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Jun
18
comment Are the fields $\mathbb{Q}(\sqrt[7]{16}+3 \sqrt[7]{8})$ and $\mathbb{Q}(\sqrt[7]{16})$ equal?
@user222801 That doesn't get you anything directly because you haven't yet shown that either of $\sqrt[7]{8}$ or $\sqrt[7]{2}$ are in your extension.
Jun
18
comment Find the sum of reciprocals of divisors given the sum of divisors
Your argument still works fine in the case where $n$ is a perfect square - in that case, the middle divisor $d_m$ won't have any 'partner', but you have $d_m=\frac{n}{d_m}$ so it is (in effect) its own partner.
Jun
18
comment Quartic Solution on Wikipedia special cases problem $S=0$ how to “change the choice of cubic root”?
@MatrixPeckham en.wikipedia.org/wiki/Root-finding_algorithm is a good starting point. I second the recommendation that you may as well use a rootfinding algorithm - especially since whatever square root/cube root algorithm you're finding is doing that behind the scenes anyway.
Jun
17
comment Quadratic Equation Recurrence?
This feels very close to the Arithmetic-Geometric mean, but I don't see any way of reducing one recurrence to the other...
Jun
16
answered Using the Limit Comparison Test on $\sum_{n=1}^{\infty} \frac{n^2} {n!}$
Jun
15
comment Why is my solution “not enough”?
While that text is legible, it's barely so. If you could transcribe the notes into text+TeX (and maybe highlight the part that your professor suggests is not enough) that would be immensely helpful.
Jun
13
comment Why every finite non-abelian simple group of order $n$ has a proper subgroup of index at most $kn^{\frac{3}{7}}$?
How much group theory do you know? This sounds like a deep research-level result, not necessarily something that can be 'clarified' with any sort of simplicity...
Jun
13
comment Is this number prime or composite?Prove your answer
...91 is divisible by 3?
Jun
12
comment Is there a “C vs NC”-problem, where C stands for “constant time”?
It's not clear what you mean by 'purely symbolically' here, or for that matter by 'verify'. The latter, particularly, has a formal meaning that I don't think you're applying here.
Jun
12
comment Very tentative proof that the terms in Beal's Conjecture must not be squares?
@HyperLuminal To reinforce Andre's point - suppose (to choose a slightly different number) that $a=10$, $x=6$ (so $e=2$). Then $a^x=a^{2(e+1)}=1000000$, $a^{e+1}=1000$, but it could easily be the case that $\sqrt{c^{g+1}+b^{f+1}}=125$, $\sqrt{c^{g+1}-b^{f+1}}=8$, and neither is a multiple of $a$.
Jun
12
comment Is symmetric group on natural numbers countable?
This is harder to prove than the original claim (and still requires showing that these sequences are distinct), but it's still a cute approach to the problem.
Jun
11
comment When is a binomial coefficient a factorial, i.e. when is $\binom{m}{j} = n!$ for some $n,m,j$?
I do wonder whether the additional structure here offers enough to conclude that there are no more solutions for this special case. It seems unlikely (since in particular you would presumably need some results about e.g. $n$ and $n+1$ not being simultaneously smooth that could be really hard) but this case does seem more within reach than the general question of products-of-factorials.
Jun
11
comment When is a binomial coefficient a factorial, i.e. when is $\binom{m}{j} = n!$ for some $n,m,j$?
These have to be fairly rare. Consider: the largest binomial coefficient in row $m$, the central coefficient, is smaller than $2^m$ (since the sum of all the coefficients in that row is of that size). So for a given value $n$, since $n!\approx (n/e)^n$ (I'm neglecting any subexponential factors here), the minimal value of the row $m$ where we can expect to find $n!$ satisfies $2^m\approx (n/e)^n$ or $m\approx n\log_2n$. But then we have to 'dodge' all of the primes between $n$ and $m\approx n\log_2n$ as potential factors for our binomial coefficients, and this is going to be very hard to do.
Jun
11
comment Rationalisation of $\frac{(1-\sqrt{1-t})(1-\sqrt[3]{1-t})…(1-\sqrt[n]{1-t})}{t^{n-1}},t\in \mathbb{R}$
Seconded - this can't be correct as it stands; the former expression is clearly irrational when e.g. $n=2, t=\frac12$ but the latter is obviously rational.
Jun
10
comment $3x + 2 > 8$ solved not using order of operation?
Order of operation is implicit here - not in the solution of the equation but in the equation itself. In algebra when we write '$3x+2$' we always mean $(3\cdot x)+2$ and not $3\cdot(x+2)$; multiplication is always given tighter 'precedence' than addition.
Jun
9
comment Is it possible to create horizontal lines of arbitrary length in match-three games?
Great question! I just fixed a small typo in your question; while I was in there, I changed it to match the current version of the Q. Feel free to edit further if you'd rather it read differently.
Jun
9
revised Is it possible to create horizontal lines of arbitrary length in match-three games?
Typo fix, changed the title to match the current question
Jun
8
comment How to explain linear algebra to someone who knows nothing about it?
That gets to an interesting question, too: the adjacency matrix is clearly a matrix, but is it really part of linear algebra or is it more a piece of combinatorics? (Not that there isn't a lot of overlap, obviously!) I think to some extent that gets to what I'm saying: if the goal is to explain linear algebra rather than matrices, then while matrices are clearly a tool that can be used for linear algebra they're neither exclusive to the subject nor the whole of it, just a concept with a lot of overlap.
Jun
8
comment How to explain linear algebra to someone who knows nothing about it?
I feel like the 'box with numbers' is more about the implementation details than it is about linear algebra proper; if someone's ever seen a matrix then they might nod their heads a little but otherwise it seems like it might only make them more confused.
Jun
5
comment Minimal Polynomial of $\sqrt{2}+\sqrt{3}+\sqrt{5}$
I didn't mean linearly independent; I've swapped that terminology out for 'mutually irreducible', which I think gets more cleanly to the heart of things.