Show that if a, b∈ R with a < b, then there exists a number x ∈ I with a < x < b

Let I be the set of real numbers that are not rational. (We call

elements of I irrational numbers. Show that if a, b are real

numbers with a < b, then there exists a number x ∈ I with

a < x < b.

Im not sure if this is right:

Using the density theorem to the real numbers a/√2 and b/√2, we obtain a rational number such that:

a/√2 < r < b/√2

where r ≠ 0

This gives ---> x = r√2, which is an irrational number because it contains √2, which itself is irrational. This satisfies b < x < a

Is this right? or do i need more in the proof?