Real numbers

In the previous chapter it was demonstrated that not every number is rational. It’s time to delve deeper into the nature of irrational.

UNDER CONSTRUCTION

Introduction

By this point, we know there exist numbers which cannot be expressed as a ratio of whole numbers. They are called irrational, meaning non-rational, although as we will discover shortly, dope would be quite accurate as well.

We have proved that 2^1/2 is irrational. Is it possible to construct a segment that would be 2^1/2 units long? It is not clear how to go about it yet so let’s try to establish whether distance can be irrational. To do that we will grab two segments of arbitrary length and figure out how many times one fits into the other. If we denote them with variables L and D, the task at hand can be expressed as: D/L=?

To begin with, it is reasonable to postulate that for every two segments either one is longer than the other or their length is exactly the same. With this in mind, we can always find such natural power n that:

2ⁿ L ≤ D < 2ⁿ⁺¹ L

In other words, the segment 2ⁿ⁺¹ L is longer than D, and 2ⁿ L is either shorter or equally long as D. The latter case D/L=2ⁿ is not irrational. We are not interested in this scenario so let’s focus on D longer than that:

2ⁿ L < D < 2ⁿ⁺¹ L

To simplify the notation, we will introduce a new unit 2ⁿ⁺¹ L denoted by the variable L’:

L’/2 < D < L’

D is somewhere between the lower: L’/2 and upper bound: L’.

To improve the accuracy of the estimation, let’s narrow down the gap between the bounds. In the situation presented in the picture we can stack L’/4 to the right of the segment L’/2: L’/2 + L’/4. The result is longer than D, however L’/2+L’/8 is shorter so:

L’/2 + L’/8 < D < L’/2 + L’/4 = L’/2 + 2 L’/8

Next, we can add to the mix another fraction of L’, namely L’/16:

L’/2 + L’/8 + L’/16 < D < L’/2 + L’/8 + 2 L’/16

The accuracy of our estimation can be expressed as:

0 < D – (L’/2 + L’/8 + L’/16) < L’/16

That is much better than our staring point:

0 < D – L’/2 < L’/2

In principle it is possible to imagine such D that we could continue this game indefinitely. The result obtained after “m” iterations of the accuracy improvement can be expressed as:

$0 < D - \sum_{i=1}^{m}d_{-i} \cdot \frac{L'}{2^i} < \frac{L'}{2^m}$

where d_-i is a digit equal to 0 or 1. In the situation described above we have:
d_-1 = 1, d_-2 = 0, d_-3 = 1, d_-4 = 1. The indices of d_-i were chosen to be negative to demonstrate that what we get is consistent with fractions expressed in the positional system described in the previous chapter.

Instead of dividing L’ by powers of two, we can use powers of any other number, for example ten (decimal system). The procedure for this case looks like this:

x on the horizontal line in the picture represents D so:
7L’/10 < D < 8L’/10

In the lower part of the picture, we zoomed in on the segment between 7L’/10 and 8L’/10. It is clear that:
7L’/10 + 2L’/100 < D < 7L’/10 + 3L’/100

Continuation of this procedure leads to the following approximation:

$0 < D - \sum_{i=1}^{m}d_{-i} \cdot \frac{L'}{10^i} < \frac{L'}{10^m}$

This time 0≤d_-i≤9. As before, we can choose D in such a way that the series never ends. Rational numbers can be represented by an infinite series. Can series be infinite and not represent a rational number? Let’s take a closer look.

Cycling fractional series

Let’s take a look at a few examples of rational numbers expressed as infinite series of digits in decimal system:

1/3 = 0.333333 …

41/333 = 0.123123…

In the first case we have 3 on every position and in the second, the sequence 123 is repeated over and over again. Does every sequence with repetitions like that represents a rational number?

Let’s grab some arbitrary number with cycling digits:

0.2538461538461538461538461… =

$= 0 + \frac{2}{10} + \frac{1}{10} \cdot \frac{538461}{10^6} \cdot (1 + 10^{-6} + 10^{-6 \cdot 2} + 10^{-6 \cdot 3} + \dots)$

Using the above as a template, we can convert any cycling series into:

$w + r + \Gamma^{-s} \cdot \frac{a}{\Gamma^k} \cdot (1 + q + q^{2} + q^{3} + \dots)$

“w” stands for the whole part, ratio “r” for the fractional part that doesn’t cycle, “s” tells us at what position after the dot the series starts to cycle and the rest is a cycling part where q=1/Γ^k and k is a number of cycling digits. For the example above, we have: w=0, r=2/10, Γ=10, s=1, a=538461, k=6 and q=1/10⁶.

Is it possible to merge an infinite series:

$1 + q + q^{2} + q^{3} + \dots$

into a finite number? Prepare for a hat-trick.

Let’s multiply a series with n-1 members by (1-q):

$(1 - q) \cdot (1 + q + q^{2} + q^{3} + \dots + q^{n-2} + q^{n-1}) =$

if 0<q<1 as it always is in our case, we have a sequence of positive numbers to add:

$= (1-q) + (q - q^2) + (q^2-q^3) + \ldots +(q^{n-2} - q^{n-1}) + (q^{n-1} - q^{n}) =$

$= (1-q^2) + (q^2-q^3) + \ldots = (1 - q^3) + \ldots = 1 - q^n$

After dividing both sides by 1-q we get:

$1 + q + q^{2} + q^{3} + \dots + q^{n-2} + q^{n-1} = \frac{1 - q^n}{1-q}$

qⁿ is getting smaller as smaller when we go with n to infinity. With big enough n, we can make qⁿ as close to 0 as we wish. This fact is expressed in maths as:

$\lim_{n \rightarrow \infty} q^n = 0$

From this it follows that:

$\lim_{n \rightarrow \infty} \frac{1 - q^n}{1-q} = \frac{1}{1-q}$

$w + r + \Gamma^{-s} \cdot a \cdot \lim_{n \rightarrow \infty} \sum_{i=0}^{n} q^i = w + r + \Gamma^{-s} \cdot \frac{a}{1-q}$

Let’s apply this to our example: a=538461, Γ=10, k=6 and q=10^-6:

$\frac{538461}{10^6} \cdot \frac{1}{1-10^{-6}}=\frac{538461}{10^6-1}=\frac{538461}{999999}$

Using technique for calculating the greatest common divisor, we can transform it into:

$\frac{538461}{999999} =\frac{7 \cdot 76923}{13 \cdot 76923} = \frac{7}{13}$

so eventually:

0.2538461538461538461538461… =

$= \frac{2}{10} + \frac{1}{10} \cdot \frac{7}{13}$

Cycling is usually denoted with an overline so we can also write:

$0.2\overline{538461}$

The conclusion from all of that is: every infinite cycling series of digits represents a rational number.

Positional series for rational numbers

In the chapter about rational numbers, it was demonstrated how to find a positional series representing such numbers. Unfortunately if number is not whole and cycle is very long, that approach is not very useful for revealing whether the series is infinite or not. We need something better.

Let’s refresh our memory first. We are after expressing a ratio of natural numbers a/b, a<b with a series:

$a/b = \sum_{i=m}^{n} d^i \cdot \Gamma^{i}$

where d_i are digits and Γ is the base of the system. For infinite series, n is equal to minus infinity.

If the series is finite, we can rewrite it as:

$a/b = \frac{r}{\Gamma^k}$

where r and k represent natural numbers such that the ratio on the right is equal to the one on the left.

Let’s take a look at a few examples in decimal system, that is with Γ equal to ten and digits running from zero to nine.

Example 1

Decimal 1/2. The base is 10 (ten) and it is divisible by two so we can write:

$\frac{1}{2}=\frac{1}{2} \cdot 1 = \frac{1}{2} \cdot \frac{10}{10} = \frac{10}{2} \cdot \frac{1}{10} = 5 \cdot \frac{1}{10} = 0.5$

That is: a digit (5) multiplied by power of 10 (10^-1=1/10).

Example 2

Decimal 3/4. 10 is not divisible by 4 but both (10 and 4) are divisible by 2 so:

$\frac{3}{4}=\frac{1}{2} \cdot \frac{3}{2} = \frac{5}{10} \cdot (\frac{3}{2} \cdot \frac{10}{10}) =\frac{5}{10} \cdot (\frac{10}{2} \cdot \frac{3}{10}) = \frac{5}{10} \cdot \frac{15}{10}=\frac{75}{100}=$

$=\frac{70+5}{100}=\frac{70}{100}+\frac{5}{100}=\frac{7}{10}+\frac{5}{100}=0.75$

Example 3

Decimal 33/130. 130 is divisible by ten so we have:

$\frac{33}{130}=\frac{1}{10} \cdot \frac{33}{13}$

33 is divisible by 13 however not without a remainder:

$\frac{33}{130} =\frac{1}{10} \cdot \frac{2 \cdot 13 + 7}{13}=\frac{1}{10} \cdot (2+\frac{7}{13})$

Using the technique presented in one of the previous chapters, it can be demonstrated that the greatest common divisor of 13 and 10 is 1. In other words, the trick we kept pulling is not going to work this time round. What can we do?

Let’s try to find some clues in the example described in the previous section:

$0.\overline{538461}=\frac{538461}{10^6-1}$

In general, if a fraction cycles, it is always possible to express it in the following, compact form:

$\frac{A}{\Gamma^k-1}$

Note that we can reverse this reasoning. If we could prove that for any number “b” there exist such “k” that Γ^k-1 is divisible by “b” then every ratio a/b must cycle:

$\frac{a}{b}=\frac{A}{\Gamma^k-1}$

Let’s try to prove it then.

Fermat’s little theorem

We want to demonstrate that for natural Γ,b>1 there exists such k that Γ^k-1 is divisible by “b”. In other words “c” in the following equation is natural:

(Γ^k-1)/b = c

To simplify the notation, we can rewrite this into:

Γ^k = c∙b+1

or using modulo shorthand:

Γ^k mod b=1

meaning: the remainder of the division Γ^k/b is one.

How can we find “k”? We need to adopt some prospective strategy. Imagine we could find “k” numbers: 0 ≤ x₁, x₂, x₃, … ,x_k-1, x_k < b such that:

x₁∙Γ mod b = x_j(1)
x₂∙Γ mod b = x_j(2)
…
x_k-1∙Γ mod b = x_j(k-1)
x_k∙Γ mod b = x_j(k)

where numbers on the right: x_j(1), x_j(2), x_j(3), … ,x_j(k-1), x_j(k) are the same x-es as those on the left just in a different order e.g. for k=3 we could for example have: x_j(1)=x₃, x_j(2)=x₁, x_j(3)=x₂.

From the properties of multiplication it follows that if we have:
x=y
z=w

then:
x∙z=y∙w.

If, by analogy, for:
x mod b=y mod b
z mod b=w mod b

we could prove that:
x∙z mod b=y∙w mod b

then we could combine the equations:

x₁∙Γ mod b = x_j(1) = x_j(1) mod b
x₂∙Γ mod b = x_j(2) = x_j(2) mod b
…
x_k-1∙Γ mod b = x_j(k-1) = x_j(k-1) mod b
x_k∙Γ mod b = x_j(k) = x_j(k) mod b

into:

$(\prod_{i=1}^{k} x_i) \cdot \Gamma^k \mod b = \prod_{i=1}^{k} x_{j(i)} \mod b$

where we adopted Π as a shorthand for multiplication:

$(\prod_{i=1}^{k} x_i) = x_1 \cdot x_2 \cdot \ldots \cdot x_{k-1} \cdot x_k$

Note that:

$\prod_{i=1}^{k} x_i = \prod_{i=1}^{k} x_{j(i)}$

because on the right the same numbers are multiplied, just in a different order. From this it follows that:

$(\prod_{i=1}^{k} x_i) \cdot \Gamma^k \mod b = \prod_{i=1}^{k} x_{i} \mod b$

Thanks to the properties of multiplication, if we have:
x∙y=z∙y

then we can divide both sides by “y” as long as “y” is not 0:
x=z

If we could do the same for:

$(\prod_{i=1}^{k} x_i) \cdot \Gamma^k \mod b = \prod_{i=1}^{k} x_{i} \mod b$

then it would be equivalent to:

$\Gamma^k \mod b = 1 \mod b = 1$

and that’s exactly what we want to prove! Let’s try to turn our wish-list into reality.

We will start with finding: x₁, x₂, x₃, … ,x_k-1, x_k for the concrete case of Γ=7 and b=4. Let’s use consecutive numbers and see what happens:

i	i∙Γ mod 4
0	0
1	3
2	2
3	1
4	0
5	3
6	2
7	1

How convenient! Consecutive numbers do work:
0∙7 mod 4 = 0
1∙7 mod 4 = 3
2∙7 mod 4 = 2
3∙7 mod 4 = 1

Same numbers on both sides, just in a different order. Now the question is whether it would work for any Γ,b>1. We have:

0∙Γ mod b = 0
1∙Γ mod b = x₁
…
(b-1)∙Γ mod b = x_b-1

and know that: x₁, … , x_b-1<b. What’s left to show is that x₁, … , x_b-1 are all different from each other and greater than 0. In other words, we need to prove that if i≠j then i∙Γ mod b ≠ j∙Γ mod b.

To start with, we will find out what happens if the remainder of x/b is the same as for y/b where x and y are some arbitrary numbers:
x mod b = y mod b

We can write:
x = x_w∙b+x_r
y = y_w∙b+y_r

and the remainders x_r, y_r are equal: x_r=x mod b=y mod b=y_r thus:
x-y=(x_w-y_w)∙b

so x-y must be divisible by b:
(x-y) mod b = 0

If instead we assume:
(x-y) mod b = 0

then we have:
(x-y)/b=(x_w-y_w)∙b+(x_r-y_r)/b

and remainders x_r, y_r have the properties:
0≤x_r, y_r<b

so we must have:
-b < x_r-y_r < b

As a consequence, (x_r-y_r)/b is whole only when x_r=y_r i.e. x mod b = y mod b.

We have just established the following equivalence:

Eqv.1: $(x-y) \mod b = 0 \Leftrightarrow x \mod b = y \mod b$

If the above is true then this one must be true as well:

$(x-y) \mod b \ne 0 \Leftrightarrow x \mod b \ne y \mod b$

Let’s apply this to the problem at hand, namely, x=i∙Γ and y= j∙Γ:

$(i-j) \cdot \Gamma \mod b \ne 0 \Leftrightarrow i \cdot \Gamma \mod b \ne j \cdot \Gamma \mod b$

The right hand side is true only when (i-j)∙Γ is not divisible by b, that is:
(i-j)∙Γ≠ A∙b

for every whole number A. Let’s transform the above into:

(i-j)/A≠b/Γ

to make the reasoning easier. If we assume that the greatest common divisor of b and Γ is one: GCD(b,Γ)=1 then b/Γ cannot be simplified into a ratio with dividend (the upper number in the ratio) smaller than b and we know for sure that:
-b<i-j<b. In other words, if GCD(b,Γ)=1 then we cannot find such A that:
(i-j)/A=b/Γ so (i-j)∙Γ must not be divisible by b for i≠j and as a consequence:

i∙Γ mod b ≠ j∙Γ mod b

and that’s exactly what we wanted to show!

In the next step, as outlined in the beginning of this section, we would like to combine these:

1∙Γ mod b = x₁ mod b
…
(b-1)∙Γ mod b = x_b-1 mod b

into a single equation:

1∙2∙ … ∙(b-1)∙Γ^b-1 mod b = x₁∙x₂∙ … ∙x_b-1 mod 1

Let’s find out whether we can.

If:
x mod b=y mod b
z mod b=w mod b

then we can write:
x = x_w ∙ b + r₁
y = y_w ∙ b + r₁
z = z_w ∙ b + r₂
w = w_w ∙ b + r₂

because x, y have equal remainders and the same can be said about z, w.

From this it follows that:
x∙z = (x_w∙z_w∙b+x_w∙r₂+z_w∙r₁)∙b + r₁∙r₂
y∙w = (y_w∙w_w∙b+y_w∙r₂+w_w∙r₁)∙b + r₁∙r₂

so:
x∙z mod b = y∙w mod b

In other words, we have the implication:

Impl.1: $(x \mod b=y \mod b) \wedge (z \mod b=w \mod b) \Rightarrow$
$\Rightarrow x \cdot z \mod b = y \cdot w \mod b$

thus the equation:

1∙2∙ … ∙(b-1)∙Γ^b-1 mod b = x₁∙x₂∙ … ∙x_b-1 mod 1

must hold.

The multiplication of consecutive numbers is denoted with the exclamation mark ! called factorial:
(b-1)! := 1∙2∙ … ∙(b-1)

We also know that x₁∙x₂∙ … ∙x_b-1 are the same numbers as: 1,…,b-1, just multiplied in a different order so our equation can be simplified into:

(b-1)!∙Γ^b-1 mod b = (b-1)! mod 1

The last challenge is to get rid of (b-1)!

Let’s take a look at a general case:

x∙z mod b = y∙z mod b

Can we simply cross out z? As we know the above is equivalent to:

(x-y)∙z mod b = 0

so there exists such whole number A that:

(x-y)∙z = A∙b

This in turn can be transformed into:

(x-y)=(A/z)∙b

and A/z is not necessarily whole, so we cannot get rid of z, unless A/z is whole. Let’s try to find out what the relationship between z and b should be so we could cancel out z. After moving z and b to one side of the equation above, we get:

z/b = A/(x-y)

After finding GCD(z,b) this ratio can be rewritten into:

z’/b’=A/(x-y)

where: z’=z/GCD(z,b) and b’=b/GCD(z,b). Now we see that x-y must be divisible by b’:

(x-y) mod (b/GCD(z,b)) = 0

thus we proved the implication:

Impl.2: $(x \cdot z) \mod b = (y \cdot z) \mod b \Rightarrow$

$\Rightarrow x \mod (b/GCD(z,b)) = y \mod (b/GCD(z,b))$

Conclusion: we can cross out z from “(x∙z) mod b = (y∙z) mod b” only when GCD(z,b)=1.

Let’s apply this knowledge to:

(b-1)!∙Γ^b-1 mod b = (b-1)! mod 1

If “b” is not divisible by: 2,3, etc. all the way up to b-1, that is: “b” is prime, then GCD(0 < any number <b, b)=1 and we can cancel out (b-1)! on both sides:

Γ^b-1 mod b = 1

This fact is named in mathematics: the Fermat’s little theorem.

In the previous section it was demonstrated that if a divider in a ratio and the base of the positional system have the greatest common divisor equal to 1, then we can’t really tell whether the ratio will cycle or not.

If divider is a prime number, thanks to the Fermat’s little theorem, we know that the ratio must cycle, irrespective of the base of the system we use.

Let’s take a look at an example: 7/13 in decimal system (base Γ=10, meaning ten). b=13 so 10¹²-1=999999999999 must be divisible by 13 and indeed:

76923076923∙13=999999999999

$\frac{7}{13} = \frac{7\cdot 76923076923}{13\cdot76923076923}=\frac{538461538461}{10^{12}-1}=\frac{538461538461}{10^{12}}\cdot \frac{1}{1-10^{-12}}=$

$=0.\overline{538461538461}=0.\overline{538461}$

What if divider is not a prime number e.g. 7/9? We have to tackle this situation too.

Euler’s theorem

In this section we will prove that if GCD(Γ,b)=1 then the ratio a/b expressed in the positional series with the base equal to Γ, must cycle.

We will use the same strategy as in the previous section, that is: we need to find a collection of numbers: 0<x₁< … < x_k<b such that:

x₁∙Γ mod b = x_j(1)
x₂∙Γ mod b = x_j(2)
…
x_k-1∙Γ mod b = x_j(k-1)
x_k∙Γ mod b = x_j(k)

where x_j(1), x_j(2), x_j(3), … ,x_j(k-1), x_j(k) are the same numbers as 0<x₁< … < x_k<b just in a different order. After that, thanks to the Eqv.1 we can multiply these equations by each other on both sides of “=” to obtain:

$(\prod_{i=1}^{k} x_i) \cdot \Gamma^k \mod b = \prod_{i=1}^{k} x_{i} \mod b$

If for every x_i GCD(x_i,b)=1, then thanks to the Impl.2 , we can cancel them all out:

Γ^k mod b = 1 mod b = 1

This in turn can be used to show cycling of ratios a/b as will be demonstrated later.

For a good start, let’s find: 0<x₁< … < x_k<b for a concrete case of Γ=4, b=9:

i	GCD(i,9)	i∙4	(i∙4) mod 9	GCD((i∙4) mod 9, 9)
1	1	4	4	1
2	1	8	8	1
3	3	12	3	3
4	1	16	7	1
5	1	20	2	1
6	3	24	6	3
7	1	28	1	1
8	1	32	5	1

In this particular example the numbers we searched for are: 1, 2, 4, 5, 7, 8 (the first column in the table). Note that for every “i” the following is also true:
GCD(i, 9)=GCD(i∙4 mod 9, 9) so after eliminating rows 3 and 6, columns 1 and 4 must contain the same numbers: 1, 2, 4, 5, 7, 8, though the numbers in the 4th column are not in the same order: 4, 8, 7, 2, 1, 5.

If in a general case of arbitrary Γ, b and 0<i<b we could prove this implication:

$GCD(i, b)=1 \Rightarrow GCD(i \cdot \Gamma \mod b, b)=1$

then our task of finding: 0<x₁< … < x_k<b would be easy. We would simply choose only these 0<i<b that have the property GCD(i, b)=1.

In the section about finding GCD we have proved that:

GCD(i, b) = GCD(i mod b, b)

so all we have to prove is the implication:

$GCD(i, b)=1 \Rightarrow GCD(i \cdot \Gamma, b)=1$

We will prove it using reductio ad absurdum method. We assume that:
GCD(Γ, b)=1 so thanks to the following equivalence:

$\sim\big(GCD(i, b)=1 \wedge GCD(\Gamma, b)=1 \Rightarrow GCD(i \cdot \Gamma, b)=1 \big) \Leftrightarrow$

$\Leftrightarrow \big(GCD(i, b)=1 \wedge GCD(\Gamma, b)=1 \wedge GCD(i \cdot \Gamma, b)>1\big)$

all we have to show is that the three conditions in the line just above cannot be all true at the same time.

If GCD(i∙Γ,b)=w>1 then both i∙Γ and b are divisible by w. Let’s denote the results of divisions by (i∙Γ)’ and b’ respectively:

i∙Γ = (i∙Γ)’∙w
b = b’∙w

From these it follows that:

$\frac{i\cdot\Gamma}{b}=\frac{(i\cdot\Gamma)'}{b'}$

which can be transformed into:

$\frac{\Gamma}{b}=\frac{(i\cdot\Gamma)'}{b'\cdot i}$

We have GDC(Γ,b)=1 so b’∙i must be divisible by b. Let’s denote the result of this division by k:

b’∙i = k∙b

The above can be transformed into:

i = k∙(b/b’)

and we know that b/b’=w so:

b = b’∙w
i = k∙w

Both b and i are divisible by w>1 thus GCD(i,b)>1, meaning:

$GCD(i, b)=1 \wedge GCD(\Gamma, b)=1 \wedge GCD(i \cdot \Gamma, b)>1$

must be always false thus:

Impl.3: $GCD(i, b)=1 \wedge GCD(\Gamma, b)=1 \Rightarrow GCD(i \cdot \Gamma, b)=1$

must be always true.

All of that leads to the conclusion the if GCD(Γ,b)=1 then:

Γ^φ(b) mod b = 1

where φ(b) is a quantity of natural numbers x for which: 0<x<b and GCD(x, b)=1. This fact is called in mathematics the Euler’s theorem.

Let’s use it to show that a ratio: 5/9 expressed in the decimal system (base Γ=10, meaning ten) must cycle.

i	GCD(i,9)
1	1
2	1
3	3
4	1
5	1
6	3
7	1
8	1

There are 6 numbers such that GCD(i,9)=1 in the table above so according to the Euler’s theorem 10⁶-1=999999 is divisible by 9:

111111∙9=999999

$\frac{5}{9} = \frac{5\cdot 111111}{9\cdot111111}=\frac{555555}{10^{6}-1}=\frac{555555}{10^{6}}\cdot \frac{1}{1-10^{-6}}=$

$=0.\overline{555555}=0.\overline{5}$ .

That was rather trivial but that’s not always the case e.g. if Γ is a prime number.

Positional series for rational vs irrational numbers

Thanks to the Euler’s theorems presented above we know that every ratio a/b expressed in the positional system must start to cycle at some point in the series irrespective of what is the base of the system. For example in the decimal system we have:

$\frac{5}{9}=0.\overline{5}$
$\frac{8}{33}=0.\overline{24}$
$\frac{33}{130}=0.2\overline{538461}$

If a infinite series:

$\sum_{i=m}^{-\infty}d_{i} \cdot \Gamma^{i}$

doesn’t cycle, it represents an irrational number.

We can also say that every cycling infinite series converges to a rational number. In the case of non-cycling series, every next digit gets us closer to some irrational number, yet it is not possible to get there exactly. All we can have is a recipe to generate the next digit in the series that never arrives.

Definition of irrational numbers

Natural numbers are so simple that on practical grounds they don’t need a definition. Rational numbers are defined as a/b where a and b are natural, and negative numbers are defined as -r:=0-r where r is rational and positive.

Can we somehow define irrational numbers without relying on a positional system? After all, the positional system is just a convenient representation of numbers and numbers are independent of representations. The length of a segment simply exists, no matter how we choose to express it.

Let’s try to extract some general principle from the series which is guaranteed to converge to some irrational number D:

$0 < D - \sum_{i=m}^{m-n}d_{i} \cdot \Gamma^i < \Gamma^{m-n}$

“m” is fixed and equal to some natural number, “n” is natural number equal or greater than 0, d_i are digits and Γ is the base of the positional system. We can rewrite the above as:

$l_n=\sum_{i=m}^{m-n}d_{i} \cdot \Gamma^i < D < \sum_{i=m}^{m-n}d_{i} \cdot \Gamma^i+\Gamma^{m-n}=u_n$

On the left we have the lower (l_n) and on the right the upper rational approximation (u_n) of D.

Example 1

Let’s take a look at an example where D is a rational number: 0.20539461

n	l_n	u_n
0	0	1
1	0.2	0.3
2	0.20	0.21
3	0.205	0.206
4	0.2053	0.2054
5	0.20539	0.20540
6	0.205394	0.205395
7	0.2053946	0.2053947
8	0.20539461	0.20539462

The properties of l_n and u_n boil down to just two inequalities:

l_n ≤ l_n+1 < u_n+1 ≤ u_n

u_n+1 – l_n+1 < u_n – l_n

The greater “n” we take, the better the approximation i.e. the smaller the difference between the upper and lower bound.

Let’s flip the reasoning and just grab some arbitrary series of rational pairs: {p_n,q_n} having the same properties:

p_n ≤ p_n+1 < q_n+1 ≤ q_n

q_n+1 – p_n+1 < q_n – p_n

We will denote the set containing every series {p_n,q_n} having these properties with the symbol 𐍂. {p’_n,q’_n}∈𐍂 is a short meaning that the series of rational {p’_n,q’_n} is a member of the set 𐍂 i.e. has the properties we are interested in.

Now the question is: does every {p_n,q_n}∈𐍂 converge to some rational or irrational number? A rational number can be found between every pair of rational p_n, q_n e.g. (p_n+q_n)/2 so it is possible that {p_n,q_n}∈𐍂 converges to a rational number. How about irrational? Can an irrational number be inserted between every pair of rational p_n, q_n?

Example 2

Consider a pair of rational numbers:
0.12345678123 <
< 0.12345679123

Can we find any irrational number residing between them? Let’s start with a rational:

0.12345678123 <
< 0.123456782 <
< 0.12345679123

If at the end of 0.123456782 we add any non cycling series of digits, we have an irrational number we are searching for.

In the same fashion we can insert an irrational number between any other two rational numbers, so every {p_n,q_n}∈𐍂 must converge to some number. The next question is: can we tell the difference between {p_n,q_n}∈𐍂 and {p’_n,q’_n}∈𐍂 converging to different numbers? For the sake of argument, let’s assume that the first series converges to {p_n,q_n}→r and the second to {p’_n,q’_n}→r’, both irrational and r<r’.

Example 3

Imagine two irrational numbers:
r=1.41421356837… <
< 1.41421356937…=r’

Can you find a rational one that is between them?
1.4142135684 is greater than the first one, smaller than the second, and is rational.

Between every two irrational numbers we can insert some rational number b:

r<b<r’

By the same token we can insert a rational number between rational and irrational number. We also know that with rising “n”, q_n-p_n and q’_n-p’_n are getting smaller and smaller so if {p_n,q_n}→r and {p’_n,q’_n}→r’, then there must exist some number n=N that:

p_N<r<q_N<b<p’_N<r<q’_N

Indeed, if we can insert a rational number between irrational r and r’, then r must be different than r’, and if {p_n,q_n}→r and {p’_n,q’_n}→r’, we can tell the difference between r and r’ using an infinite rational series {p_n,q_n} and {p’_n,q’_n}. In other words, we have found a way to define irrational numbers without relying on positional systems.