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 1d comment Of which sets does the $\sigma$-algebra generated by the first $n$ one-point sets of $\mathbb{N}$ consist? I was just pointing out that your claim could be simplified to just saying that $\sigma(\mathcal{E}_n)$ was the power set of $\{1,...,n\}.$ Making the claim simpler seems better proof format, in my opinion. 1d comment Sum of two normal numbers need not be a normal one Should there be a division by $n$ somewhere before the sum, so that one is averaging the digits up to the $n$th? (and then taking the limit) 1d comment Of which sets does the $\sigma$-algebra generated by the first $n$ one-point sets of $\mathbb{N}$ consist? You don't need the $\mathcal (P)^c$ since as $A$ ranges over the subsets of $B$ so does $A^c.$ What I mean is your definition of $\sigma(\mathcal (E)_n)$ could instead be simply the power set of $\{1,...,n\}.$ 1d comment The equivalence relation generated by a relation @Zuriel I found the problem interesting. At first I thought it wouldn;t be true by looking at the relation $r$ on $\mathbb{Z}$ consisting of all pairs $(t,t+1).$ Here $R$ includes all pairs of integers, and I thought it might not be possible to have the third bullet. But for any specific pair $(x,y)$ one needs only a chain of finite length., 1d comment pos of matrix is close Please at least include an example or two of such "pos" matrices, or pos's of matrices if that makes more sense. This is an unfamiliar definition so examples might improve the question. 1d comment Show that point you found for $x^2 - 2y^2 = 1$ using line through$(1, 0)$ has integer coordinates. You have the one point $(1,0)$ and the other found by intersecting $y=m(x-1)$ with the curve, whose coordinates are certain fractional expressions involving the given slope $m.$ Then if one puts $m=u/v$ there are then still only two points so far, so it's unclear what the phrase "the point found previously" refers to. 2d revised The equivalence relation generated by a relation added 124 characters in body 2d answered The equivalence relation generated by a relation 2d comment Prove that $T$ has an orbit of period 3 There remains one too many copies of $T(x_2)=x_3$ in your explanation of what you started to try doing. 2d accepted Groups with “few” subgroups 2d comment Groups with “few” subgroups Derek-- Nice proof (seems to me). +1 on it for now, and likely I'll "accept" after I get a chance to go over it in detail. 2d comment Groups with “few” subgroups @j.p. That idea looks promising. One thing, should the phrase "is less than $d$" be replaced by "is at most $d$" in your comment? [though that would imply your version, I seem to recall one approach to doing the counting argument for showing some group is cyclic as involving the number being at most $d$] And +1 on the comment. 2d awarded Nice Question 2d asked Groups with “few” subgroups Apr21 comment Prove this inequality $25ab+25a+10b\le38$ With $(a,b)=(\cos t, \sin t)$ the left side of your expression is a trig function $f(t)$ which has a graph with max $38$. Using $c,s$ for the cosine and sine, the derivative of $f(t)$ isn't that bad: $50c^2+10c-25s-25.$ It should be relatively easy from here to solve where $f'(t)=0.$ Apr20 answered Perceptrons that recognize AND, OR, NOT Apr20 comment What is the explicit formula (solution) to this recursively defined binary matrix? The sibling matrix formula looks right, and interesting. Likely one can get the other one from it. +1 on question... Apr17 comment Perceptrons that recognize AND, OR, NOT What's a perceptron weight? Might be good to include that in post... Apr17 comment Partial Ordering and Hasse Diagram. This needs a line from 2 to 42 because 2|42. Also 42|210 so another line needed. Apr16 comment Prove that $x^2 - 2013^2 \le y \le 2013^2 - x^2$ has an odd number of solutions @achillehui Yes I agree with you that given there are solutions, the $x$ ends up bounded, so a finite number of $y$ in such a case. Of course if $x^2<2013^2$ there are no solutions to the OP inequality, and to be picky $0$ is even not odd, so I assume that wasn't the case.