Irrational Numbers from Adjacent Natural Numbers (Part IVa)

An Inductive Proof and the Sequence of Irrational Numbers

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 ( ) 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 + y)y] ∕ 2 is an irrational number between √3 < d < √7.5 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 then we can show by induction that

To prove the chain notation for all numbers we add the equation (2x + 1) to all three parts of the inequality as follows:

The sequence generated from √[(x + y)y] ∕ 2, is shown in Table I where the values of x ranging from 1 to 10 and those of x+1 = y from 2 to 11 are multiplied. The value for each √[(x + y)y] ∕ 2, to six decimal places, is irrational and each of these values indeed falls between two consecutive natural numbers. In addition, two larger values for x and x+1 are shown in the last row with a d value even closer to 1. The third column heading SQRT is equal to √((x+y)(y)) ∕ 2 which in turn is equal to √((2x+1)(x+1)) ∕ 2.

Part IVb is the second part where the equation √[(x + y)x] ∕ 2 has x instead of y as a variable.

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