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Proof of The Integers and the Rational Numbers are Countable

lemmalem:rationals-countable-2026a
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Reason: First published proof of lem:rationals-countable-2026a: the integers as the image of the pairs of natural numbers, the rationals as the image of a product of integer sets.

Proof

Throughout, N\mathbb{N} is the set of natural numbers, ι:NR\iota:\mathbb{N}\to\mathbb{R} is the canonical map of R\mathbb{R}, claim numbers for ι\iota refer to Properties of the Canonical Map from the Natural Numbers to an Ordered Field, and claim numbers for Z\mathbb{Z} refer to Arithmetic, Order and Discreteness of the Integers. For s,tRs,t\in\mathbb{R}, sts-t abbreviates s+(t)s+(-t) in the ordered field R\mathbb{R}.

Claim 1. Define

g:N×NR,g(m,n)=ι(m)ι(n).g:\mathbb{N}\times\mathbb{N}\to\mathbb{R},\qquad g(m,n)=\iota(m)-\iota(n).

Each value of gg is an integer: ι(m),ι(n)Z\iota(m),\iota(n)\in\mathbb{Z} by the definition of Z\mathbb{Z}, and Z\mathbb{Z} is closed under the difference of two of its elements by claim 2.

Conversely, every integer is a value of gg. By claim 1, ι(1)=1\iota(1)=1 and ι(n+1)=ι(n)+1\iota(n+1)=\iota(n)+1 for nNn\in\mathbb{N}. Hence

g(1,1)=11=0,g(n+1,1)=(ι(n)+1)1=ι(n),g(1,n+1)=1(ι(n)+1)=ι(n)g(1,1)=1-1=0,\qquad g(n+1,1)=\bigl(\iota(n)+1\bigr)-1=\iota(n),\qquad g(1,n+1)=1-\bigl(\iota(n)+1\bigr)=-\iota(n)

for every nNn\in\mathbb{N}. By the definition of Z\mathbb{Z}, every integer is 00, or ι(n)\iota(n), or ι(n)-\iota(n) for some nNn\in\mathbb{N}, so Z={g(w):wN×N}\mathbb{Z}=\{g(w):w\in\mathbb{N}\times\mathbb{N}\}. Since N×N\mathbb{N}\times\mathbb{N} is countable by The Set of Pairs of Natural Numbers is Countable, claim 4 of Basic Properties of Countable Sets shows that Z\mathbb{Z} is countable.

Claim 2. The set Z{0}\mathbb{Z}\setminus\{0\} is a subset of Z\mathbb{Z}, hence countable by claim 1 above and claim 3 of Basic Properties of Countable Sets. By claim 1 of Products and Powers of Countable Sets, the Cartesian product Z×(Z{0})\mathbb{Z}\times(\mathbb{Z}\setminus\{0\}) is countable. Define

q:Z×(Z{0})R,q(a,b)=ab1,q:\mathbb{Z}\times(\mathbb{Z}\setminus\{0\})\to\mathbb{R},\qquad q(a,b)=a\,b^{-1},

which is well defined because b0b\ne0 has a multiplicative inverse in the field R\mathbb{R}. By the definition of Q\mathbb{Q}, the set of values of qq is exactly Q\mathbb{Q}, so claim 4 of Basic Properties of Countable Sets shows that Q\mathbb{Q} is countable.

Claim 3. Immediate from claim 2 and claim 2 of Products and Powers of Countable Sets, applied with X=QX=\mathbb{Q}.

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