Irrational Numbers from Adjacent Natural Numbers (Part IIb)

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

This is a continuation of Case 1 from Part IIa.

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 2 where y is odd we can show by induction that the following inequality holds:

x < √n(x + y) < y   (1)

where x,y,n ∈ ℕ

Case 2:

When y is odd substitute x ∕ 2 for y in n, which affords

x < √x ∕ 2) (2x + 1) < x+1   (2)

since y is odd when x = 2, the basis step, affords

2 < √5 ≈ 2.24 < 3 which is true.

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

x2 < (x ∕ 2) (2x + 1)< (x + 1)2  (3)

We'll proceed by adding the equation (2x + 1) to all three parts of the inequality as follows:

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

Multiplying the second 2x + 1 term by 2 ∕ 2 and expanding the right part of the inequality

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

In the center part of the inequality factor out (2x + 1) ∕ 2

x2 + 2x + 1 < (2x2 + 5x+ 1) ∕ 2 < x2 + 4x + 2     (6)

Adding ½(2x + 5) + (2x2 + 5x+ 1) ∕ 2 = (2x2 + 7x+ 6) ∕ 2  (7)

since

½(2x2 + 5x+ 1) ∕ 2 < (2x2 + 7x+ 6) ∕ 2 and

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

x2 + 2x + 1 < ½(2x2 + 5x+ 1) ∕ 2 < (2x2 + 7x+ 6) ∕ 2 < x2 + 4x + 2 < x2 + 4x + 4  (9)

which can be reduced to

x2 + 2x + 1 < (2x2 + 7x+ 6) ∕ 2 < 2x2 + 4x + 4   (10)

Multiplying the inequality thru by 2

2x2 + 4x + 2 < 2x2 + 7x+ 6 < 2x2 + 8x + 8   (11)

which is true for all x

Factoring inequality (10)

(x + 1)2 < ½(x + 2) (2x + 3) < (x + 2)2    (12)

Taking the square roots of each part in the inequality, affords

(x + 1) < √(½(x + 2) (2x + 3) < (x + 2)    (13)

and substituting y back into x + 2 with rearrangement of 2x + 3

(x + 1) < √½(y + 1) ((x + 1) + (x + 2)) < (y + 1) (14)

and a final substitution of y for x + 2

(x + 1) < √½(y + 1) ((x + 1) + (y + 1)) < (y + 1) (15)

Thus, we have proven it for the inequality x < √½(y − 1) (x + y) < y for all x when y is odd.

Part III will use the equation √(n(x + y) to generate a staircase sequence of irrational numbers.

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Go back to Part IIa. Go back to homepage.

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