Irrational Numbers from Adjacent Natural Numbers (Part XIa)
A Staircase Sequence of Irrational Numbers from √(x + 1)y2 ∕ n
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 the initial common difference (d) between two irrational numbers generated from √(x + 1)y2 ∕ n is an irrational number between √8 ∕ 3 < d < √27 ∕ 4 which approaches d=1 as the number of natural numbers grows without bound. Therefore, we can proceed as follows:
If x and y are consecutive natural numbers where y = x + 1 and n = x + 2, then we can show by induction that:
x < √(x + 1)y2 ∕ n < y
where x,y ∈ ℕ and √(x + 1)y2 ∕ n is an irrational number.
First substituting x + 1 for y and x + 2 for n
x < √(x + 1)3∕ (x + 2) < x + 1
(1)
then substituting x = 1 in the basis step
1 < √8 ∕ 3 ≈ 1.633 < 2 which is true.
Squaring the left, center and right arts of the inequality (1)
x2 < (x + 1)3∕ (x + 2) < (x + 1)2 (2)
To prove the chain notation for all numbers we add the equation (2x + 1) to all three parts in the inequality as follows:
x2 + 2x + 1 < (x + 1)3 ∕ (x + 2) + 2x + 1 < (x + 1)2 + 2x + 1 (3)
and then expanding
x2 + 2x + 1 < (x3 + 3x2 + 3x + 1) ∕ (x + 2) + 2x + 1 < x2 + 2x + 1 + 2x + 1 (4)
To the middle part of the inequality multiplying
(2x + 1)(x + 2) ∕ (x + 2) = (2x2 + 5x + 2) ∕ (x + 2) (4a)
adding up all the terms
x2 + 2x + 1 < (x3 + 5x2 + 8x + 3) ∕ (x + 2) < x2 + 4x + 2 (5)
Multiplying the middle part of the inequality by (x + 3) ∕ (x + 3)
and since
x2 + 4x + 2 < x2 + 4x + 4 then
x2 + 2x + 1 < (x4 + 8x3 + 23x2 + 27x + 9) ∕ (x + 2)(x + 3) < x2 + 4x + 2 < x2 + 4x + 4 (6)
which can be reduced to
x2 + 2x + 1 < (x4 + 8x3 + 23x2 + 27x + 9) ∕ (x + 2)(x + 3) < x2 + 4x + 4 (7)
Adding 1 ∕ (x + 2)(x + 3) to the middle part of the inequality (7)
x2 + 2x + 1 < (x4 + 8x3 + 23x2 + 27x + 10) ∕ (x + 2)(x + 3) < x2 + 4x + 4 (8)
Dividing the middle part of the inequality (8) by x + 2
x2 + 2x + 1 < (x3 + 6x2 + 11x + 5) ∕ (x + 3) < x2 + 4x + 4 (9)
Adding (x + 3) ∕ (x + 3) to the middle part of the inequality (9)
x2 + 2x + 1 < (x3 + 6x2 + 12x + 8) ∕ (x + 3) < x2 + 4x + 4 (10)
Multiplying the whole inequality by x + 3
x3 + 5x2 + 6x + 3 < x3 + 6x2 + 12x + 8 < x3 + 7x2 + 16x + 12 (11)
which is true for all x
Factoring all parts of the inequality (10)
(x + 1)2 < (x + 2)3 ∕ (x + 3) < (x + 2)2 (12)
Taking the square roots of all the parts, affords:
(x + 1) < √(x + 2)3∕ (x + 3) < (x + 2) (13)
Substituting x = y − 1 for two of the (x + 2) terms under the square root and for the
rightmost part of the inequality and n for x + 2 in denominator (finally affords:
(x + 1) < √x + 2)(y + 1)2∕ (n + 1) < (y + 1) (14)
and proves the chain inequality: x < √(x + 1)y2 ∕ n < y for all x and for y = x + 1.
Go to Part XIb for the staircase sequence.
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