To show: There's at least one rational number between any two different irrationals.
Outline of proof: Consider the non-repeating binary expansions of any two irrationals and then construct a rational that lies between them.
Proof (by construction): Let s and t be irrationals with s < t. We construct a number r as follows:
(1) Beginning with the highest-order expansion term in t, assign t's coefficient to the expansion of r so long as t's coefficient is equal to the corresponding coefficient in s.
(2) Due to the stipulation that s < t, we're guaranteed that we will eventually reach a 1 in t that corresponds to a 0 in s. Assign a 1 to r in that place.
(3) Assign a 0 to each subsequent term in r.
Claim that r lies in the interval (s, t) and is rational.
s < r by (2); r < t by (3), and r is rational also by (3).
To show: There's at least one irrational number between any two different rationals.
Discussion: The key to the proof is finding a pair of rationals between which there lies at least one irrational. It is then easy to show that the interval between any two rationals will contain at least one irrational.
Proof: Observe that √2 lies in the interval (1, 2). Next consider the more general interval (a/b, c/d), where a, b, c, and d are integers with b and d not zero and a/b < c/d.
Claim that v = a/b + (c/d − a/b)(√2/2) is irrational and lies in (a/b, c/d).
To show that a/b < v, we utilize the fact that √2 > 0.
a/b < c/d ⇒ 0 < c/d − a/b
⇒ 0 < (c/d − a/b)(√2/2)
⇒ a/b < a/b + (c/d − a/b)(√2/2) = v
To show that v < c/d, we utilize the fact that 2 > √2.
a/b < c/d ⇒ a/b − c/d < 0
⇒ (a/b − c/d)(1 − √2/2) < 0
⇒ a/b − c/d + (c/d − a/b)(√2/2) < 0
⇒ a/b + (c/d − a/b)(√2/2) < c/d
⇒ v < c/d
We are done, except for one small point. To show that v, as the sum of a rational and an irrational, is irrational. Let w be irrational, and let e, f, g, and h be integers, f and h not zero. By contradiction, let w + e/f = g/h. It follows that w = g/h − e/f = (gf − eh)/fh, which is rational.