Proof of The Negative of Zero in ℝ, and Multiplying an Inequality on the Left by a Nonnegative Real

The negative of $0$. By The Structure of ℝ as a Field and an Ordered Ring (i), $\mathbb{R}$, with $+$, $\cdot$, $0$ and $1$, is a field, hence by Fields (i) a commutative ring. So Negatives, Differences, Reciprocals and Quotients (i) applies to $\mathbb{R}$: $-0$ is the unique $w\in\mathbb{R}$ with $0+w=0$, the uniqueness being Additive and Multiplicative Inverses Are Unique (i) with $x=0$. By Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (v) with $x=0$, $0+0=0$. So $w=0$ satisfies $0+w=0$, and by this uniqueness $-0=0$, which is (i).

Scaling on the left. By The Structure of ℝ as a Field and an Ordered Ring (ii), $\mathbb{R}$, with $+$, $\cdot$, $0$, $1$ and $\le$, is an ordered ring; negatives in $\mathbb{R}$ are those of Negatives, Differences, Reciprocals and Quotients (i), as in the statement. So Rules of Order in an Ordered Ring: Differences, Sums, Negatives, Products and Squares applies with $R=\mathbb{R}$. Let $x,y,z\in\mathbb{R}$ with $x\le y$ and $0\le z$. By Rules of Order in an Ordered Ring: Differences, Sums, Negatives, Products and Squares (iv), $x\cdot z\le y\cdot z$. By Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (ii), $x\cdot z=z\cdot x$ and $y\cdot z=z\cdot y$. Hence $z\cdot x\le z\cdot y$, which is (ii).