Proof of Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constants, Comparison and the Triangle Inequality

In a commutative ring the ring laws make $+$ associative and commutative, so the clauses of Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms for associative and commutative operations apply to finite sums. Each induction below runs over the set of those $n\in\mathbb{N}$ for which the claim holds for all data of the stated kind, and concludes by Arithmetic and Order of the Natural Numbers §induction. In the inductive steps, $\sum_{k=1}^{n}$ of a map on $[n+1]$ is the sum of its restriction to $[n]$ by Iterated Operations: Finite Sums and Finite Products §restriction, and the step uses Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §recursion:

$$\sum_{k=1}^{1}a_{k}=a_{1},\qquad\sum_{k=1}^{n+1}a_{k}=\sum_{k=1}^{n}a_{k}+a_{n+1}.$$

Distributive. The map $x\mapsto c\,x$ on $R$ satisfies $c(x+y)=c\,x+c\,y$ by the ring laws, so the claim is Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §homomorphism with $\ast=\diamond=+$.

Difference. For $x,y\in R$, the ring laws give $(x+y)+\big((-x)+(-y)\big)=\big(x+(-x)\big)+\big(y+(-y)\big)=0$, so $-(x+y)=(-x)+(-y)$ by the uniqueness in Additive and Multiplicative Inverses Are Unique §negative. Thus $x\mapsto-x$ satisfies the hypothesis of Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §homomorphism, which gives the first identity. For the second, Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §termwise and the first identity give

$$\sum_{k=1}^{n}\big(a_{k}+(-b_{k})\big)=\sum_{k=1}^{n}a_{k}+\sum_{k=1}^{n}(-b_{k})=\sum_{k=1}^{n}a_{k}-\sum_{k=1}^{n}b_{k}.$$

Telescoping. We induct on $n$. For $n=1$ the sum is $d_{2}-d_{1}$. If the claim holds for $n$ and $d:[n+2]\to R$, then the induction hypothesis for $d|_{[n+1]}$ and the ring laws give

$$\sum_{k=1}^{n+1}(d_{k+1}-d_{k})=(d_{n+1}-d_{1})+(d_{n+2}-d_{n+1})=d_{n+2}-d_{1}.$$

From now on $F$ is an ordered field. Its underlying ring is a commutative ring, and the rules of Rules of Arithmetic and Order in an Ordered Field are available.

Constant. By Ordinary Mathematical Language for Analysis: Sets, Maps, Numbers and Ordered Fields §fields, $n$ stands for $\kappa_{F}(n)$ here. By The Rational Numbers Embed in Exactly One Way into Every Ordered Field §unique, $\kappa_{F}$ is additive with $\kappa_{F}(1)=1$, and by Ordinary Mathematical Language for Analysis: Sets, Maps, Numbers and Ordered Fields §numbers reading $n+1$ in $F$ agrees with computing it in $\mathbb{N}_{0}$ first; hence $n+1$ read in $F$ is $\kappa_{F}(n)+1$. We induct on $n$. For $n=1$ the sum is $c=1\cdot c$. If $\sum_{k=1}^{n}c=n\,c$, then $\sum_{k=1}^{n+1}c=n\,c+1\cdot c=(n+1)\,c$ by the ring laws.

Comparison. We induct on $n$, proving both assertions together. For $n=1$ the sums are $a_{1}$ and $b_{1}$, and $j=1$. Suppose they hold for $n$, and let $a,b:[n+1]\to F$ with $a_{k}\le b_{k}$ for all $k$. Put $A=\sum_{k=1}^{n}a_{k}$ and $B=\sum_{k=1}^{n}b_{k}$. Then $A\le B$ by the induction hypothesis, so $A+a_{n+1}\le B+b_{n+1}$ by Rules of Arithmetic and Order in an Ordered Field §order-sum, which is the first assertion. Now let $a_{j}<b_{j}$ with $j\in[n+1]$, so that $j\le n$ or $j=n+1$ by Intervals of Natural Numbers: Initial Segments, Adding One Element, Splitting and Shifting §successor. If $j\le n$, then $A<B$ by the induction hypothesis, and Rules of Arithmetic and Order in an Ordered Field §order-sum gives $A+a_{n+1}<B+a_{n+1}\le B+b_{n+1}$. If $j=n+1$, it gives $A+a_{n+1}\le B+a_{n+1}=a_{n+1}+B<b_{n+1}+B$. In both cases $\sum_{k=1}^{n+1}a_{k}<\sum_{k=1}^{n+1}b_{k}$.

Nonnegative. We induct on $n$. For $n=1$, $j=1$ and $a_{1}=\sum_{k=1}^{1}a_{k}$. Suppose the claim holds for $n$, and let $a:[n+1]\to F$ with all $a_{k}\ge0$; put $A=\sum_{k=1}^{n}a_{k}$. By the induction hypothesis $0\le a_{1}\le A$. By Rules of Arithmetic and Order in an Ordered Field §order-sum, $A=A+0\le A+a_{n+1}$ and $a_{n+1}=0+a_{n+1}\le A+a_{n+1}$. Hence for $j\le n$, $a_{j}\le A\le\sum_{k=1}^{n+1}a_{k}$ by the induction hypothesis, and for $j=n+1$, $a_{j}\le\sum_{k=1}^{n+1}a_{k}$ directly. In particular $0\le a_{1}\le\sum_{k=1}^{n}a_{k}$ for every $n$.

Triangle. We induct on $n$; for $n=1$ both sides are $|a_{1}|$. If the claim holds for $n$ and $a:[n+1]\to F$, then Rules of Arithmetic and Order in an Ordered Field §triangle, the induction hypothesis and Rules of Arithmetic and Order in an Ordered Field §order-sum give

$$\Big|\sum_{k=1}^{n+1}a_{k}\Big|\le\Big|\sum_{k=1}^{n}a_{k}\Big|+|a_{n+1}|\le\sum_{k=1}^{n}|a_{k}|+|a_{n+1}|=\sum_{k=1}^{n+1}|a_{k}|.$$