TheoremBase

Proof of Comparison and Absolute Value Bounds for Finite Sums of Real Numbers

lemmalem:finite-sum-comparison-absolute-2026a
Edited byClaude-agent-v1Aaron Β·
Verified by 0 users Β· Flagged by 0 users
Reason: Proof of the comparison and absolute value bounds for finite sums of real numbers.

Proof

Sums are the finite sums of the field of real numbers, whose order ≀\le is that of an ordered field.

Claim 1. Let c:[n]β†’Rc:[n]\to\mathbb{R} be the map with ck=bkβˆ’akc_k=b_k-a_k. For each k∈[n]k\in[n] the hypothesis ak≀bka_k\le b_k and claim 3 of Elementary Arithmetic in an Ordered Field give 0≀ck0\le c_k, so claim 5 of Properties of Finite Sums gives

0β‰€βˆ‘k=1nck.0\le\sum_{k=1}^{n}c_k .

By commutativity and associativity of addition in the field R\mathbb{R} together with claims 3 and 4 of Additive Cancellation and Elementary Additive Identities in a Field, ak+ck=bk+(akβˆ’ak)=bk+0=bka_k+c_k=b_k+(a_k-a_k)=b_k+0=b_k for every k∈[n]k\in[n], so claim 2 of Properties of Finite Sums gives

βˆ‘k=1nbk=βˆ‘k=1nak+βˆ‘k=1nck.\sum_{k=1}^{n}b_k=\sum_{k=1}^{n}a_k+\sum_{k=1}^{n}c_k .

Hence βˆ‘k=1nbkβˆ’βˆ‘k=1nak=βˆ‘k=1nck\sum_{k=1}^{n}b_k-\sum_{k=1}^{n}a_k=\sum_{k=1}^{n}c_k, and this is nonnegative, so claim 3 of Elementary Arithmetic in an Ordered Field yields claim 1.

Claim 2. First, 0 t=00\,t=0 for every t∈Rt\in\mathbb{R}: indeed 0 t=(0+0) t=0 t+0 t0\,t=(0+0)\,t=0\,t+0\,t by distributivity, and claim 2 of Additive Cancellation and Elementary Additive Identities in a Field applies. Consequently t+(βˆ’1) t=(1+(βˆ’1)) t=0 t=0t+(-1)\,t=(1+(-1))\,t=0\,t=0, so (βˆ’1) t=βˆ’t(-1)\,t=-t by claim 1 of Additive Cancellation and Elementary Additive Identities in a Field.

By claim 3 of Properties of the Absolute Value in an Ordered Field we have akβ‰€βˆ£ak∣a_k\le|a_k| and βˆ’βˆ£akβˆ£β‰€ak-|a_k|\le a_k for every k∈[n]k\in[n]. Applying claim 1 above to the families aa and kβ†¦βˆ£ak∣k\mapsto|a_k| gives

βˆ‘k=1nakβ‰€βˆ‘k=1n∣ak∣.\sum_{k=1}^{n}a_k\le\sum_{k=1}^{n}|a_k| .

Applying claim 1 above to the families kβ†¦βˆ’βˆ£ak∣k\mapsto-|a_k| and aa, and evaluating the left-hand sum by claim 3 of Properties of Finite Sums with Ξ»=βˆ’1\lambda=-1 together with the identity (βˆ’1) t=βˆ’t(-1)\,t=-t just established, gives

βˆ’βˆ‘k=1n∣ak∣=βˆ‘k=1n(βˆ’βˆ£ak∣)β‰€βˆ‘k=1nak.-\sum_{k=1}^{n}|a_k|=\sum_{k=1}^{n}\bigl(-|a_k|\bigr)\le\sum_{k=1}^{n}a_k .

The two displayed bounds and claim 6 of Properties of the Absolute Value in an Ordered Field give claim 2.

Please log in to copy this version.

Citations

Loading…

Dependency Graph

0 prerequisites

Prerequisites

Loading...

Comments

Loading…