Irrational Numbers from Adjacent Natural Numbers (Part IXa)

Proofs for the General Irrational Numbers √(n((xk ∕ yk) + 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((xk ∕ yk) + x + y). The exponent k in this general equation, which can take on values greater than zero, has already been used in Part VIIa and Part VIIb where k = 1. This will be discussed in more detail 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((xk ∕ yk) + x + y) < y     (1)

where x,y ∈ ℕ

Case 1:

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

x < √(y ∕ 2) (xk + xyk + yk+1) ∕ yk  < y   (2)

Cancelling out y, affords,

x < √½ (xk + xyk + yk+1) ∕ yk−1  < y   (3)

Although it was easy to obtain an inductive proof for Part Va when k = 1, large values of k, however, make this path extremely difficult to attain. Instead we can show that by substituting x + 1 and y + 1 for all x and y values, respectively, produces the desired the inequality if we were able to proceed via that route.

x + 1 < √½ [(x + 1)k + (x + 1)(y + 1)k + (y + 1)k+1] ∕ (y + 1)k−1  < y + 1   (4)

Case 2:

x < √n((xk ∕ yk) + x + y) < y     (1)

Substituting (y − 1) ∕ 2 for n and multiplying thru by yk affords:

x < √(½(y − 1) (xk + xyk + yk+1) ∕ yk  < y   (5)

Although it was easy to obtain an inductive proof for Part VIa when k = 1, again large values of k, however, make this path difficult to attain. Instead we can show again that by substituting x + 1 and y + 1 for all x and y values, respectively, produces the desired the inequality if we were able to proceed via that route.

x + 1 < √½y [(x + 1)k + (x + 1)(y + 1)k + (y + 1)k+1] ∕ (y + 1)k  < y + 1   (6)

Part IXb will use the equation √n((xk ∕ yk) + x + y) to generate two staircase sequences of irrational numbers.

Go to Part I. Go back to homepage.

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