Is this sufficient to prove that the positive rational number is countable?

Asked Jun 11 '20 at 08:35

Active Jun 11 '20 at 08:44

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n∈N-{0} , m∈N , n/m

f(n,1)=1(n/1) is bijetive and countable...the cardinality is n

f(n,2)=2(n/2) is ibjetive and countable...the cardinality of f(n,2) plus the previous cardinality is 2n-floor(n/2)

f(n,3)=3(n/3) is ibjetive and countable...the cardinality of f(n,3) plus the previous cardinality is 3n-floor(n/3)

The cardinality of f(n,m) is nm+1-$\sum_{i=2}^{m} $$⌊\frac{n}{i}⌋$ and it is countable.

+1 because of 0/m

My question is :

1.Did it prove postive rational number is countable?

2.How to change $\sum_{i=2}^{m} $$-⌊\frac{n}{i}⌋$ to the closed form?

edited Jun 11 '20 at 08:44

asked Jun 11 '20 at 08:35

1

The argument is correct. There should be a concrete bijection somewhere on this site. – Peter Jun 11 '20 at 08:41
@Peter Such a concrete bijection is given, for example, by the Calkin-Wilf sequence. A proof that such a sequence enumerates the positive rationals can be found here. – Rick Jun 11 '20 at 13:13

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