Which set of real numbers is bigger set of integers or set of rational numbers explain?

Answer 1

The question is not well-posed, in that the term "bigger" can be understood in different ways.

If A is a subset of B, we can call B bigger than A.

However, in set theory, the cardinality of a set is defined as the class of sets with the "same number" of elements:

Two sets A and B have the same cardinality if there exists a bijection f:A->B.

Theorem: If there is an injection i:A->B and an injection i:B->A, then there is a bijection f:A->B. Not proved here.

The set of integers and the set of rational numbers can be mapped as follows.

Since the natural numbers are a subset of the rational numbers by i:N->R: n-> n/1, we have half of the proof.

Now, order the rational numbers as follows:

- assign to each rational number p/q (p,q > 0) the point (p,q) in the plane.

Next, consider that you can assign a natural number to each rational number by walking through them in diagonals:

(1,1) -> 1; (2,1) -> 2; (1,2) -> 3; (3,1) ->4 ; (2,2) ->5; (1,3) -> 6; (4,1) -> 7; (3,2) -> 8, (2,3) -> 9; (1,4) -> 10, etc. (make a drawing).

It is clear that in this way you can assign a unique natural number to EACH rational number. This means that you have an injection from the rational numbers to the natural numbers.

Now you have two injections, from the natural numbers to the rational numbers and from the rational numbers to the natural numbers.

By the theorem, there is a bijection, which means that the natural numbers and the rational numbers have the same cardinality. Neither of them is "bigger" than the other in this sense. The cardinality of these two sets is called Aleph-zero, and the sets are also called countable (because the elements can be counted with the natural numbers).