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Aug
21
comment Intuition behind $(-\frac{1}{2})! = \sqrt{\pi}$
I'm definitely not sure, but this seems strongly related to this question: math.stackexchange.com/questions/3444/…
Aug
21
comment Intuition behind $(-\frac{1}{2})! = \sqrt{\pi}$
@YellowSkies: wikiwand.com/en/Gamma_function
May
27
comment Solve this equation: $(x+2)(\sqrt{2x+3}-2\sqrt{x+1})+\sqrt{2x^2+5x+3}=1$
$-1$ seems to work...
May
12
comment Quadratic solutions puzzle
(-1/2, -1/2) is a solution, but it is also said in the constraints that $a \neq b$.
May
11
comment Confused about the prime elements of a ring.
@BillDubuque: Ok, thanks! It is just hard to wrap my head around. A lot of things in abstract algebra are requiring some serious rewiring of my brain. sigh
May
11
comment Confused about the prime elements of a ring.
@BillDubuque: Ok, so then we can construct any numerator as a unique combination of primes (well known property of primes) and choose a specific bottom element as a particular unit so this would imply that R is a UFD then?
May
11
comment Confused about the prime elements of a ring.
@BillDubuque: When you say $2$ is a unit, are you referring to the $p \times 2^{-1}$ is unnecessary just as $-1$ is unnecessary in prime factorization?
May
10
comment Need help determining what this center is isomorphic to
Oh, whoops my bad, but $Z$ should have order $2$. That was me slipping up there, I didn't mean to write that, just mixed things around sorry about that. Thanks! (Modified original comment.)
May
10
comment Need help determining what this center is isomorphic to
Ok, so I think the problem is I didn't know about p-group. In this case the argument should go: Since it is non-trivial, we know that the order is $2$, $4$ or $8$, but G is not abelian so the order cannot be $8$ and $2$ is guaranteed cyclic hence $Z$ has order $2$ and $|G/Z| = 4$
May
8
comment Help show the following isomorphism cannot exist.
Ok, so I think I get it, I made a small typo in my question, but was able to follow through and get $v^2 - v + 1 = 0$. But still, $x^2 - x + 1 \not\in \langle g(x) \rangle$. Thanks!
May
8
comment Help show the following isomorphism cannot exist.
I still applied simplifications as if it was $\mathbb{Z}_3$ if that is what you are wondering...
May
8
comment Help show the following isomorphism cannot exist.
Oops, sorry about that.
May
5
comment a question about abstract algebra,the order of $\Bbb Z_{5}[x]/ (x^3+x+1)$
Beat me too it too. +1
May
4
comment Finding an ideal such that $\mathbb{Z}[x]/I \cong \mathbb{Z}[i]$.
If understand this correctly: $f$ is the evaluation homomorphism, if we restrict the domain of $f : \mathbb{Z}[x] \to \mathbb{Z}[i]$ the same still holds. However, the first isomorphism theorem implies that the quotient group is isomorphic to this subgroup. However, $f$ is also surjective, so the quotient group is isomorphic to $\mathbb{Z}[i]$?
May
1
comment Show that if $x ≡ 1 (\text{mod } λ)$, then $x^3 ≡ 1 (\text{mod } λ^3)$…
@Ebearr: Sorry my bad. (Referring to deleted answer) I guess it seemed a little too simple. Haha. Should of thought of that.
Mar
28
comment Proving that a sequence is increasing
Cleaner than what I had wrote. +1 :) Also, I had a small hole in my writing now that I think about it...
Mar
20
comment Can I use Burnside's Theorem? Or should I take a different approach for this proof?
Thank you for the response. Just to make sure I am understanding this: $3|[G : G_x]$ since $|G|= 3^3$ And we know that $|G|/|G_x|$ must be an integer, resulting in either $3^3$, $3^2$ or $3$, hence $|X'| \equiv 32 \mod 3$ meaning $|X'|$ has a strong lower bound of at least $2$ (since 32 mod 3 = 2)?
Mar
19
comment How do you apply an element on the left of a permutation?
@ZachGershkoff: I was testing some small cases for a larger problem. I do not believe I am allowed to discuss the question that motivated this question, but I believe there was probably a flaw in the original question.
Mar
19
comment How do you apply an element on the left of a permutation?
Hmm, I was testing out small cases for a larger problem. After reviewing your answer and the comments, I believe the question that motivated this question has a flaw in it. I would post the original but, I do not believe I am allowed to do so. Thanks for the help!
Mar
18
comment What about my proof is “nonsense”?
@MikeMiller: I was thinking, that $x = ah$ for some $a \in G$ and $h \in H (\leq G)$, by closure of multiplication it has to be a member of $G$? Does this not make that much sense?