Irrational Numbers from Adjacent Natural Numbers (Part IVb)

An Inductive Proof and the Sequence of Irrational Numbers

This is a continuation of Part IVa which uses the equation [(x + y)y] ∕ 2.

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)x] ∕ 2 is an irrational number between 1.5 < d < 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

x < [(x + y)x] ∕ 2 < y   (1)

where x,y ∈ ℕ and [(x + y)x] ∕ 2 is an irrational number.

First substituting x + 1 for y, affords

x < ((2x + 1)x) ∕ 2 < x + 1   (2)

and distributing

x < ((2x2 + x) ∕ 2 < x + 1   (3)

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

1 < 3 ∕ 2 < 2 which is true.

Squaring the the parts of the inequality (3)

x2 < (2x2 + x) ∕ 2 < (x + 1)2  (4)


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


x2 + 2x + 1 < (2x2 + x) ∕ 2 + 2x + 1 < (x+1)2 + 2x + 1   (5)

Multiplying the central part 2x + 1 by 2 ∕ 2, expanding adding terms

x2 + 2x + 1 < (2x2 + 5x + 2) ∕ 2 < x2 + 4x + 2     (6)

since

(2x2 + 5x + 2) ∕ 2 < (2x2 + 5x + 3) ∕ 2 and

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

x2 + 2x + 1 < (2x2 + 5x + 2) ∕ 2 < (2x2 + 5x + 3) ∕ 2 < x2 + 4x + 2 < x2 + 4x + 4  (7)

which can be reduced to

x2 + 2x + 1 < (2x2 + 5x + 3) ∕ 2 < x2 + 4x + 4   (8)

Multiplying the inequality thru by 2

2x2 + 4x + 2 < 2x2 + 5x + 3 < 2x2 + 8x + 8    (9)

which is true for all x

Factoring of (8), affords

(x + 1)2 < ½(x + 1)(2x + 3) < (x + 2)2    (10)

Taking the square roots of all the terms and rearranging

½(2x + 3)(x + 1)     (11)

and then to

½[(x + 1) + (x + 2)](x + 1)  (12)

Substituting x = y − 1 for the (x + 2) term, finally affords

(x + 1) < ½[(x + 1) + (y + 1)](x + 1) < (y + 1)  (13)

and proves the inequality x < [(x + y)x] ∕ 2 < y for all x and for y = x + 1.

The sequence generated from [(x + y)x] ∕ 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)x] ∕ 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)(x)) ∕ 2 .

Table I Irrational (Sequence)
x x+1 SQRTd
121.224745
1.011323
232.236068
1.004302
343.240370
1.002271
454.242641
1.001403
565.244044
1.000954
676.244998
1.000690
787.245688
1.000523
898.246211
1.000410
9109.246621
1.000330
101110.246951
9910099.249685
1.000003
100101100.249688

In the next two sections, Part V and Part VI it will be shown that between two consecutive natural numbers there exists an irrational number n(xy) ∕ (x + y). Part V will be a proof by induction and Part VI a second staircase sequence.

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