Proof of Natural Numbers Read in the Integers and in the Rationals Agree with Their Embeddings

Each result cited below is universally quantified over the data in its own statement and is applied to the data indicated where it is cited.

By The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §ring, $\mathbb{Z}$, with $+$, $\cdot$, $0_{\mathbb{Z}}$ and $1_{\mathbb{Z}}$, is a commutative ring. By The Rational Numbers Form an Archimedean Ordered Field Containing the Integers §ordered-field, $\mathbb{Q}$, with $+$, $\cdot$, $0_{\mathbb{Q}}$, $1_{\mathbb{Q}}$ and $\le$, is an ordered field; so it is a field by Ordered Fields §ordered-field, and hence a commutative ring, with $+$, $\cdot$, $0_{\mathbb{Q}}$ and $1_{\mathbb{Q}}$, by Fields §field. For $n\in\mathbb{N}_{0}$ write $\nu_{\mathbb{Z}}(n)$ and $\nu_{\mathbb{Q}}(n)$ for the images $n_{\mathbb{Z}}$ and $n_{\mathbb{Q}}$ of $n$ in these two rings; the new letter keeps them apart from the constants $0_{\mathbb{Z}},1_{\mathbb{Z}}$ of The Integers §constants and $0_{\mathbb{Q}},1_{\mathbb{Q}}$ of The Rational Numbers §constants. By Commutative Rings, Fields and Ordered Fields: Standard Notation §numerals, the two clauses state that $\nu_{\mathbb{Z}}(n)=\iota_{0}(n)$ and $\nu_{\mathbb{Q}}(n)=j(\iota_{0}(n))$ for every $n\in\mathbb{N}_{0}$.

Facts used. By The Image of the Natural Numbers with Zero in a Commutative Ring Respects Zero, One, Sums, Products, Differences, Powers, and Finite Sums and Products §constants, $\nu_{\mathbb{Z}}(0)=0_{\mathbb{Z}}$, $\nu_{\mathbb{Z}}(1)=1_{\mathbb{Z}}$, $\nu_{\mathbb{Q}}(0)=0_{\mathbb{Q}}$ and $\nu_{\mathbb{Q}}(1)=1_{\mathbb{Q}}$, these being the zeros and units of the two rings. Hence, for every $n\in\mathbb{N}_{0}$, The Image of the Natural Numbers with Zero in a Commutative Ring Respects Zero, One, Sums, Products, Differences, Powers, and Finite Sums and Products §sum gives

$$\nu_{\mathbb{Z}}(n+1)=\nu_{\mathbb{Z}}(n)+1_{\mathbb{Z}},\qquad\nu_{\mathbb{Q}}(n+1)=\nu_{\mathbb{Q}}(n)+1_{\mathbb{Q}}.$$

By The Natural Numbers with Zero and Their Embedding into the Integers §embedding, $\iota_{0}(0)=0_{\mathbb{Z}}$ and $\iota_{0}(m+n)=\iota_{0}(m)+\iota_{0}(n)$ for all $m,n\in\mathbb{N}_{0}$. Since $1\in\mathbb{N}$ by The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §sets, the same clause gives $\iota_{0}(1)=\iota(1)$, and $\iota(1)=1_{\mathbb{Z}}$ by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §embedding; so $\iota_{0}(n+1)=\iota_{0}(n)+1_{\mathbb{Z}}$ for every $n\in\mathbb{N}_{0}$. Finally, by The Rational Numbers Form an Archimedean Ordered Field Containing the Integers §embedding, $j(0_{\mathbb{Z}})=0_{\mathbb{Q}}$, $j(1_{\mathbb{Z}})=1_{\mathbb{Q}}$ and $j(x+y)=j(x)+j(y)$ for all $x,y\in\mathbb{Z}$.

Integers. Let $A$ be the set of those $n\in\mathbb{N}_{0}$ with $\nu_{\mathbb{Z}}(n)=\iota_{0}(n)$. Then $0\in A$, since $\nu_{\mathbb{Z}}(0)=0_{\mathbb{Z}}=\iota_{0}(0)$. If $n\in A$, then

$$\nu_{\mathbb{Z}}(n+1)=\nu_{\mathbb{Z}}(n)+1_{\mathbb{Z}}=\iota_{0}(n)+1_{\mathbb{Z}}=\iota_{0}(n+1),$$

so $n+1\in A$. By induction from $0$ on $\mathbb{N}_{0}$ (The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §induction), $A=\mathbb{N}_{0}$.

Rationals. Let $B$ be the set of those $n\in\mathbb{N}_{0}$ with $\nu_{\mathbb{Q}}(n)=j(\iota_{0}(n))$. Then $0\in B$, since $\nu_{\mathbb{Q}}(0)=0_{\mathbb{Q}}=j(0_{\mathbb{Z}})=j(\iota_{0}(0))$. If $n\in B$, then

$$\nu_{\mathbb{Q}}(n+1)=\nu_{\mathbb{Q}}(n)+1_{\mathbb{Q}}=j(\iota_{0}(n))+j(1_{\mathbb{Z}})=j\big(\iota_{0}(n)+1_{\mathbb{Z}}\big)=j(\iota_{0}(n+1)),$$

so $n+1\in B$. By induction from $0$ on $\mathbb{N}_{0}$ (The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §induction), $B=\mathbb{N}_{0}$.