Proof of The Quotient 1/x, Order Rules for Sums, and Nesting of Closed Intervals in ℝ

Throughout, $\mathbb{R}$ is an ordered field, so the rules for ordered fields apply; we use reflexivity and transitivity of $\le$ and the neutral elements, Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (viii), Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (v), without further mention.

For (i), let $x\neq0$. Then $1/x=1\cdot x^{-1}=x^{-1}$ by Negatives, Differences, Reciprocals and Quotients (ii) and Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (v) (here $x^{-1}\in\mathbb{R}$, as $\mathbb{R}$ is a field by The Structure of ℝ as a Field and an Ordered Ring (i)).

For (ii), let $1\le p$. By Rules of Arithmetic and Order in an Ordered Field (vii), $0<1$; together with $1\le p$, Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (i) gives $0<p$.

For (iii), let $0\le x$ and $0\le y$. Adding these inequalities by Rules of Arithmetic and Order in an Ordered Field (iv) gives $0+0\le x+y$, that is, $0\le x+y$.

For (iv), let $0\le y$. Adding $x\le x$ and $0\le y$ by Rules of Arithmetic and Order in an Ordered Field (iv) gives $x+0\le x+y$, that is, $x\le x+y$.

For (v), let $x\le y$. Adding $z\le z$ and $x\le y$ by Rules of Arithmetic and Order in an Ordered Field (iv) gives $z+x\le z+y$.

For (vi), let $a\le c$ and $d\le b$, and let $u\in[c,d]$. By Intervals of the Real Line (ii), $u$ is a real number with $c\le u$ and $u\le d$. By transitivity, $a\le c$ and $c\le u$ give $a\le u$, and $u\le d$ and $d\le b$ give $u\le b$; so $u\in[a,b]$ by Intervals of the Real Line (ii). As $u\in[c,d]$ was arbitrary, $[c,d]\subseteq[a,b]$ by Subclasses and Subsets (i).