Irrational Numbers from Adjacent Natural Numbers (Part VIIa2)
A First Inductive Proof for the Irrational Number √(n((x2 ∕ y2) + 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((x2 ∕ y2) + x + y). Although a general equation √n((xk ∕ yk) + x + y) was proposed in Part IXa no definitive conclusion since complicated math is required. Therefore, it was easier to look at a less complicated system such as when k = 2. The data for this system is already tabulated and discussed in Part IXb. 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((x2 ∕ y2) + x + y) < y (1)
where x,y ∈ ℕ
Case 1:
Substituting y ∕ 2 for n and multiplying thru by y2 affords:
x < √(y ∕ 2) ((x2 + xy2 + y3) ∕ y2) < y (2)
Cancelling out y, affords,
x < √ ½ ((x2 + xy2 + y3) ∕ y) < y (3)
Replacing y by x + 1 affords
x < √ ½ (x2 + x(x + 1)2 + (x + 1)3) ∕ (x + 1) < x + 1 (4)
then substituting x = 1 in the basis step, affords:
1 < √3.25 ≈ 1.80 < 2 which is true.
Squaring the left, center and right parts of inequality (4), affords:
x2 < ½ (x2 + x(x + 1)2 + (x + 1)3 < (x + 1)2 (5)
Multiplying and expanding affords
x2 < ½ (x2 + x3 + 2x2+ x + x3 + 3x2 + 3x + 1 ) ∕ (x + 1) < x2 + 2x + 1 (6)
Adding like terms
x2 < ½ (2x3 + 10x2 + 4x + 1) ∕ (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 + 10x2 + 4x + 1) ∕ (x + 1)] + 2x + 1 < (x + 1)2 + 2x + 1 (8)
Expanding
x2 + 2x + 1 < ½ (2x3 + 10x2 + 4x + 1 + 4x2 + 6x + 2 ) ∕ (x + 1) < x2 + 4x + 2 (9)
Adding terms
x2 + 2x + 1 < ½ (2x3 + 10x2 + 10x + 3) ∕ (x + 1) < x2 + 4x + 2 (10)
Multiplying the middle term by (x + 2) ∕ (x + 2)
x2 + 2x + 1 < ½ (2x4 + 14x3 + 30x2 + 23x + 6) ∕ (x + 1)(x + 2) < x2 + 4x + 2 (11)
and since
x2 + 4x + 2 < x2 + 4x + 4 then
x2 + 2x + 1 < ½ (2x4 + 14x3 + 30x2 + 23x + 6) ∕ (x + 1)(x + 2) < x2 + 4x + 4 (12)
Adding (2(x + 1)(x + 2) − 1) ∕ (x + 1)(x + 2) to the middle term
x2 + 2x + 1 < ½ (2x4 + 14x3 + 34x2 + 35x + 13) ∕ (x + 1)(x + 2) < x2 + 4x + 4 (13)
and after multiplying the inequality by the denominator
2x4 + 10x3 + 18x2 + 14x + 4 < 2x4 + 14x3 + 34x2 + 35x + 13 < 2x4 + 14x3 + 36x2 + 40x +16 (14)
which is true for all x when y is even.
Dividing the middle part of the inequality (13) by x + 1 affords
x2 + 2x + 1 < ½ (2x3 + 12x2 + 22x + 13) ∕ (x + 2) < x2 + 4x + 4 (15)
Working backwards by taking apart the equation and grouping terms
x2 + 2x + 1 < ½ [(x2 + 2x + 1) + (x3 + 5x2 + 8x + 4) + (x3 + 6x2 + 12x + 8)] ∕ (x + 2) < x2 + 4x + 4 (16)
Factoring the groups and taking the square roots of all three parts of the inequality affords
x + 1 < √ ½ ((x + 1)2 + (x + 1)(x + 2)2 + (x + 2)3) ∕ (x + 2) < x + 2 (17)
Substituting y + 1 for all x + 2 back into the inequality
x + 1 < √ ½ [(x + 1)2 + (x + 1)(y + 1)2 + (y + 1)3] ∕ (y + 1) < y + 1 (18)
Multiplying the equation under the square root by (y + 1) ∕ (y + 1)
x + 1 < √ ½ (y + 1) [(x + 1)2 + (x + 1)(y + 1)2 + (y + 1)3] ∕ (y + 1)2 < y + 1 (19)
then dividing by (y + 1)2
x + 1 < √ ½ (y + 1) [(x + 1)2 ∕ (y + 1)2 + (x + 1) + (y + 1)] < y + 1 (20)
Thus we have proven the inequality x < √n((x2 ∕ y2) + x + y) < y for all x when y is even and n = ½y
Part VIIb2 will continue with a proof for Case 2
of √n((x2 ∕ y2) + x + y).
Part IXb will use the equation √n((x2 ∕ y2) + x + y) to generate a staircase sequence of irrational numbers.
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