Proof of Arithmetic, Order and Discreteness of the Integers

Throughout we use the properties of the canonical map $\iota$, cited by claim number, together with the elementary order arithmetic in an ordered field, which is what justifies every step below in which the same quantity is added to both sides of an inequality, an inequality is negated, or both sides are multiplied by a positive quantity.

Claim 1. By claim 3 of the canonical-map lemma, $0<\iota(n)$ for every $n\in\mathbb{N}$; adding $-\iota(n)$ to both sides gives $-\iota(n)<0$. Hence every element of $\iota(\mathbb{N})$ is positive and every element of $-\iota(\mathbb{N})$ is negative, so neither set contains $0$ and the two sets are disjoint from each other and from $\{0\}$. Their union is $\mathbb{Z}$ by the definition of $\mathbb{Z}$.

For the second assertion, $\iota(\mathbb{N})\subseteq\{x\in\mathbb{Z}:0<x\}$ has just been shown. Conversely let $x\in\mathbb{Z}$ with $0<x$. By the first assertion $x$ lies in $\{0\}$, in $\iota(\mathbb{N})$ or in $-\iota(\mathbb{N})$; the first is excluded because $x\ne0$ and the third because elements of $-\iota(\mathbb{N})$ are negative. So $x\in\iota(\mathbb{N})$. The description of $\{x\in\mathbb{Z}:x<0\}$ is obtained in the same way.

Claim 2. $0\in\mathbb{Z}$ by the definition of $\mathbb{Z}$, and $1=\iota(1)\in\iota(\mathbb{N})\subseteq\mathbb{Z}$ by claim 1 of the canonical-map lemma.

Negation. If $x=0$ then $-x=0\in\mathbb{Z}$. If $x=\iota(n)$ then $-x=-\iota(n)\in\mathbb{Z}$. If $x=-\iota(n)$ then $-x=\iota(n)\in\mathbb{Z}$, using $-(-a)=a$ in $\mathbb{R}$.

Addition. Let $x,y\in\mathbb{Z}$. If $x=0$ then $x+y=y\in\mathbb{Z}$, and similarly if $y=0$. Otherwise, by claim 1 we may write $x$ as $\iota(m)$ or $-\iota(m)$, and $y$ as $\iota(n)$ or $-\iota(n)$, for some $m,n\in\mathbb{N}$. There are four cases.

If $x=\iota(m)$ and $y=\iota(n)$, then $x+y=\iota(m)+\iota(n)=\iota(m+n)\in\mathbb{Z}$ by claim 4 of the canonical-map lemma.

If $x=-\iota(m)$ and $y=-\iota(n)$, then $x+y=-\bigl(\iota(m)+\iota(n)\bigr)=-\iota(m+n)\in\mathbb{Z}$, again by claim 4.

If $x=\iota(m)$ and $y=-\iota(n)$, then by the properties of the order on the natural numbers exactly one of $m<n$, $m=n$, $n<m$ holds. If $m=n$ then $x+y=0\in\mathbb{Z}$. If $n<m$ then, by the definition of the order on $\mathbb{N}$, there is $k\in\mathbb{N}$ with $m=n+k$; then $\iota(m)=\iota(n)+\iota(k)$ by claim 4, so $x+y=\iota(k)\in\mathbb{Z}$. If $m<n$ then symmetrically $n=m+k$ for some $k\in\mathbb{N}$, so $\iota(n)=\iota(m)+\iota(k)$ and $x+y=-\iota(k)\in\mathbb{Z}$.

If $x=-\iota(m)$ and $y=\iota(n)$, the previous case applies to $y+x$, which equals $x+y$ by commutativity of addition.

Subtraction. $x-y=x+(-y)$, which lies in $\mathbb{Z}$ by closure under negation and addition.

Multiplication. If $x=0$ or $y=0$ then $xy=0\in\mathbb{Z}$. Otherwise write $x=\varepsilon\,\iota(m)$ and $y=\eta\,\iota(n)$ where each of $\varepsilon,\eta$ is $1$ or $-1$. Then $xy=(\varepsilon\eta)\,\iota(m)\iota(n)=(\varepsilon\eta)\,\iota(mn)$ by claim 5 of the canonical-map lemma, and $\varepsilon\eta$ is $1$ or $-1$; in the first case $xy=\iota(mn)\in\mathbb{Z}$ and in the second $xy=-\iota(mn)\in\mathbb{Z}$.

Claim 3. Let $x\in\mathbb{Z}$ with $0<x$. By claim 1, $x=\iota(n)$ for some $n\in\mathbb{N}$, and $1\le\iota(n)$ by claim 2 of the canonical-map lemma; so $1\le x$.

If some $x\in\mathbb{Z}$ satisfied $0<x<1$, the previous paragraph would give $1\le x$ together with $x<1$, hence $1<1$, which is false.

Finally let $x,y\in\mathbb{Z}$ with $x<y$. By claim 2 the real number $y-x$ is an integer, and $0<y-x$ because $x<y$. Hence $1\le y-x$, and adding $x$ to both sides gives $x+1\le y$.