Why is the set of Rational numbers countably infinite?

Question

Why is the set of Rational numbers ,$\mathbb Q$, a countably finite set?

I think that - if we assign $n$ to a rational number, and $n+1$ to another rational number, Then I can surely find a rational number in between these two, which is not accounted for.

I using the definition - If a set is countably infinite, then each element of the set can be mapped to the set of natural numbers.

Another question - Is the cardinal product of countably infinite set of countably infinite sets uncountable or countable?

Answer 1

A rational number is of the form $\frac pq$ . Associate the set with natural numbers, in this order $(1,\frac 21,\frac 12,\frac 31,\frac 22,\frac 13,\frac 41,....)$ This set is a super set of the rational numbers. This set is clearly countable. So, the set of rational numbers is countable.

Yes, the cardinal product of countably infinite set of countably infinite sets is uncountable, where as the cardinal product of countably finite set of countably infinite sets is countable.

Answer 2

Just try to think a rational number i.e $$p/q$$ $$(p,q)=1$$ where p,q belong to integers As an ordered pair of integers. And try to define a map from $Q$ to $Z×Z$ as $$f (p/q)=(p,q)$$ where Z is the set of integers. You'll see that this is a one one onto map .Hence follows the conclusion..

Answer 3

You say

I think that - if we assign $n$ to a rational number, and $n+1$ to another rational number, Then I can surely find a rational number in between these two, which is not accounted for.

Well, it depends on how you assign $n$ to a rational number. There exists a way in which you can create a mapping $\mathbb N\to \mathbb Q$ such that you cannot find a rational number in between:

$$1\to \frac11\\ 2\to\frac21\\ 3\to\frac12\\ 4\to\frac31\\ 5\to\frac22\\ 6\to\frac13\\ 7\to\frac41\\ 8\to\frac32\\ 9\to\frac23\\ 10\to\frac14\\ \vdots$$