Proof of The Number of Integers between Two Integers

Each result cited below is universally quantified over the data in its own statement, and is applied to the data named where it is cited. Throughout, an element $k$ of $\mathbb{N}_{0}$ standing where an integer is required, in particular as an operand next to an integer, stands for $\iota_{0}(k)$, as in The Integers and the Rational Numbers, with the Natural Numbers and the Integers Identified with Subsets of the Rationals §identification and The Integers and the Rational Numbers, with the Natural Numbers and the Integers Identified with Subsets of the Rationals §numerals. By The Integers and the Rational Numbers, with the Natural Numbers and the Integers Identified with Subsets of the Rationals §embeddings, $\iota_{0}:\mathbb{N}_{0}\to\mathbb{Z}$ is injective and preserves $0$, $1$, sums and the order in both directions, and its image is the set of nonnegative integers, that is $\{x\in\mathbb{Z}:0_{\mathbb{Z}}\le x\}$ by Positive, Nonnegative, Negative and Nonpositive Elements of an Ordered Ring §sign. By The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §ring, $\mathbb{Z}$ is a commutative ring with $u+(-u)=0_{\mathbb{Z}}$, and $u-v=u+(-v)$ by The Integers §operations. By the ring laws we mean the identities of Commutative Rings §ring (associativity and commutativity of $+$, and $u+0_{\mathbb{Z}}=u$) together with $u+(-u)=0_{\mathbb{Z}}$. They give, for all $u,v\in\mathbb{Z}$,

$$u-u=0_{\mathbb{Z}},\qquad(u+v)-v=u,\qquad(v+u)-v=u,\qquad(u-v)+v=u,\qquad v+(u-v)=u,$$

since $u-u=u+(-u)=0_{\mathbb{Z}}$, $(u+v)-v=u+(v+(-v))=u+0_{\mathbb{Z}}=u$, $(v+u)-v=(u+v)-v$, $(u-v)+v=u+((-v)+v)=u+(v+(-v))=u$, and $v+(u-v)=(u-v)+v$. By The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §ordered-ring, $\le$ is a total order on $\mathbb{Z}$ and $\mathbb{Z}$ is an ordered ring, so for $u,v,w\in\mathbb{Z}$ with $u\le v$ we have $u+w\le v+w$ by Ordered Rings §ordered-ring, hence $w+u\le w+v$ by commutativity, and $u-w\le v-w$, taking $-w$ in place of $w$.

The difference $b-a$. From $a\le b$, subtracting $a$ gives $a-a\le b-a$, that is $0_{\mathbb{Z}}\le b-a$ by the ring laws. Hence $b-a$ lies in the image of $\iota_{0}$, and we choose $d\in\mathbb{N}_{0}$ with $\iota_{0}(d)=b-a$.

The number $b-a+1$. Let $n=d+1$. Here $\mathbb{N}_{0}$ is the set $\omega$ of The Class Omega of Natural Numbers with Zero §omega, as in The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §sets, and $\mathbb{N}=\mathbb{N}_{0}\setminus\{0\}$ with $1\in\mathbb{N}$ by the same clause, so $1\neq0$. Hence $d+1\neq0$ by Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §zero-sum, and so $n\in\mathbb{N}$, again by The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §sets. As $\iota_{0}$ preserves sums and $1$, $\iota_{0}(n)=\iota_{0}(d)+\iota_{0}(1)=(b-a)+1_{\mathbb{Z}}$. Thus the integer $b-a+1$ is the image under $\iota_{0}$ of the natural number $n$; by The Integers and the Rational Numbers, with the Natural Numbers and the Integers Identified with Subsets of the Rationals §identification it stands for $n$, and in this sense $b-a+1\in\mathbb{N}$.

The map. Let $I=\{x\in\mathbb{Z}:a\le x\le b\}$, a set as a subset of $\mathbb{Z}$ formed by Sets and Maps: Ordinary Notation §set-builder. By Intervals of Natural Numbers §segment and Intervals of Natural Numbers §interval, $[n]=\{k\in\mathbb{N}_{0}:1\le k\le n\}$. Let $k\in[n]$. Then $1\le k\le n$ in $\mathbb{N}_{0}$, hence $1_{\mathbb{Z}}\le k\le\iota_{0}(n)$ in $\mathbb{Z}$, as $\iota_{0}$ preserves $1$ and the order. Adding $a$ on the left and then subtracting $1_{\mathbb{Z}}$ gives

$$(a+1_{\mathbb{Z}})-1_{\mathbb{Z}}\le(a+k)-1_{\mathbb{Z}}\le(a+\iota_{0}(n))-1_{\mathbb{Z}}.$$

Here $(a+1_{\mathbb{Z}})-1_{\mathbb{Z}}=a$ by the ring laws, and, as $\iota_{0}(n)=(b-a)+1_{\mathbb{Z}}$, associativity gives $a+\iota_{0}(n)=(a+(b-a))+1_{\mathbb{Z}}=b+1_{\mathbb{Z}}$ by the ring laws, so $(a+\iota_{0}(n))-1_{\mathbb{Z}}=b$. Hence $a\le(a+k)-1_{\mathbb{Z}}\le b$ and $(a+k)-1_{\mathbb{Z}}\in I$. By Sets and Maps: Ordinary Notation §maps, $f:[n]\to I$, $k\mapsto(a+k)-1_{\mathbb{Z}}$, is a map.

Injectivity. Let $k,k'\in[n]$ with $f(k)=f(k')$. Adding $1_{\mathbb{Z}}$ gives $a+k=a+k'$ by the ring laws, as $((a+k)-1_{\mathbb{Z}})+1_{\mathbb{Z}}=a+k$; subtracting $a$ gives $\iota_{0}(k)=\iota_{0}(k')$, as $(a+k)-a=\iota_{0}(k)$ by the ring laws. Hence $k=k'$, $\iota_{0}$ being injective, and $f$ is injective.

Surjectivity. Let $x\in I$. From $a\le x$, subtracting $a$ gives $0_{\mathbb{Z}}=a-a\le x-a$, so we may choose $e\in\mathbb{N}_{0}$ with $\iota_{0}(e)=x-a$. From $x\le b$, subtracting $a$ gives $\iota_{0}(e)=x-a\le b-a=\iota_{0}(d)$, so $e\le d$ in $\mathbb{N}_{0}$, as $\iota_{0}$ reflects the order. Let $k=e+1\in\mathbb{N}_{0}$. As $0\le e$, $0$ being the least element of $\mathbb{N}_{0}$ by The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §order, Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §order and Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §commutative give $0+1\le e+1$ and $e+1\le d+1$, and $0+1=1$ by Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §zero; so $1\le k\le n$ and $k\in[n]$. Finally $\iota_{0}(k)=\iota_{0}(e)+1_{\mathbb{Z}}=(x-a)+1_{\mathbb{Z}}$, as $\iota_{0}$ preserves sums and $1$, so by associativity and the ring laws

$$f(k)=(a+((x-a)+1_{\mathbb{Z}}))-1_{\mathbb{Z}}=((a+(x-a))+1_{\mathbb{Z}})-1_{\mathbb{Z}}=(x+1_{\mathbb{Z}})-1_{\mathbb{Z}}=x.$$

Hence $f$ is surjective, and so a bijection from $[n]$ onto $I$.

Clause count. As $n\in\mathbb{N}\subseteq\mathbb{N}_{0}$ and $f$ is a bijection from $[n]$ onto $I$, the set $I$ is finite. By The Number of Elements of a Finite Set §cardinality, $\#I$ is the unique element of $\mathbb{N}_{0}$ for which there is a bijection from its segment onto $I$, unique by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §unique; as $f$ is such a bijection for $n$, $\#I=n$. Since $b-a+1$ stands for $n$, as shown above, $\#I=b-a+1$; read in $\mathbb{Z}$, this says $\iota_{0}(\#I)=\iota_{0}(n)=(b-a)+1_{\mathbb{Z}}$, equality agreeing in $\mathbb{N}_{0}$ and in $\mathbb{Z}$ by The Integers and the Rational Numbers, with the Natural Numbers and the Integers Identified with Subsets of the Rationals §agreement. We already showed $b-a+1\in\mathbb{N}$, which completes the clause.