12,827 reputation
11433
bio website math.brown.edu/~lubinj
location Minnesota
age 78
visits member for 2 years, 10 months
seen 12 hours ago

Aged mathematician


Jul
24
comment Class group of $\mathbb{Q}(\sqrt[4]{-2})$
@AdamHughes, there’s something wrong with your analysis, since $\sqrt2+1$ is a unit of $\mathbb Z[\sqrt2\,]$
Jul
24
answered How do we know that $\mathbb{Q}$ is the initial field of characteristic $0$?
Jul
23
comment Quick question on the basis of subset of polynomals
I like this answer better than my own more complete one.
Jul
23
comment Quick question on the basis of subset of polynomals
Because for the function $p(x)=1$ identically, $p(6)\ne0$.
Jul
23
answered Quick question on the basis of subset of polynomals
Jul
23
comment How to solve for $x$ for $\frac{1}2 x^{-1/2}- \frac14x^{-3/4}$
Strictly speaking, it’s only an equation that may be solved for one of the variables. Without an equals sign, it’s not automatically clear what you’re asking for. In fact, I assumed that you had $y$ equaling the expression you wrote, and wanted to solve that for $x$ (in terms of $y$).
Jul
22
comment $\Bbb Z^\ast$ What is this notation?
For an algebraist, $\mathbb Z^*$ would certainly be the set of invertible elements in the multiplicative monoid, i.e. $\{\pm1\}$, but that interpretation seems excluded by the context. Perhaps the notation is peculiar to the Brazilian educational system.
Jul
22
comment How are $\pi/4$ and $3\pi/4$ solutions to $\sec^2 \theta -2 = 0$?
It certainly isn’t true that $\sec\theta$ never equals $2$. Indeed, that’s the same as cosine equaling $1/2$, which you know all about.
Jul
21
comment Polynomial division problem
In other words, add up all the even-degree coeffs and you get the constant term; add up all the odd-deg coeffs and you get the linear term.
Jul
20
comment Mistake in a question in Fulton's algebraic curves book?
Just by substitution. Can’t you express every polynomial in the $X$s as, instead, a polynomial in the $Y$s? And vice versa?
Jul
20
comment Intuition behind topological spaces
Just because every neighborhood of $0$, and in particular each neighborhood $\langle-\varepsilon,\varepsilon\rangle$ has nonempty intersection with that open set.
Jul
20
comment Fibonacci number that ends with 2014 zeros?
All of this must be known to Fib specialists, but that term doesn’t apply to me. By a cumbersome $p$-adic computation ($p=2,5$), I seem to have shown that for $m\ge3$, $2^m|F_k$ if and only if $3\cdot2^{m-2}|k$. And that $5^m|F_k$ if and only if $5^m|k$. This would show that your $750$, etc. are the smallest numbers satisfying the desired conditions.
Jul
19
comment Thinking of sohcahtoa with 90 in a triangle.
Of course the sine and cosine of a right angle can’t be defined using sohcahtoa, but one you believe the laws of sines and cosines, this is the best way.
Jul
19
answered Mistake in a question in Fulton's algebraic curves book?
Jul
18
comment x≡1 (mod 8) x≡9 (mod 12) has solution x = x_0. How many solutions mod 24 are there to the system of congruences?
It is easy: you look at all the numbers from $0$ to $23$ inclusive, and see which of them satisfy the given conditions. That’s pretty much it.
Jul
17
comment Inverted Circle?
I like your second formulation because it allows quick and easy hand computation of loads of rational points.
Jul
16
comment The $i^{th}$ prime in a given ring R
In spite of what both @PatrickDaSilva and I have said, I think the rings that come closest to your ideal are the rings of integers in quadratic imaginary fields which are unique factorization domains. Unfortunately, there are only finitely many of these. But as for $\mathbb Z[i]$, which is more or less typical, you at least have only at most two primes dividing each ordinary prime, and you can, again in spite of what we said, make a reasonable choice between each one of them and between the prime and it times a unit (which is usually just $\pm1$).
Jul
16
answered The $i^{th}$ prime in a given ring R
Jul
16
answered Help with composite functions?
Jul
12
comment $E \subset \mathbb R$ is an Interval $\iff E$ Is connected
It seems to me that the text’s proof is not completely right. Disconnectedness of $E$ means that there are disjoint open subsets of $E$ with the specified property. So there would be open subsets $A$ and $B$ of $\mathbb R$ with $A\cap B\cap E=\emptyset$, not merely $A\cap B=\emptyset$.