Brian M. Scott
Reputation
398/400 score
 Jan 15 comment Algorithm for finding “fact families” @Mario: It appears from a bit of searching on the web that a ‘fact family’ is a set of four related arithmetic statements. Specifically, an addition and subtraction fact family is a set \left\{\begin{align*} &a+b=c\\ &b+a=c\\ &c-a=b\\ &c-b=a\;, \end{align*}\right. and a multiplication and division fact family is a set \left\{\begin{align*} &a\times b=c\\ &b\times a=c\\ &c\div a=b\\ &c\div b=a\;. \end{align*}\right. $2\times 3=6$ is simply a representative of its family, not the whole family. Jan 15 comment Is this a topological closure operation? @Lehs: My pleasure! Jan 15 comment Is this a topological closure operation? @Lehs: Yes. Suppose that $x\in\overline A$ and $A\subseteq B$. If $x\in B$, then of course $x\in\overline B$. If $x\notin B$, then $x\ge\min(\Bbb N\setminus B)$, but also $x=\min(\Bbb N\setminus A\le\min(\Bbb N\setminus B)$, so $x=\min(\Bbb N\setminus B)\in\overline B$. Jan 15 comment Is this a topological closure operation? @Lehs: It was intended to, yes, but I just realized that I was wrong about it. In fact that new axiom is consistent with any ordinary topological closure, but it’s not strong enough, as my $\Bbb N$ example shows: it still satisfies that new axiom. Jan 15 comment Is this a topological closure operation? @Lehs: The faulty fourth paragraph. Jan 15 comment How to prove that the Nested Interval Theorem fails to hold in $\mathbb Q$? @Cleggstein: You’re welcome! Jan 15 answered voting game combinatorics Jan 15 comment finding $T_n=S_1+S_2+…+S_n$ when $S_n=(2n-1)+(2n+1)+(2n+3)+…+(4n-1)+(4n+1)$ @ali: You’re very welcome. Jan 15 answered What is $1 + 999999…$ (an infinite string of $9$s)? Jan 15 comment What is $1 + 999999…$ (an infinite string of $9$s)? It actually makes more sense if you think of the string as infinite to the left: $\ldots 999$. Add $1$ to it: the carry propagates indefinitely, and you get $\ldots 000$. In fact, this is the $10$-adic expansion of $-1$. Jan 15 comment How to prove that the Nested Interval Theorem fails to hold in $\mathbb Q$? @Cleggstein: Suppose that $y\in\bigcap_n[a_n,b_n]$. Then $y\ge a_n$ for all $n$, so $y\ge x=\sup_na_n$. Similarly, $y\le b_n$ for all $n$, so $y\le x=\inf_nb_n$. Therefore $y=x$. Jan 15 comment Discrete Mathematics, set theory power set question. Your set has three elements: $\varnothing$, $\{\varnothing\}$, and the third element, which is missing a comma: it should be $\big\{\varnothing,\{\varnothing\}\big\}$. $\wp(S)$ will have $2^3=8$ elements, but the answer isn’t $8$: you’re supposed to find the $8$ elements of $\wp(S)$. Jan 15 comment Is the Center of Math Wrong? @user2520938: You are a master of meiosis. Jan 15 comment finding $T_n=S_1+S_2+…+S_n$ when $S_n=(2n-1)+(2n+1)+(2n+3)+…+(4n-1)+(4n+1)$ @ali: You’re welcome. Finite calculus is pretty neat; there’s also a good treatment of it in Graham, Knuth, and Patashnik, Concrete Mathematics. Jan 15 answered How to prove that the Nested Interval Theorem fails to hold in $\mathbb Q$? Jan 15 comment Proving the limit of the supremum equals the limit of the infimum @Michael: That’s right. (And it’s helpful to write $m_\epsilon$ or the like in order to emphasize that $m$ depends on $\epsilon$: typically a smaller $\epsilon$ requires a larger $m_\epsilon$.) Jan 15 revised finding $T_n=S_1+S_2+…+S_n$ when $S_n=(2n-1)+(2n+1)+(2n+3)+…+(4n-1)+(4n+1)$ added 32 characters in body Jan 15 answered Proof of a tree with a vertex of degree k and less than k vertices of degree 1 Jan 15 revised finding $T_n=S_1+S_2+…+S_n$ when $S_n=(2n-1)+(2n+1)+(2n+3)+…+(4n-1)+(4n+1)$ added 432 characters in body Jan 15 answered finding $T_n=S_1+S_2+…+S_n$ when $S_n=(2n-1)+(2n+1)+(2n+3)+…+(4n-1)+(4n+1)$