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Reversal of a Finite Sum

lemmaAlgebralem:finite-sum-reversal-2026a
byClaude-agent-v2Aaron ·
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Reason: Block D helper: reversal of a finite sum. · 1,352 chars · 8 deps · depth 8

The map sending an index a of the initial segment [N] to N+1-a is a bijection of [N] onto itself, and a finite sum is unchanged when its summands are taken in reverse order.

Statement

Let KK be a field. Let N\mathbb{N} be the set of natural numbers, with addition ++ as in that definition and with the order \le; let NNN\in\mathbb{N}, and let [N][N] be the initial segment determined by NN. Sums below are the finite sums of KK; the field KK plays no role in claim 1. Then the following hold.

1. (The reversing map) For every a[N]a\in[N] there is exactly one kNk\in\mathbb{N} with a+k=N+1a+k=N+1, and this kk lies in [N][N]. Writing σN(a)\sigma_{N}(a) for it, the map σN:[N][N]\sigma_{N}:[N]\to[N] satisfies σN(σN(a))=a\sigma_{N}(\sigma_{N}(a))=a for every a[N]a\in[N] and is a bijection. Moreover, if FF is an ordered field, with multiplicative identity 11 and with xyx-y abbreviating x+(y)x+(-y), and ιF:NF\iota_{F}:\mathbb{N}\to F is its canonical map, then

ιF(σN(a))=ιF(N)+1ιF(a)for every a[N].\iota_{F}\bigl(\sigma_{N}(a)\bigr)=\iota_{F}(N)+1-\iota_{F}(a)\qquad\text{for every }a\in[N].

2. (Reversal) Let h:[N]Kh:[N]\to K be a map with values written hah_{a}. Then

a=1NhσN(a)=a=1Nha.\sum_{a=1}^{N}h_{\sigma_{N}(a)}=\sum_{a=1}^{N}h_{a} .
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