Talk:Basis Representation Theorem

Possible simple clarification needed in proof?
If I'm reading the proof right, then shouldn't the following statement ...

---snip---

So this inequality implies the following:


 * $\forall m, n: s_b \left({m}\right) \le s_b \left({m - 1}\right) \le \ldots \le s_b \left({n + 1}\right) \le s_b \left({n}\right)$

---snip---

... specify that m >= n? Perhaps like this?

---edit---

So this inequality implies the following:


 * $\forall m \ge n: s_b \left({m}\right) \le s_b \left({m - 1}\right) \le \ldots \le s_b \left({n + 1}\right) \le s_b \left({n}\right)$

---end---

...or maybe this is clearer...

---edit---

So this inequality implies the following:


 * $\forall m, n : m \ge n : s_b \left({m}\right) \le s_b \left({m - 1}\right) \le \ldots \le s_b \left({n + 1}\right) \le s_b \left({n}\right)$

---end---

However I'm unsure about my proofwiki style ...

That's all! BTW - pretty nifty proof.


 * Possibly. The source work containing this proof (Andrews) is arbitrarily fussy: it goes to $s_b \left({m - 2}\right) \le \cdots \le s_b \left({n + 2}\right)$ and specifies $m \ge n + 4$, but I found that unnecessarily inelegant. --prime mover (talk) 16:31, 27 November 2016 (EST)


 * Thanks for the reply. Curious - why do you suppose Andrews was so fussy? It doesn't seem necessary, the edit you just made to improve the proof seems both necessary and sufficient.


 * The following question reveals my lack of mathematical depth (i.e.; I'm sure this is common knowledge) but it seems that our use of ellipsis in a statement such as $m, m-1, m-2, ..., n+2, n+1, n$ after you've specified $m \ge n$ shouldn't preclude the sequence from collapsing down to the boundary case when $m = n$ so long as it doesn't create some kind of logical quandary elsewhere no? Am I off track in thinking that's why Andrews was being fussy?  --Jrowellfx (talk) 19:30, 27 November 2016 (EST)


 * The point is that you can make $m$ and $n$ whatever you want them to be. The line is, itself, simply an explanatory amplification of the fact that "smaller numbers have no fewer representations than larger numbers" and the line could be replaced with "$m \ge n \implies s_b \left({m}\right) \le s_b \left({n}\right)$" -- which, if you want to make that ultra-rigorously explicit, can of course be hammered out in an induction proof. Could be done, but this particular presentation of this proof keeps close to the spirit of the original. --prime mover (talk) 01:09, 28 November 2016 (EST)