4,139 reputation
822
bio website shapetales.wordpress.com
location Prague, Czech Republic
age 28
visits member for 3 years, 11 months
seen 13 hours ago

Math PhD. student and functional programming enthusiast.


Jan
7
comment What's the explanation for why n^2+1 is never divisible by 3?
@fretty: it seems to me OP understands the proof perfectly. But knowledge of a topic doesn't come from perfect understanding of one proof but from knowing multiple proofs and how the topic connects with rest of mathematics. As the many answers show, there is much more going on here than meets the eye. Kudos to OP for asking this question.
Jan
7
comment What's the explanation for why n^2+1 is never divisible by 3?
@fretty: calm your temper. There are many ways to prove a theorem and not all proofs are equally good. Some are easy to explain, some are short, some are clever, some are easy to generalize in different directions. Asking for a different proof actually shows a potential in OP. I know great many a student that are satisfied with reading one proof, thinking they know everything there is to know.
Jan
7
comment What's the explanation for why n^2+1 is never divisible by 3?
Let's generalize the result to $p \not | n^{p-1} + 1$ when $p > 2$ and $p \not | n^{(p-1)/2} + 2$ when $p > 3$.
Jan
7
comment What's the explanation for why n^2+1 is never divisible by 3?
Yeah, but the cost of not using that language is that the proof is not a one-liner anymore. Sometimes a little different packaging makes for a very different product :)
Jan
7
comment What's the explanation for why n^2+1 is never divisible by 3?
I'll just add that the simplest way to see this is of course by using modular arithmetic, since $2^2 \equiv 1^2 \equiv 1 \pmod{3}$. But I guess that doesn't satisfy the requirement of assuming OP doesn't know algebra.
Jan
7
comment What's the explanation for why n^2+1 is never divisible by 3?
Beaten to the same answer. Good for you, +1 :)
Jan
7
answered Proof of the correctness of Taylor series
Jan
7
comment Point set topology from an algebraic perspective?
Have you seen this article? en.wikipedia.org/wiki/Kuratowski_closure_axioms
Jan
7
comment The set of all sections of a vector bundle
@MontyGill: great! You're welcome.
Jan
7
comment The set of all sections of a vector bundle
@MontyGill The map $s: U \to U \times {\bf R}^n$ is continuous. Since a map to a product $s: x \mapsto (x, t(x))$ is continuous precisely when both component maps are it follows that the map $t: U \to {\bf R}^n$ is continuous.
Jan
7
answered The set of all sections of a vector bundle
Jan
7
comment An involutive property of the quotient bundle coming from a Lie group action.
I see no reason why $Q$ should be involutive in general. $Q/G$ is involutive but that's only because it has full rank, so that's not a good argument.
Jan
7
comment Why $I=\left\{p(x)\in \mathbb{Z}\left[X\right]:2\mid p(0)\right\}$ is not a principal ideal?
@Yoav: ${\bf Z}[X]$ is two-dimensional (since $\bf Z$ is one-dimensional and taking polynomials adds a dimension). Maximal ideals correspond to points, hence have dimension zero. Principal ideals on the other are generated by a single element, hence correspond to one-dimensional curves. This is an intuition for why a maximal ideal is not principal (it would be a full proof in ${\bf C}[X, Y]$, say, since maximal ideals correspond to points in $\bf C$ and principal ideals to curves). But these are just aside comments, you have complete answer given, so try to understand it before asking more.
Jan
7
comment Why is Euler's totient function equal to $(p-1)(q-1)$ when $N=pq$ and $p$ and $q$ are prime in a clean intuitive way?
The number is coprime to $pq$ when $n \not \equiv 0 \pmod{p}$ and $n \not \equiv 0 \pmod{q}$. There are $p-1$ solutions of the first equation in integers mod $p$ and $q-1$ solutions of the second equation in integers mod $q$. So there must be $(p-1)(q-1)$ solutions in integers mod $pq$. Alternatively, a number is a unit in integers mod $pq$ if and only if it is a unit both in mod $p$ and mod $q$. All of this is just rephrasing Isaac's comment on Chinese remainder theorem.
Jan
7
answered Fundamental group of projective plane is $C_{2}$???
Jan
6
comment A “complementary” topology
You probably know that Alexandrov topologies correspond to preordered sets. So the dual topology corresponds to reversing the preorder. I'm not sure what kind of properties you are looking for but obviously in general there can be as little or as much relation between the duals as you wish. E.g. on one hand you can consider spaces having only clopen sets, so these spaces are self-dual; on the other hand, you can consider spaces with no proper clopen sets which can have very different duals. Also, remember that finite topological spaces are all Alexandrov, so try to look at those as well.
Jan
6
answered $V = W\oplus W^\perp$
Jan
6
comment Integral domain whose irreducible elements are not prime
Ah, a domain without non-trivial prime ideals must contain no non-trivial ideals at all (since every ideal is contained in a maximal ideal which is prime) and thus be a field. So we're indeed looking for rings having only non-principal prime ideals.
Jan
6
comment Integral domain whose irreducible elements are not prime
It's true that an integral domain with no prime ideals is a field (actually, I don't even have proof of this much weaker statement but it seems plausible). But can't it happen that a ring has only non-principal prime ideals therefore having no primes?
Jan
6
comment Integral domain whose irreducible elements are not prime
@Gaffney: can you give a proof?