Irrational Numbers from Adjacent Natural Numbers (Part Vb)
An Inductive Proof for the Irrational Number √2xy2 ∕ (x + y)
This is a continuation of Part Va a similar type of equation.
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 the initial common difference (d) between two irrational numbers generated from √2xy2 ∕ (x + y) is an irrational number between 1 < d < 2 which approaches d=1 as the number of natural numbers grows without bound and n is equal to 2x. 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 < √2xy2 ∕ (x + y) < y (1)
where √2xy2 ∕ (x + y) is an irrational number.
First substituting x + 1 for y, affords
x < √2x (x + 1)2 ∕ (x + x + 1) < x + 1 (2)
and addition of terms affords
x < √2x(x + 1)2 ∕ (2x + 1) < x + 1 (3)
then substituting x = 1 in the basis step, affords:
1 < √8 ∕ 3 ≈ 1.632.. < 2 which is true.
Squaring the left, center and right parts of the inequality (3), affords
x2 < 2x(x + 1)2 ∕ (2x + 1) < (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 + 2
x + 1 < 2
x(
x + 1)
2 ∕ (2
x + 1) + 2
x + 1 < (
x+1)
2 + 2
x + 1 (5)
Multiplying the second 2
x + 1 term by ( 2
x + 1) ∕ ( 2
x + 1) and expanding
x2 + 2
x + 1 < (2
x3 + 4
x2 + 2
x) ∕ (2
x + 1) + (2
x + 1)
2 ∕ (2
x + 1) <
x2 + 4
x + 2 (6)
and expanding
x2 + 2
x + 1 < [(2
x3 + 8
x2 + 2
x) + (4
x2 + 4
x + 1)] ∕ (2
x + 1) <
x2 + 4
x + 2 (7)
Adding up all the terms
x2 + 2
x + 1 < (2
x3 + 8
x2 + 6
x + 1) ∕ (2
x + 1) <
x2 + 4
x + 2 (8)
since
x2 + 4
x + 2 <
x2 + 4
x + 4
so that
x2 + 2
x + 1 < (2
x3 + 8
x2 + 6
x + 1) ∕ (2
x + 1) <
x2 + 4
x + 2 <
x2 + 4
x + 4 (9)
which can be reduced to
x2 + 2
x + 1 < (2
x3 + 8
x2 + 6
x + 1) ∕ (2
x + 1) <
x2 + 4
x + 4 (10)
Multiplying thru by 2
x + 1, affords
(
x2 + 2
x + 1)(2
x + 1) < (2
x3 + 8
x2 + 6
x + 1) < (
x2 + 4
x + 4)(2
x + 1) (11)
and on expanding
2
x3 + 5
x2 + 4
x + 1 < 2
x3 + 8
x2 + 6
x + 1 < 2
x3 + 9
x2 + 12
x + 4 (12)
where it can be seen that the left part < the middle part < the right part of the inequality
In order to complete the proof we must be able to convert the denominator from 2x + 1 to 2x + 3
which is required in (22) and (23) of the proof.
Multiplying [(2
x3 + 8
x2 + 6
x + 1) ∕ (2
x + 1)] by (2
x + 3) ∕ (2
x + 3) affords
(4
x4 + 22
x3 + 36
x2 + 20
x + 3) ∕ (2
x + 1)(2
x + 3) (13)
Adding (6
x2 + 12
x + 5) ∕ (2
x + 1)(2
x + 3) from (14) affords
[4
x4 + 22
x3 + 42
x2 + 32
x + 3] ∕ (2
x + 1)(2
x + 3) (14)
Division of (14) by (2
x + 1) affords
(2
x3 + 10
x2 + 16
x + 8) ∕ (2
x + 3) (15)
Continuing with the proof determining which of the two is the smallest
(2
x3 + 8
x2 + 6
x + 1) ∕ (2
x + 1) < (2
x3 + 10
x2 + 16
x + 8) ∕ (2
x + 3) (16)
Multiplying thru by (2
x + 1)(2
x + 3)
(2
x3 + 8
x2 + 6
x + 1)(2
x + 3) < (2
x3 + 10
x2 + 16
x + 8)(2
x + 1) (17)
Expanding the terms of the inequality affords:
4
x4 + 22
x3 + 36
x2 + 20
x + 1 < 4
x4 + 22
x3 + 42
x2 + 32
x + 8 (18)
which is indeed true.
Since the right part of the inequality is greater than the left part, we can rewrite inequality (10) as
x2 + 2
x + 1 < (2
x3 + 10
x2 + 16
x + 8) ∕ (2
x + 3) <
x2 + 4
x + 4 (19)
Multiplying thru by 2
x + 3 we get;
(
x2 + 2
x + 1)(2
x + 3) < 2
x3 + 10
x2 + 16
x + 8 < (
x2 + 4
x + 4)(2
x + 3) (20)
which on multiplication and expansion affords
2
x3 + 7
x2 + 8
x + 3 < 2
x3 + 10
x2 + 16
x + 8 < 2
x3 + 11
x2 + 20
x + 12 (21)
which again is indeed true.
Since (21) is true, we can reduce this inequality after factoring all three parts of the inequality to
(
x + 1)
2 < (2(
x + 1)(
x + 2)
2 ∕ (2
x + 3) < (
x + 2)
2 (22)
Taking the square roots
x + 1 <
√2(x + 1)(x + 2)2 ∕ (2x + 3) <
x + 2 (23)
and so the original inequality (3)
x <
√2x(x + 1)2 ∕ (2x + 1) <
x + 1 is true for all
x
(23) can be expanded to
x + 1 <
√2(x + 1)(x + 2)2 ∕ (x + 1 + x + 2) <
x + 2 (24)
Substituting
x =
y − 1 for the three (
x + 2) terms
x + 1 <
√2(x + 1)(y + 1)2 ∕ (x + 1 + y + 1) <
y + 1 (25)
which proves the inequality:
√2xy2 ∕ (x + y) <
y for all
x, for
y =
x + 1.
Go to Part VIb shows the sequence derived from the equation √2xy2 ∕ (x + y).
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