Sequences for Pythagorean Multiples

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) + x + y). This will be discussed in more detail in Part VIIIa. 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. Case 1 when y is even is discussed in Part VIIa.

x < √n((x ∕ y) + x + y) < y (1)

where x,y ∈ ℕ

Case 2:

Substituting ½(y − 1) = ½x for n, affords:

x < √ ½x ((x + xy + y²) ∕ y) < y (2)

and distributing the x

x < √ ½ (x² + x²y + xy²) ∕ y < y (3)

Replacing y by x + 1 affords

x < √ ½ [(x² + x²(x + 1) + x(x + 1)²] ∕ (x + 1) < x + 1 (4)

Multiplying and expanding

x < √ ½ (2x³ + 4x² + x) ∕ (x + 1) < x + 1 (5)

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

1 < √3.5 ≈ 1.87 < 2 which is true.

Squaring the left, center and right parts of the inequality (5), affords:

x² < ½ (2x³ + 4x² + x) ∕ (x + 1) < (x + 1)² (6)

Multiplying and expanding affords

x² < ½ (2x³ + 4x² + x) ∕ (x + 1) < x² + 2x + 1 (7)

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

x² + 2x + 1 < ½(2x³ + 4x² + x) ∕ (x + 1) + 2x + 1 < (x + 1)² + 2x + 1 (8)

Multiplying in the middle part of the inequality the second term by (2(x + 1)) ∕ (2(x + 1))

x² + 2x + 1 < ½(2x³ + 4x² + x) ∕ (x + 1) + (2x + 1)(2(x + 1)) ∕ (2(x + 1)) < (x + 1)² + 2x + 1 (9)

Expanding and adding all the terms

x² + 2x + 1 < ½(2x³ + 8x² + 7x + 2) ∕(x + 1) < x² + 4x + 2 (10)

At this point we have to transform the x + 1 into x + 2

Multiplying the middle part of (10) by (x + 2) ∕ (x + 2)

½(2x⁴ + 12x³ + 23x² + 16x + 4) ∕ (x + 1)(x + 2) (11)

Subtracting 1 ∕ (x + 1)(x + 2) from (11)

½(2x⁴ + 12x³ + 23x² + 16x + 3) ∕ (x + 1)(x + 2) (12)

Dividing ½(2x⁴ + 12x³ + 23x² + 16x + 3) by (x + 1)

½(2x³ + 10x² + 13x + 3) ∕ (x + 2) (13)

Adding 2(x + 2) ∕ 2(x + 2) to (13)

½(2x³ + 10x² + 15x + 7) ∕ (x + 2) (14)

Since

x² + 4x + 2 < x² + 4x + 4 then

We can replace the left part of this inequality by the right part and rearrange the inequality as

x² + 2x + 1 < ½(2x³ + 8x² + 7x + 2) ∕(x + 1) < ½(2x³ + 10x² + 15x + 7) ∕ (x + 2) < x² + 4x + 4 (15)

Multiplying thru by 2(x + 1)(x + 2) to afford

2x⁴ + 10x³ + 18x² + 14x + 2 < 2x⁴ + 10x³ + 23x² + 16x + 4 < 2x⁴ + 12x³ + 25x² + 22x + 7 < 2x⁴ + 14x³ + 36x² + 40x + 16 (16)

which is true for all x when y is odd and the inequality can now be replaced by

x² + 2x + 1 < ½(2x³ + 10x² + 15x + 7) ∕ (x + 2) < x² + 4x + 4 (17)

Taking the square roots of each term in the inequality, affords:

(x + 1) < √½(2x³ + 10x² + 15x + 7) ∕ (x + 2) < (x + 2) (18)

and working backwards by factoring and grouping terms

(x + 1) < √½[(x² + 2x + 1) + (x³ + 4x² + 5x + 2) + (x³ + 5x² + 8x + 4)] ∕ (x + 2) < (x + 2) (19)

and factoring each group

(x + 1) < √½[(x + 1)² + (x + 1)²(x + 2) + (x + 1)(x + 2)² )] ∕ (x + 2) < (x + 2) (20)

Factoring out x + 1 from the equation inside the square root

(x + 1) < √½(x + 1)[x + 1 + (x + 1)(x + 2) + (x + 2)²] ∕ (x + 2) < (x + 2) (21)

Dividing the terms within the brackets by x + 2

(x + 1) < √½(x + 1)[(x + 1) ∕ (x + 2) + (x + 1) + (x + 2)] < (x + 2 (22)

and for under the square root substituting back y + 1 for all x + 2 as well as y for the first x + 1

(x + 1) < √½y[(x + 1) ∕ (y + 1) + (x + 1) + (y + 1)] < (y + 1) (23)

Thus we have proven the inequality x < √n((x ∕ y) + x + y) < y for all x when y is odd and n = ½(y − 1)

Part VIII will use the equation√n((x ∕ y) + x + y) to generate a staircase sequence of irrational numbers. Go back to Part VIIa.

Irrational Numbers from Adjacent Natural Numbers (Part VIIb)

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