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| bio | website | |
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| location | ||
| age | ||
| visits | member for | 8 months |
| seen | 3 hours ago | |
| stats | profile views | 46 |
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1d |
answered | A question about direct sums of subspaces… |
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1d |
answered | Find $u\in\mathbb{R}$ such that $\mathbb{Q}(u) = \mathbb{Q}(2^{1/2}, 5^{1/3})$. |
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May 14 |
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$i^{th}$ root(s) of unity Whatever the i-th roots of unity are, $e^{2\pi}$ is one. Do you count this as meaningful? :) |
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May 14 |
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Calculating upper eccentricity in a graph Please link us to the actual paper. (Google results change depending on who is searching and where they're searching from!) |
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May 14 |
answered | How do you graph $x + y + z = 1$ without using graphing devices? |
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May 14 |
answered | Infinitely many primes of the form $4n+3$ |
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May 14 |
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if $\sin24^\circ = p$ what is $\cos24^\circ$? I downvoted for the line $\cos A = \sqrt{1 - \sin^2 A}$, which is false, e.g. when $A = 270^\circ$. It should read $\cos A = \pm\sqrt{1 - \sin^2 A}$ - you have to do a little more work to find out which of $+$ or $-$ is appropriate. |
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May 14 |
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dimension of a quotient space Yes. This is true in the second case too. (Can you prove it?) That's enough of a reason. (But if you like, you might want to think about this: if $V$ is the space of polynomials of degree $\leq 50$, $W$ is the space of polynomials in $V$ divisible by $x^4$, and $U$ is the space of polynomials divisible by $x^{51}$, can you interpret $V' = V\oplus U$ as the space of all polynomials and $W' = W\oplus U$ as the space of polynomials in $V'$ divisible by $x^4$? Then, when we try to work out $V'/W'$, the summand $U$ simply cancels: informally, degree $\geq 51$ terms are irrelevant in the quotient.) |
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May 14 |
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dimension of a quotient space Well, why is dim(V/W) = 4 in the first case? (Can you pick a basis for V/W?) |
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May 14 |
answered | Homomorphism from $\mathbb{Z}/n\mathbb{Z}$ |
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May 14 |
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which of the following statements are true in ring theory (a) Okay - so, can you give me a commutative ring with unity, and a non-prime ideal? If so, calculate R/I, and see what you get. (b) What's your definition of R[x]? Can x ever be a unit, for example? |
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May 14 |
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Determine no. elements in $ \operatorname{Aut}(H)$ where $H$ is the 6 point/5line graph in the shape of H (By the way: don't worry about it too much. Groups are weird. No matter how much you think you understand them, someone else can prove to you that you don't. :)) |
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May 14 |
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Determine no. elements in $ \operatorname{Aut}(H)$ where $H$ is the 6 point/5line graph in the shape of H Are you sure? I get (1 6 3 4)(2 5), which isn't the same as $\sigma_3$. (Alternative 1: draw pictures and do this multiplication geometrically. Alternative 2: notice that the identity and your $\sigma$s are all rigid motions, whereas my (1 3) is a 'twist', as it flips one leg and leaves the other alone. Doing a rigid motion followed by a twist can't give you a rigid motion. Alternative 3 (a translation of 2): call the group of size 4 I mentioned "G". Then (1 3) and the identity are in different cosets (of G in Aut(H)), so $\sigma_2(1\;3) = \sigma_3$ (and so $G(1\;3)=G$) is impossible.) |
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May 13 |
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Determine no. elements in $ \operatorname{Aut}(H)$ where $H$ is the 6 point/5line graph in the shape of H It's definitely not the case that there are no others. (For a group-theoretic reason: the group {identity, $\sigma_1$, $\sigma_2$, $\sigma_3$} is a subgroup of Aut(H), so Aut(H) must have size divisible by 4. Practically: try multiplying (1 3) by $\sigma_2$ and see whether you get anything new - I suspect you will.) When you think you're done, try to prove there's nothing else that could possibly happen. (Here's a purely combinatorial answer: let $f$ be an automorphism of $H$. How many choices are there for $f(1)$? How many choices are there for $f(2)$ if you know $f(1)$? etc.) |
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May 13 |
answered | Is iteration process through equal sign possible? |
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May 13 |
revised |
Determine no. elements in $ \operatorname{Aut}(H)$ where $H$ is the 6 point/5line graph in the shape of H edited body |
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May 13 |
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Determine no. elements in $ \operatorname{Aut}(H)$ where $H$ is the 6 point/5line graph in the shape of H By the way, if you're just looking for the number of elements in Aut(H), it might be a little easier to apply the orbit-stabiliser theorem. But I think this is good practice too. |
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May 13 |
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Real Projective Space Well, what have you tried? |
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May 13 |
answered | Determine no. elements in $ \operatorname{Aut}(H)$ where $H$ is the 6 point/5line graph in the shape of H |
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May 13 |
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Property of Hom-functor Is $\mathbb{Z}$ an $\mathbb{F}_2$-module? Or a $\mathbb{C}$-module? (Note that the identity elements $1_{\mathbb{F}_2}$ and $1_{\mathbb{C}}$ must act trivially on $\mathbb{Z}$.) |
