Prove if a b element of R then for all n element of N there

Prove if a, b element of R, then for all n element of N, there are numbers x_1, x_2, ...,x_n-1, x_n element of Q, all different, so that for all i = 1, 2, 3...,n, we have a

Solution

Let a < b, Consider the set of numbers of the form x_i= p_i/q_i with q_i fixed, and p_i any

integer. Assume that there are no such numbers between a and b. Let x₁= p₁/q₁ be the number immediately before a. Then p₁+1/q₁ is the number immediately after b. We necessarily have

p₁+1/q₁ - p/q >= b-a

which implies that 1/q >= b-a.

If we choose q sufficiently large, then the above inequality is wrong. Then there is at least one rational number between a and b.

We know that there must be a rational number, say x₁, between a and b. Then there is another rational number, say x₂, between x₁ and b. Then there is x₃, etc... All those numbers are distinct and they are between a and b.

Hence if a and b are real numbers then for all n belongs to N, there are numbers x_1, x_2, x_3, x_4,... x_n belongs to rational number so that we have a < x _i< b.