No. That is the usual proof. It works.
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?