Irrational Numbers from Adjacent Natural Numbers (Part VIIc)

A First Inductive Proof for the Irrational Number √(n((−x ∕ y) + 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) + x + y). This will be discussed in more detail in Part VIIIb. 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) + x + y) < y     (1)

where x,y ∈ ℕ

Case 1:

Substituting y ∕ 2 for n and multiplying thru by y affords:

x < √(y ∕ 2) (((−x + xy + y2) ∕ y) < y   (2)

Cancelling out y, affords,

x < √ ½ ((−x + xy + y2) < y      (3)

Replacing y by x + 1 affords

x < √ ½ ((−x + x(x + 1) + (x + 1)2) < x + 1  (4)

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

1 < √2.5 ≈ 1.581.. < 2 which is true.

Squaring the left, center and right parts of inequality (4), affords:

x2 < ½ ((−x + x(x + 1) + (x + 1)2 < (x + 1)2  (5)

Multiplying and expanding affords

x2 < ½((−x + x2 + x + x2 + 2x + 1) < x2 + 2x + 1 (6)

Adding like terms

x2 < ½(2x2 + 2x + 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(2x2 + 2x + 1) + 2x + 1< (x + 1)2 + 2x + 1  (8)

Expanding and adding

x2 + 2x + 1 < ½(2x2 + 6x + 3) < x2 + 4x + 2   (9)

Since

½(2x2 + 6x + 3) < ½(2x2 + 6x + 5) and

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

x2 + 2x + 1 <½(2x2 + 6x + 5) < x2 + 4x + 4   (10)

which is true for all x when y is even.

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

(x + 1) < √½(2x2 + 6x + 5) < (x + 2)    (11)

Working backwards by taking apart the equation and grouping terms

(x + 1) < √½[−(x + 1) + (x2 + 3x + 2) + (x2 + 4x + 4)] < (x + 2)  (12)

and factoring the groups

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

Multiplying the equation inside the square root by (x + 2) ∕ (x + 2)

(x + 1) < √½(x + 2)[−(x + 1) + (x + 1)(x + 2) + (x + 2)2] ∕ (x + 2) < (x + 2) (14)

Dividing the terms within the brackets by x + 2

(x + 1) < √½(x + 2)[−(x + 1) ∕ (x + 2) + (x + 1) + (x + 2)] < (x + 2  (15)

and substituting y + 1 for all x + 2 back into the inequality

(x + 1) < √½(y + 1)[−(x + 1) ∕ (y + 1) + (x + 1) + (y + 1)] < (y + 1)  (16)

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

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

Go to Part I. Go back to homepage.

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