Question

If x and y are arbitrary real numbers such that x < y, prove that there exists at least one rational number r satisfying x < r < y, and hence infinitely many.

Answer 1

Since (y − x) > 0, by the Archimedean property, there exists a positive integer n such that

n(y − x) > 1. Now, let m = [nx] (the greatest integer ≤ nx). Then m ≤ nx < m + 1. Also, we must have

m + 1 < ny. To see this, assume that m + 1 ≥ ny. Then, ny − 1 ≤ m ≤ nx, which is a contradiction since

n(y − x) > 1. Thus, we have m ≤ nx < m + 1 < ny. Dividing by n, we get (m/n) ≤ x < (m + 1)/n < y.

We have thus proved that the rational number r = (m + 1)/n lies between x and y. We can now apply this result to r and y.

By the Archimedian property we havem ≤ nx < m + 1 < ny. Dividing by n, we get (m/n) ≤ x < (m + 1)/n < y.

We have thus proved that the rational number r = (m + 1)/n lies between x and y. We can now apply this result to r and y.

Hence result.