Proof of Rules of Arithmetic in a Commutative Ring: Zero, Signs, Squares, Images of Natural Numbers, and No Zero Divisors in a Field

Throughout we use the axioms of a commutative ring freely. By commutativity, $0+x=x$, $1\cdot x=x$ and $(y+z)x=yx+zx$.

Zero. Let $a=x\cdot0$. Then $a=x\cdot(0+0)=a+a$. Add $-a$ to both sides (Negatives, Differences, Reciprocals and Quotients §negative): $0=a+(-a)=(a+a)+(-a)=a+(a+(-a))=a+0=a$. Hence $x\cdot0=0$, and $0\cdot x=x\cdot0=0$ by commutativity.

Signs. By Additive and Multiplicative Inverses Are Unique §negative, for each $u\in R$ the negative $-u$ is the only $w$ with $u+w=0$, so to prove $-u=v$ it suffices to show $u+v=0$. First, $(-x)+x=x+(-x)=0$, so $-(-x)=x$. Second, $(x+y)+((-x)+(-y))=(x+(-x))+(y+(-y))=0+0=0$ by associativity and commutativity, so $-(x+y)=(-x)+(-y)$. Third, $xy+(-x)y=(x+(-x))y=0\cdot y=0$ by Zero, so $(-x)y=-(xy)$. Fourth, by the third rule and commutativity, $x(-y)=(-y)x=-(yx)=-(xy)$, so $(-x)(-y)=-(x(-y))=-(-(xy))=xy$. Fifth, $-(x-y)=-(x+(-y))=(-x)+(-(-y))=(-x)+y=y-x$. Sixth, $(x-y)z=xz+(-y)z=xz+(-(yz))=xz-yz$.

Naturals. By The Image of a Natural Number in a Commutative Ring §image and Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §recursion, $1_{R}=\sum_{k=1}^{1}1=1$. For additivity, let $m,n\in\mathbb{N}_{0}$. If $m=0$, then $m+n=n$ by Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §zero, and $(m+n)_{R}=n_{R}=0+n_{R}=m_{R}+n_{R}$; the case $n=0$ is the same. Otherwise $m,n\in\mathbb{N}$ by The Natural Numbers with Zero and Their Embedding into the Integers §naturals, so $m+n\in\mathbb{N}$, and since $+$ on $R$ is associative, Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §splitting applied to the constant map $k\mapsto1$ on $[m+n]$ gives $(m+n)_{R}=\sum_{k=1}^{m+n}1=\sum_{k=1}^{m}1+\sum_{j=1}^{n}1=m_{R}+n_{R}$.

For multiplicativity, if $m=0$ or $n=0$, then $mn=0$ by Arithmetic of Multiplication on Omega: Recursion Rules, Distributivity, Associativity, Commutativity, No Zero Divisors, Cancellation and Compatibility with the Order §zero, so $(mn)_{R}=0$, while $m_{R}n_{R}=0$ by Zero, since one factor is $0$. Otherwise $m,n\in\mathbb{N}$. Fix $m$, and let $A$ be the set of $n\in\mathbb{N}$ with $(mn)_{R}=m_{R}n_{R}$. We show $A=\mathbb{N}$ by induction over $\mathbb{N}$ (Arithmetic and Order of the Natural Numbers §induction). First, $1\in A$, since $m\cdot1=m$ (Arithmetic and Order of the Natural Numbers §one) and $m_{R}1_{R}=m_{R}\cdot1=m_{R}$. Now let $n\in A$. Then $m(n+1)=mn+m$ by Arithmetic and Order of the Natural Numbers §distributive and Arithmetic and Order of the Natural Numbers §one. By additivity and $1_{R}=1$, we get $(m(n+1))_{R}=(mn)_{R}+m_{R}=m_{R}n_{R}+m_{R}\cdot1=m_{R}(n_{R}+1_{R})=m_{R}(n+1)_{R}$, so $n+1\in A$.

Finally let $n\ge1$, so $n\in\mathbb{N}$. By commutativity and Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality §distributive, $n_{R}\,x=x\sum_{k=1}^{n}1=\sum_{k=1}^{n}x\cdot1=\sum_{k=1}^{n}x$.

Squares. By Arithmetic and Order of the Natural Numbers §digits, $2=1+1$. Therefore $2_{R}=1_{R}+1_{R}=1+1$ by Naturals. Also, by Powers with Exponents in the Natural Numbers with Zero §power and Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §recursion, $u^{2}=\prod_{k=1}^{1+1}u=\big(\prod_{k=1}^{1}u\big)\cdot u=u\cdot u$ for every $u\in R$. Hence $u+u=1\cdot u+1\cdot u=(1+1)u=2_{R}\,u$. By distributivity, associativity and commutativity, $(x+y)^{2}=(x+y)x+(x+y)y=xx+(xy+xy)+yy=x^{2}+2_{R}\,xy+y^{2}$. Apply this to $x$ and $-y$. By Signs, $x(-y)=-(xy)$, so $2_{R}\,x(-y)=-(2_{R}\,xy)$, and $(-y)(-y)=yy$. This gives $(x-y)^{2}=x^{2}-2_{R}\,xy+y^{2}$. Lastly, by Signs, $(x-y)(x+y)=x(x+y)+(-y)(x+y)=xx+xy+(-(yx))+(-(yy))=x^{2}+(xy+(-(xy)))-y^{2}=x^{2}-y^{2}$.

Field. Let $R$ be a field, let $xy=0$, and suppose $x\neq0$. By Fields §field there is $w\in R$ with $xw=1$. Then $y=1\cdot y=(wx)y=w(xy)=w\cdot0=0$ by Zero.

Ordered. Let $R$ with $\le$ be an ordered field. Write $\kappa=\kappa_{R}$, the map of The Rational Numbers Embed in Exactly One Way into Every Ordered Field §unique. This map satisfies $\kappa(1_{\mathbb{Q}})=1$ and $\kappa(u+v)=\kappa(u)+\kappa(v)$ for $u,v\in\mathbb{Q}$, and $\kappa(0_{\mathbb{Q}})=0$ by The Rational Numbers Embed in Exactly One Way into Every Ordered Field §zero.

By Ordinary Mathematical Language for Analysis: Sets, Maps, Numbers and Ordered Fields §numbers, the element $n$ of $\mathbb{N}_{0}$ read as a rational number is $j(\iota_{0}(n))$. Here $\iota_{0}$ is the map of The Natural Numbers with Zero and Their Embedding into the Integers §embedding, and $j$ is the embedding of $\mathbb{Z}$ into $\mathbb{Q}$, applied by The Integers and the Rational Numbers, with the Natural Numbers and the Integers Identified with Subsets of the Rationals §identification. We must therefore show $n_{R}=\kappa(j(\iota_{0}(n)))$.

If $n=0$, then $\iota_{0}(0)=0_{\mathbb{Z}}$ by The Natural Numbers with Zero and Their Embedding into the Integers §embedding, and $j(0_{\mathbb{Z}})=0_{\mathbb{Q}}$ by The Rational Numbers Form an Archimedean Ordered Field Containing the Integers §embedding. So $\kappa(j(\iota_{0}(0)))=\kappa(0_{\mathbb{Q}})=0=0_{R}$.

If $n\in\mathbb{N}$, then $\iota_{0}(n)=\iota(n)$ by The Natural Numbers with Zero and Their Embedding into the Integers §embedding, so we must show $n_{R}=\kappa(j(\iota(n)))$. Let $B$ be the set of $n\in\mathbb{N}$ for which this holds; we show $B=\mathbb{N}$ by induction over $\mathbb{N}$ (Arithmetic and Order of the Natural Numbers §induction). First, $j(\iota(1))=1_{\mathbb{Q}}$ by The Natural Numbers and the Integers inside the Rational Numbers §naturals, so $\kappa(j(\iota(1)))=1=1_{R}$ by Naturals; thus $1\in B$. Now let $n\in B$. The natural number $n+1$ is read as the rational number $j(\iota(n+1))$, and by The Natural Numbers and the Integers inside the Rational Numbers §naturals this equals $j(\iota(n))+j(\iota(1))=j(\iota(n))+1_{\mathbb{Q}}$, a sum computed in $\mathbb{Q}$. Hence $\kappa(j(\iota(n+1)))=\kappa(j(\iota(n)))+\kappa(1_{\mathbb{Q}})=n_{R}+1=n_{R}+1_{R}=(n+1)_{R}$, by additivity of $\kappa$, the induction hypothesis and Naturals. So $n+1\in B$.