Irrational Numbers from Adjacent Natural Numbers (Part IIa)

A First Inductive Proof for the Irrational Number √(n(x + y).

It has been shown (Between_two_Rational_Numbers on Wiki) that between two real numbers (ℝ) there exists an irrational number. Accordingly, it will be shown here that between two adjacent natural numbers (ℕ) there exists an irrational number (ℝ\ℚ). In addition, an arithmetic progression or sequence of irrational numbers will be generated where there are two common differences (Δ1) and (Δ2) between the irrational numbers generated from √n(x + y). This will be discussed in more detail in Part III. So, therefore, let's proceed as follows:

If x and y are consecutive natural numbers where y = x + 1 and n may have two values: n = y ∕ 2 when y is even and n = (y − 1) ∕ 2 when y is odd. So in Case 1 where y is even we can show by induction that the following inequality holds

x < √n(x + y) < y     (1)

where x,y ∈ ℕ

Case 1:

When y is even substitute (x + 1) ∕ 2 for y in n which affords:

x < √((x + 1) ∕ 2) (2x + 1) < x+1   (2)

then substituting x = 1 in the basis step, affords:

1 < √3 < 2 which is true.

Squaring the left, center and right parts of inequality (2)

x2 < ((x + 1) ∕ 2) (2x + 1)< (x + 1)2  (3)

Adding the equation (2x + 1) to all three parts of the inequality as follows:

x2 + 2x + 1 < (½(x + 1)) (2x + 1) + 2x + 1< (x + 1)2 + 2x + 1  (4)

multiplying out terms in the middle part the inequality and multiplying (2x + 1) by 2 ∕ 2

x2 + 2x + 1 < ½(2x2 + 3x + 1) + 2(2x + 1) ∕ 2 < x2 + 2x + 1 + 2x + 1  (5)

Adding up the terms

x2 + 2x + 1 < ½(2x2 + 7x + 3) < x2 + 4x + 2   (6)

since

½(2x2 + 7x + 3) < ½(2x2 + 7x + 6) and

x2 + 4x + 2 < x2 + 4x + 4 then

x2 + 2x + 1 < ½(2x2 + 7x + 3) < ½(2x2 + 7x + 6) <x2 + 4x + 2 < x2 + 4x + 4   (7)

which can be reduced to

x2 + 2x + 1 < ½(2x2 + 7x + 6) < x2 + 4x + 4   (8)

Multiplying thru the inequality by 2

2x2 + 4x + 2 < 2x2 + 7x + 6 < 2x2 + 8x + 8   (9)

which is true for all x

Factoring all the terms in the inequality (8)

(x + 1)2 < ½(x + 2) (2x + 3) < (x + 2)2    (10)

and taking the square roots of each term in the inequality

(x + 1) < √((x + 2) ∕ 2) (2x + 3) < (x + 2)    (11)

and substituting y back into the inequality with a rearrangement of last term in the middle part of the inequality

(x + 1) < √((y + 1) ∕ 2) ((x + 1) + (x + 2)) < (y + 1) (12)

and a last substitution of y for x + 2

(x + 1) < √((y + 1) ∕ 2) ((x + 1) + (y + 1)) < (y + 1) (13)

Thus we have proven the inequality x < √(y ∕ 2) (x + y) < y for all x when y is even.

Part IIb will continue with a proof for Case 2 of √(n(x + y).
Part III will use the equation √(n(x + y) to generate a staircase sequence of irrational numbers.

Go to Part I. Go back to homepage.

Copyright © 2026 by Eddie N Gutierrez. E-Mail: enaguti1949@gmail.com