Proof of Difference and Quotient Identities in ℝ

$\mathbb{R}$, with $+$, $\cdot$, $0$, $1$ and $\le$, is an ordered field by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (ii), hence, by Ordered Fields (i), an ordered ring with a total order; so Rules of Arithmetic and Order in an Ordered Field, Rules of Order in an Ordered Ring: Differences, Sums, Negatives, Products and Squares and Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field apply to it. Rearrangements by the commutative, associative, distributive, neutral and negative laws of $\mathbb{R}$ are made without further mention.

Proof of (i). The identity $x-y=x+(-y)$ is Negatives, Differences, Reciprocals and Quotients (i). Hence $(x+h)-x=(x+h)+(-x)=h+(x+(-x))=h+0=h$, and $x-x=x+(-x)=0$. By Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field (ii), $-(x-y)=y-x$; in particular $-(x-h)=h-x$, so $x-(x-h)=x+(-(x-h))=x+(h+(-x))=h+(x+(-x))=h$. Finally, $x<y$ if and only if $0<y-x$, by Rules of Order in an Ordered Ring: Differences, Sums, Negatives, Products and Squares (i).

Proof of (ii). Let $b\neq0$. By Negatives, Differences, Reciprocals and Quotients (ii), $b^{-1}$ is the element of $\mathbb{R}$ with $b\cdot b^{-1}=1$, and $x/b=x\cdot b^{-1}$. Hence $(x/b)\cdot b=x\cdot(b^{-1}\cdot b)=x\cdot1=x$, and $b\cdot(x/b)=(x/b)\cdot b=x$. For all $u,v\in\mathbb{R}$, $(-u)\cdot v=-(u\cdot v)$ by Rules of Arithmetic and Order in an Ordered Field (ii). So, with $u=c$ and $v=b^{-1}$,

$$(a-c)/b=(a+(-c))\cdot b^{-1}=a\cdot b^{-1}+(-c)\cdot b^{-1}=a/b+\bigl(-(c\cdot b^{-1})\bigr)=a/b-c/b .$$

Taking $c=m\cdot b$, and using $(m\cdot b)/b=(m\cdot b)\cdot b^{-1}=m\cdot(b\cdot b^{-1})=m$, we get $(a-m\cdot b)/b=a/b-(m\cdot b)/b=a/b-m$.

Proof of (iii). Positive quotients. Let $0<x$ and $0<b$. Then $b\neq0$, as $<$ is the strict relation of $\le$ (Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (ii)), and $0<b^{-1}$ by Rules of Arithmetic and Order in an Ordered Field (viii). By Rules of Arithmetic and Order in an Ordered Field (vi), applied with $z=b^{-1}$, $0<x$ gives $0\cdot b^{-1}<x\cdot b^{-1}$; and $0\cdot b^{-1}=0$ by Rules of Arithmetic and Order in an Ordered Field (i). So $0<x\cdot b^{-1}=x/b$.

Negative reciprocals. Let $b<0$. Then $b\neq0$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (ii), so $b^{-1}$ is defined. As $-0=0$, Rules of Order in an Ordered Ring: Differences, Sums, Negatives, Products and Squares (iii) turns $b<0$ into $0<-b$; so $-b\neq0$, and $0<(-b)^{-1}$ by Rules of Arithmetic and Order in an Ordered Field (viii). By Rules of Arithmetic and Order in an Ordered Field (ii), $(-b)\cdot(-(b^{-1}))=b\cdot b^{-1}=1$, so $(-b)^{-1}=-(b^{-1})$, the reciprocal being unique by Negatives, Differences, Reciprocals and Quotients (ii). Hence $0<-(b^{-1})$, that is $-0<-(b^{-1})$, and $b^{-1}<0$ by Rules of Order in an Ordered Ring: Differences, Sums, Negatives, Products and Squares (iii).

Absolute values. Let $b\neq0$. Then $|b|\neq0$ by Rules of Arithmetic and Order in an Ordered Field (x). By the same clause, $|b|\cdot|b^{-1}|=|b\cdot b^{-1}|=|1|$, and $|1|=1$ by Absolute Value in an Ordered Field (i), as $0\le1$ by Rules of Order in an Ordered Ring: Differences, Sums, Negatives, Products and Squares (vi). So $|b^{-1}|$ is the element $z$ of $\mathbb{R}$ with $|b|\cdot z=1$, which is $|b|^{-1}$ by Negatives, Differences, Reciprocals and Quotients (ii). Finally, by Rules of Arithmetic and Order in an Ordered Field (x) again, $|x/b|=|x\cdot b^{-1}|=|x|\cdot|b^{-1}|=|x|\cdot|b|^{-1}=|x|/|b|$.