Jun
17
comment Fundamental polygon square $abab$
you could look at the euler characteristic, which is zero, so if it's actually a surface, then it would be either the klein bottle or torus. unless i'm mistaken, calculating the homology (using the cellular boundary maps) makes it clear that this isn't the case, so it's not a surface.
Jun
11
asked $d \log f'$ is a compactly supported radon measure for $f\in C_c^{1}$ with bounded variation
Jun
11
asked Multiple of $p$ in first $p+1$ Fibonacci Numbers
Jun
4
comment A logic problem about set theory
fixed, thank you
Jun
4
revised A logic problem about set theory
deleted 8 characters in body
Jun
4
answered A logic problem about set theory
Jun
4
answered Prove that $n(r) < 2\pi \sqrt[3]{r^{2}}$
May
24
revised Sophomore's dream changing “x”
added 349 characters in body
May
24
answered Sophomore's dream changing “x”
May
23
comment Find $\int_c \bar z$ along the parabola $y=x^2$ from $(0,0)$ to $(1,1)$
when parametrizing your curve by $x+ix^2$ from $0$ to $1,$ the derivative is $1+2ix$ so your integral should be $\int_{0}^{1} (x-ix^2)(1+2ix) \mathrm{d}x$
May
22
comment simplicial homology definition
the simplex $\Delta^n$ lives in $\mathbb{R^{n+1}}$ so when he says $e_{\alpha}^n$ is an $n$-simplex of $X$ this is really the canonical map $\sigma_\alpha: \Delta^{n}\to X$
May
22
comment A consequence of the inequality $\pi(x)+\pi(y)\ge\pi(x+y)$
that's what i'm saying. Your "proof" doesn't work by my comments above
May
22
comment A consequence of the inequality $\pi(x)+\pi(y)\ge\pi(x+y)$
Also, note that it's possible that $\pi(p_n-p_m+1) = \pi(p_n - p_m) + 1,$ so then the left hand side becomes $\pi(p_m) + \pi(p_n-p_m),$ which doesn't tell you anything new.
May
22
comment A consequence of the inequality $\pi(x)+\pi(y)\ge\pi(x+y)$
That's not strong enough to give strict inequality. Try, for example, $m=2$ so $p_m = 3$ (in particular then your inequality is not true from the hypothesis). Then the question becomes to determine when $\pi(p_n-3) > \pi(p_n) - 2.$ If $p_n - 2$ and $p_n$ are twin primes, then you're in trouble.
May
22
comment Tangent bundle of manifold with no odd dimensional sub-bundles
@JoeS the orientable double cover of a manifold is connected only when the manifold is not orientable, otherwise you get two disjoint copies. I'll think about what you wrote more closely but that's what i meant
May
18
answered A consequence of the inequality $\pi(x)+\pi(y)\ge\pi(x+y)$