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Oct
13
comment Prove that in $\mathbb{R}$, if $|a-b|>\alpha$ for all $a\in A$ and $b\in B$, then outer measure $m^*(A\cup B)=m^*(A)\cup m^*(B)$
I can see it's not, and I feel like your getting at compactness here... I thought about this one all last night though and I still can't figure it out. You think you could help me out a bit more?
Oct
13
asked Prove that in $\mathbb{R}$, if $|a-b|>\alpha$ for all $a\in A$ and $b\in B$, then outer measure $m^*(A\cup B)=m^*(A)\cup m^*(B)$
Oct
12
revised $\int_{\gamma}\frac{dz}{\sqrt{1-z^2}}=2\pi$ Along the path $\gamma(t)=2e^{it}$ for $0\leq t\leq 2\pi$ Implies Sine is $2\pi$ periodic
added 277 characters in body
Oct
12
asked $\int_{\gamma}\frac{dz}{\sqrt{1-z^2}}=2\pi$ Along the path $\gamma(t)=2e^{it}$ for $0\leq t\leq 2\pi$ Implies Sine is $2\pi$ periodic
Oct
6
accepted Comparing the homomorphism $\mathbb{Z}[x]\rightarrow\mathbb{Z}_n[x]$ with the homomorphism $\mathbb{Z}[x]\rightarrow\mathbb{F}_p[x]$
Oct
6
comment Comparing the homomorphism $\mathbb{Z}[x]\rightarrow\mathbb{Z}_n[x]$ with the homomorphism $\mathbb{Z}[x]\rightarrow\mathbb{F}_p[x]$
lol, we were talking about roots, but we were also talking about reducibility more generally I thought.
Oct
6
comment Comparing the homomorphism $\mathbb{Z}[x]\rightarrow\mathbb{Z}_n[x]$ with the homomorphism $\mathbb{Z}[x]\rightarrow\mathbb{F}_p[x]$
But what about $4x^4-1 = (2x^2 +1)(2x^2-1)$. Over $\mathbb{Z}_4$ that becomes $3$. Yet $3$ is irreducible over $\mathbb{Z}_4$, in fact how can we even make sense of things when the degree drops? Or are you just talking about roots?
Oct
6
comment Comparing the homomorphism $\mathbb{Z}[x]\rightarrow\mathbb{Z}_n[x]$ with the homomorphism $\mathbb{Z}[x]\rightarrow\mathbb{F}_p[x]$
Sorry if I came off as curt, I wasn't intending to be. Ok so what you're saying is that there is nothing special about $n$ being prime, and whether $n$ is prime or not, any level of irreducibility over $\mathbb{Z}_n$ must be mimicked over $\mathbb{Z}$, (assuming the degree of the poly doesn't drop under the homomorphism).
Oct
6
comment Comparing the homomorphism $\mathbb{Z}[x]\rightarrow\mathbb{Z}_n[x]$ with the homomorphism $\mathbb{Z}[x]\rightarrow\mathbb{F}_p[x]$
Ok cool, so it's for reasons of efficiency that we choose $p$ prime, good to see that I'm not going crazy here.
Oct
6
comment Comparing the homomorphism $\mathbb{Z}[x]\rightarrow\mathbb{Z}_n[x]$ with the homomorphism $\mathbb{Z}[x]\rightarrow\mathbb{F}_p[x]$
Yah I know that $\mathbb{F}_p\cong\mathbb{Z}_p$, I was just criticized previously for using the latter notation when I'm looking at it as a field instead of a group. I'm confused by your mention of quartics however, it's for polynomials of all degree correct?
Oct
6
asked Comparing the homomorphism $\mathbb{Z}[x]\rightarrow\mathbb{Z}_n[x]$ with the homomorphism $\mathbb{Z}[x]\rightarrow\mathbb{F}_p[x]$
Oct
1
accepted Trying to prove that $p$ prime divides $\binom{p-1}{k} + \binom{p-2}{k-1} + \cdots +\binom{p-k}{1} + 1$
Oct
1
comment Trying to prove that $p$ prime divides $\binom{p-1}{k} + \binom{p-2}{k-1} + \cdots +\binom{p-k}{1} + 1$
Ah! Pascal's Rule setting $1$ = $p-k$ choose $0$, thanks!
Oct
1
comment Trying to prove that $p$ prime divides $\binom{p-1}{k} + \binom{p-2}{k-1} + \cdots +\binom{p-k}{1} + 1$
Thank you for your explanation Hagen =] +1.
Oct
1
comment Trying to prove that $p$ prime divides $\binom{p-1}{k} + \binom{p-2}{k-1} + \cdots +\binom{p-k}{1} + 1$
@Gerry: Ah! I see, you have to write 1 as $p-k$ choose $0$! Nice thanks.
Sep
30
comment Trying to prove that $p$ prime divides $\binom{p-1}{k} + \binom{p-2}{k-1} + \cdots +\binom{p-k}{1} + 1$
Hey Hagen could you show me how that whole thing sums up to p choose k? Thanks.
Sep
30
revised Trying to prove that $p$ prime divides $\binom{p-1}{k} + \binom{p-2}{k-1} + \cdots +\binom{p-k}{1} + 1$
edited body
Sep
30
comment Trying to prove that $p$ prime divides $\binom{p-1}{k} + \binom{p-2}{k-1} + \cdots +\binom{p-k}{1} + 1$
Yes you're right I should have put "any".
Sep
30
revised Trying to prove that $p$ prime divides $\binom{p-1}{k} + \binom{p-2}{k-1} + \cdots +\binom{p-k}{1} + 1$
deleted 1 characters in body
Sep
30
comment Trying to prove that $p$ prime divides $\binom{p-1}{k} + \binom{p-2}{k-1} + \cdots +\binom{p-k}{1} + 1$
@Martin: Thanks I'll look into that.