Dense Property of the Rational numbers
Theorem
Given any two real numbers 𝑎 and 𝑏, with 𝑎 < 𝑏 there exists a rational number 𝑟 such that
𝑎<𝑟<𝑏
𝑚
Proof. We can assume that every rational number is of the form where 𝑛 > 0
𝑛
Given that 𝑏 − 𝑎 > 0 then in the Archimedean property, taking 𝑦 = 1, there exists an integer 𝑛
such that 𝑛 𝑏 − 𝑎 > 1. Hence 𝑛𝑏 − 𝑛𝑎 > 1 so that the interval 𝑛𝑎, 𝑛𝑏 will contain an integer ,
𝑚
say 𝑚. then we have 𝑛𝑎 < 𝑚 < 𝑛𝑏 or 𝑎 < < 𝑏. Thus proving the result.
𝑛
Remark. This property of rational numbers is usually stated as the set of rational numbers is dense in
ℝ. This means that where ever in the real line we consider an interval, there is a rational number in
it. or every interval contains a rational number. Now it is easy to see that this implies that in
between two real numbers 𝑎 and 𝑏, there exist infinitely many rational numbers
Corollary
Given any two real numbers 𝑎 and 𝑏, with 𝑎 < 𝑏 there exists an irrational number 𝑡 such that
𝑎<𝑡<𝑏
Proof. Since 𝑎 < 𝑏, we have 𝑎 − 2 < 𝑏 − 2 then by above theorem, there is a rational number 𝑟
with 𝑎 − 2 < 𝑟 < 𝑏 − 2. Then we get 𝑎 < 𝑟 + 2 < 𝑏. Clearly 𝑟 + 2 is an irrational number
Problems
1. For 𝑎 ∈ ℝ, let 𝐶𝑎 = 𝑟 ∈ ℚ ∶ 𝑟 < 𝑎 then show that 𝑙𝑢𝑏𝐶𝑎 = 𝑎
2. Given any two real numbers 𝑎 and 𝑏, with 𝑎 < 𝑏 then ℚ ∩ (𝑎, 𝑏) is infinite
3. Given any two real numbers 𝑎 and 𝑏, with 𝑎 < 𝑏 and 𝑡 > 0. Show that there exists rational
number 𝑟 such that 𝑎 < 𝑡𝑟 < 𝑏