"a class is a collection of sets (or sometimes other mathematical objects) which can be unambiguously defined by a property that all its members share."
eh? few groups have translation invariant total orders. without translation invariance of the order, the order restricted to one coset may look totally different from that of another
@PeterTamaroff A group, like the set of all symmetries of a square ,might contain a subgroup, like the set consisting of just the identity symmetry and the reflection across the vertical axis. You get a coset of this subgroup when you take each element of it and then multiply by some group element, say a quarter clockwise turn, giving a set { quarter clockwise turn, diagonal reflection } which is the "right coset" of the original subgroup.
There's a bit in Winning Ways about a game in which one player is a "square-eater", which they dub a "quadraphage", and then diffidently remark that that mixes two languages, and perhaps should be "tesseravore" instead.
I'm sure I remember this, because I remember someone answering about factorial-base representation, and then someone else said no, they have to be distinct.
it's given to you as homework. why would that be given in a calc class? it isn't. why would an e.g. elem NT class require a calc solution? it wouildn't.
@PeterTamaroff My formula also works, where $\sigma_p(x)$ is the sum of the base-$p$ digits of $x$.
Is your question, "why would you write for $\sigma_p$ for something when it is already standard that it denotes something else"? the answer is: because he felt like it.
@PeterTamaroff my formula works much better for binomial coefficients. Things cancel and all you need to look at is the sum of the digits divided by $p-1$
@PeterTamaroff The UK is GMT+1. New York is GMT-5 or -6. Thus it makes sense that GMT-2 would be right between them, geographically speaking- which is right in the middle of the Atlantic Ocean.
@anon Consider an order relation $\leq$, and its strict order $<$, and a poset $(X,\leq)$. It is correct to say $a\in X$ is minimal if $b<a$ is false for each choice of $b$?
I guess that, on second thought, the Altantic is much squarer and doesn't cover anywhere near as much as I thought at the top of the northern hemisphere, or the South Atlantic.
Suppose $$ n=\sum_kd_kp^k $$ Then look at the contribution of each digit from your method: $$ d_kp^{k-1}+d_kp^{k-2}+d_kp^{k-3}+\dots+d_k=d_k\frac{p^k-1}{p-1}=\frac{d_kp^k-d_k}{p-1} $$ Now sum in $k$.
