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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 $S_n=(2n-1)+(2n+1)+(2n+3)+…+(4n-1)+(4n+1)$