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 Jan8 answered Nets and Convergence: Why directed indices? Jan8 comment Hilbert's syzygy theorem in the analytic setting I don't know much about this topic but I think you mean finite resolutions. Infinite ones should always exist. Anyway, nice question. Jan8 comment Finding conjugacy classes of $D_{10}$ $(g\tau g^{-1})^2 = g\tau g^{-1}g\tau g^{-1} = g \tau^2 g^{-1} = e$. Jan7 comment What's the explanation for why n^2+1 is never divisible by 3? This is the best answer given yet in my opinion. Not necessarily for the OP, but certainly for me because it strikes right at the heart of the problem. Jan7 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. Jan7 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. Jan7 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 $p > 3$. Jan7 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 :) Jan7 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. Jan7 comment What's the explanation for why n^2+1 is never divisible by 3? Beaten to the same answer. Good for you, +1 :) Jan7 answered Proof of the correctness of Taylor series Jan7 comment Point set topology from an algebraic perspective? Have you seen this article? en.wikipedia.org/wiki/Kuratowski_closure_axioms Jan7 comment The set of all sections of a vector bundle @MontyGill: great! You're welcome. Jan7 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. Jan7 answered The set of all sections of a vector bundle Jan7 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. Jan7 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. Jan7 answered Fundamental group of projective plane is $C_{2}$??? Jan6 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. Jan6 answered $V = W\oplus W^\perp$