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.

AWAITS A REVIEW

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 (not in the picture to avoid too much clutter):

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

The accuracy of this 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:

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:

This time 0≤d_-i≤9 are digits of decimal system. As before, we can choose D in such a way that the series never ends. Rational numbers can be represented by an infinite series too. 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 an 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 represent a rational number?

Let’s grab some arbitrary number with cycling digits:

0.2538461538461538461538461… =

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

“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:

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

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

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

After dividing both sides by 1-q we get:

qⁿ is getting smaller and 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 mathematics as:

From this it follows that:

so

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

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

so eventually:

0.2538461538461538461538461… =

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

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, this 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:

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:

where r and k represent natural numbers such that the ratio on the right is equal to the one on the left e.g. 26839/50000=53678/100000=53678/10⁵=0.53678

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 a divisor of our ratio, namely two, so we can write:

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 (in fact 2 is the greatest common divisor of 10 and 4) so:

Example 3

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

33 is divisible by 13 however not without a remainder:

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 in the examples above is not going to work for 7/13. What can we do?

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

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

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:

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:

where we adopted Π as a shorthand for multiplication:

Note that:

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

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:

then it would be equivalent to:

and that’s exactly what we want to prove! Let’s try to turn our wish 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∙7	i∙7=w∙4+r	i∙7 mod 4
0	0	0=0∙0+0	0
1	7	7=1∙4+3	3
2	14	14=3∙4+2	2
3	21	21=5∙4+1	1
4	28	28=7∙4+0	0
5	35	35=8∙4+3	3
6	42	42=10∙4+2	2
7	49	49=12∙4+1	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: 0,1,2,3, just in a different order. Now the question is whether this scheme 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 we also 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 do this, first, 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

(x_r-y_r)/b must be 0 according to the initial assumption and it is possible only when: x_r=y_r i.e. x mod b = y mod b.

We have just established the following equivalence:

Eqv.1:

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

We want to prove that if i≠j then i∙Γ mod b ≠ j∙Γ mod b. Let’s use the equivalence above:

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 where b and Γ are replaced with smaller numbers (see this section). We know for sure that: -b<i-j<b so 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 we must have:

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 (∧ is a logical conjunction operator):

Impl.1:


thus the equation above that we wished to be true:

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 in:
x∙z mod b = y∙z mod b

unless A/z is whole. Let’s try to find out what relationship between z and b we need to cancel out z. Let’s transform (x-y)=(A/z)∙b into:

z/b = A/(x-y)

to have only z and b on the left. 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

This and Eqv.1 leads to the implication:

Impl.2:

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 called 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 what the base is equal to.

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

so

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 Impl.1 we can multiply these equations by each other on both sides of “=” to obtain:

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. To make the reasoning convenient, we will do it in the decimal system.

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 the previous chapter we have proved that if GCD(Γ,b)=1 then column 4 must contain the same numbers as column 1, just in a different order. The problem we have now is that we are interested only in these “i” for which GCD(i,9)=1. This is not the case for 3 and 6. Thankfully, for i≠3 and i≠6:
(i∙4) mod 9

is neither equal to 3 nor to 6, so the numbers we searched for are: x₁=1, x₂=2, x₃=4, x₄=5, x₅=7, x₆=8.

Note that for every “i” the following is also true:
GCD(i, 9)=GCD(i∙4 mod 9, 9)

and this is the reason why for i such that GCD(i,9)=1 the formula:
(i∙4) mod 9

must produce the same sequence, just in a different order.



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

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(z, b) = GCD(z mod b, b)

so all we have to prove is the implication:

We will prove it using reductio ad absurdum method. We know that:
GCD(i, b)=1 and GCD(Γ, b)=1 so thanks to the following equivalence:
(negation operator: ~, conjunction operator: ∧)

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:

which can be transformed into:

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:

must be always false thus:

Impl.3:

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 such natural numbers x for which: 0<x<b and GCD(x, b)=1 e.g. φ(9)=6 in the example above. 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. GCD(9, 10)=1 and there are 6 numbers such that GCD(i,9)=1 so according to the Euler’s theorem 10⁶-1=999999 is divisible by 9:

111111∙9=999999

so

.

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 theorem 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:

Reminder: overline indicates that the digits below it do cycle.

If an infinite series:

doesn’t cycle, it represents an irrational number.

Every cycling infinite series converges to a rational number. In the case of non-cycling series, every next digit gets us closer to a number that cannot be expressed as a ratio thus called irrational.

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:

“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:

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 some irrational number 0<D<1:

n	ln	un	un-ln
0	0	1	1
1	0.2	0.3	0.1
2	0.20	0.21	0.01
3	0.205	0.206	0.001
4	0.2053	0.2054	0.0001
5	0.20539	0.20540	0.00001
6	0.205394	0.205395	0.000001
7	0.2053946	0.2053947	0.0000001
8	0.20539461	0.20539462	0.00000001

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 narrower the gap between the lower and upper bound.

Let’s flip the reasoning and just grab some arbitrary sequence of rational pairs: {p_n,q_n} constituting a contracting interval (CI), that is having the properties:

1) p_n ≤ p_n+1 < q_n+1 ≤ q_n

2) q_n+1 – p_n+1 < q_n – p_n

The sequence of pairs having these two properties will be denoted with the subscript CI: {p_n,q_n}_CI. Now the question is: does every {p_n,q_n}_CI 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}_CI 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 in-betweener:

   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 were searching for.

In the same fashion we can insert an irrational number between every two rational numbers, so every {p_n,q_n}_CI must converge to some rational or irrational number.

The next question is: can we tell the difference between {p_n,q_n}_CI and {p’_n,q’_n}_CI converging to different numbers? For the sake of argument, let’s assume that the first sequence converges to {p_n,q_n}_CI→r and the second to {p’_n,q’_n}_CI→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 it is rational.

From example 3 it follows that between every two irrational numbers we can insert some rational 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 there must exist some natural n=N that:

p_N<r<q_N<b<p’_N<r<q’_N

If on the other hand for every n there exist such m that:

p_n<p’_m
q’_m<q_n

then {p’_n,q’_n}_CI must converge to the same number as {p_n,q_n}_CI. This reasoning can be repeated for every combination of rational and irrational r and r’. In other words, we have found a way to define irrational numbers without relying on positional systems.

Initially, the fact that {p_n,q_n}_CI doesn’t exclusively converge to irrational numbers may sound like a problem but thanks to this feature we will be able to come up with a uniform procedure for performing mathematical operations on every type of numbers introduced thus far. Collectively rational and irrational numbers are called the real numbers, and the set containing them all is denoted with the symbol ℝ.

Addition of positive real numbers

The starting point of our study was a segment that had a length equal to a unit U multiply by some irrational number. Imagine we have two segments, both with lengths given in terms of real numbers: r and r’. After we stack one on top of the other, what will be the length of the resulting segment?

Let’s grab any two sequences having the contracting interval properties (CI), and converging to r and r’ respectively:
{p_n,q_n}_CI → r
{p’_n,q’_n}_CI → r’

It is obvious that the length of r’ stacked up on top of r must be somewhere between: p_n+p’_n and q_n+q’_n for every natural n:

p_n+p’_n < r+r’ < q_n+q’_n

Also, because both: {p_n,q_n}_CI and {p’_n,q’_n}_CI are CI sequences then we must have:

p_n ≤ p_n+1 < q_n+1 ≤ q_n

p’_n ≤ p’_n+1 < q’_n+1 ≤ q’_n

If p_n+1 is greater or equal to p_n, and p’_n+1 is greater or equal to p’_n, then p_n+1+p’_n+1 must be greater or equal to p_n+p’_n. By the same token we can combine all the inequalities above into:

p_n + p’_n≤ p_n+1+ p’_n+1< q_n+1+ q’_n+1≤ q_n+ q’_n

which means that {p_n+p’_n,q_n+q’_n} has the first of the CI properties. The second CI property for {p_n,q_n}_CI and {p’_n,q’_n}_CI are:

q_n+1 – p_n+1 < q_n – p_n

q’_n+1 – p’_n+1 < q’_n – p’_n

and these can be combined into:

(q_n+1 + q’_n+1) – (p_n+1 + p’_n+1) < (q_n + q’_n) – (p_n + p’_n)

so {p_n+p’_n,q_n+q’_n} is an CI sequence.

Thanks to all of this, we can define addition of r and r’ as:

{p_n,q_n}_CI+ {p’_n,q’_n}_CI= {p_n+p’_n,q_n+q’_n}_CI

where {p_n+p’_n,q_n+q’_n}_CI converges to r+r’.

Negative real number

On a straight line we can move only in two directions. One of them can be designated as positive and the other as negative. Let’s move r units in the positive direction where r is some irrational number. If we want to move the same distance in the negative direction, we simply flip the direction of movement:

If we grab a sequence {p_n,q_n}_CI convergent to r, its flipped version would be:
{-q_n,-p_n}. Does it converge to anything? If it has CI properties it must. Let’s find out whether it is the case.

{p_n,q_n}_CI is an CI sequence so we must have:

p_n ≤ p_n+1 < q_n+1 ≤ q_n

After multiplying it times -1, we get (see the properties of inequalities here):

-p_n ≥ -p_n+1 > -q_n+1 ≥ -q_n

so {-q_n,-p_n} has the first CI property.

The second CI property of {p_n,q_n}_CI reads:

q_n+1 – p_n+1 < q_n – p_n

which can be rearranged into:

-p_n+1 – (-q_n+1) < -p_n – (-q_n)

so {-q_n,-p_n} has the second and final CI property, meaning: it must converge to something. What is it then? {-q_n,-p_n}_CI is a flipped version of {p_n,q_n}_CI so it must converge to -r. Let’s see whether it follows from our definition of addition.

We will add two sequences: {p_n,q_n}_CI → r and {p’_n,q’_n}_CI → -r using the procedure introduced in the previous section:

{p_n,q_n}_CI+ {p’_n,q’_n}_CI= {p_n+p’_n,q_n+q’_n}_CI

It is fairly easy to show that the sequence on the right has CI properties despite the fact that this time round p’_n,q’_n<0. It is a CI sequence , so it must converge to something. Let’s take a look at p_n+p’_n. p_n is smaller than r and positive and p’_n is smaller than negative -r.

After adding them together, the result must be smaller than 0 for every n. A similar reasoning leads to the conclusion that q_n+q’_n must be greater than 0 for every n, so altogether {p_n+p’_n,q_n+q’_n}_CI→ 0. In other words, we have:

r+(-r)=0

Addition of real numbers and its properties

How shall we define an addition of some arbitrary real numbers r₁ and r₂ ? First, we need to find any two sequences {a_n,b_n}_CI and {c_n,d_n}_CI that are convergent to r₁ and r₂ respectively. Their sum will be given as:

{a_n,b_n}_CI + {c_n,d_n}_CI = {a_n+c_n,b_n+d_n}_CI

The result is an CI sequence, so it must be convergent to some real number. In agreement we everything said so far, we are going to define r₁ + r₂ as a number to which {a_n+c_n,b_n+d_n}_CI converges. Note that due to the properties of numbers and definition of CI sequence, if we take any other two sequences {e_n,f_n}_CI→ r₁ and {g_n,h_n}_CI→ r₂ then:

{e_n,f_n}_CI+{g_n,h_n}_CI={e_n+g_n,f_n+h_n}_CI→ r₁+r₂

It is easy to check that addition of real numbers defined in this manner must have the properties of addition for rational numbers i.e. addition is commutative:

r₁+r₂=r₂+r₁

and associative:

(r₁+r₂)+r₃=r₁+(r₂+r₃)

Multiplication of positive real numbers

Imagine we have a segment L with length r in some unit U:
L=r∙U

If U has length r’ in some other unit U’: U=r’∙U’, we can write:
L=r∙(r’∙U’)

Now we can define the multiplication of real numbers r and r’ as:
L/U’=r∙(r’∙U’)/U’=r∙r’

so if know that: L/U=r, and U/U’=r’, and there exist a procedure to calculate the result of r∙r’, then we also know how long L is in terms of unit U’ without measuring it directly.

Real r and r’ are defined in terms of contracting intervals (CI):
{p_n,q_n}_CI → r
{p’_n,q’_n}_CI → r’

so the question is how to find a CI sequence that is convergent to r∙r’. To simplify things, let’s focus on positive r and r’ to begin with.

We know that:
p_n<r<q_n

and p_n, r, q_n are numbers that multiplied by a unit produce a segment. The inequality above is true for every unit including U=r’∙U’ so:

p_n∙(r’∙U’)<r∙(r’∙U’)<q_n∙(r’∙U’)

For the sequence convergent to r’ we have:

p’_n∙U’<r’∙U’
r’∙U’<q’_n∙U’

so we can combine our inequalities into:

p_n∙(p’_n∙U’)<p_n∙(r’∙U’)<r∙(r’∙U’)<q_n∙(r’∙U’)<q_n∙(q’_n∙U’)

From the chapter about rational numbers we know that:
p_n∙(p’_n∙U’)=(p_n∙p’_n)∙U’
q_n∙(q’_n∙U’)=(q_n∙q’_n)∙U’

so:
(p_n∙p’_n)∙U'<r∙(r’∙U’)<(q_n∙q’_n)∙U’

After dividing this by U’, we get:

p_n∙p’_n<r∙(r’∙U’)/U’=r∙r'<q_n∙q’_n

namely the number r∙r’ is between p_n∙p’_n and q_n∙q’_n for every index n.
If {p_n∙p’_n, q_n∙q’_n} had CI properties, then it would have to converge to r∙r’ and we would have our definition of multiplication! Let’s check whether it is the case.

If p_n+1 is greater or equal to p_n, and p’_n+1 is greater or equal to p’_n then:

p_n∙p’_n ≤ p_n+1∙p’_n+1

By the same token:

q_n+1∙q’_n+1 ≤ q_n∙q’_n

so {p_n∙p’_n, q_n∙q’_n} has the first CI property. Does it have the second? The left side of the second CI property for {p_n∙p’_n, q_n∙q’_n} reads:

q_n+1∙q’_n+1 – p_n+1∙p’_n+1

Let’s see how the above depends on p_n, p’_n, q_n, q’_n:

q_n+1∙q’_n+1 – p_n+1∙p’_n+1 =

=(q_n+(q_n+1-q_n))∙(q’_n+(q’_n+1-q’_n)) – (p_n+(p_n+1-p_n))∙(p’_n+(p’_n+1-p’_n))=

= q_n∙q’_n + q_n∙(q’_n+1-q’_n) + (q_n+1-q_n)∙(q’_n+(q’_n+1-q’_n))

– p_n∙p’_n – p_n∙(p’_n+1-p’_n) – (p_n+1-p_n)∙(p’_n+(p’_n+1-p’_n))=

= (q_n∙q’_n – p_n∙p’_n) – q_n∙(q’_n-q’_n+1) – (q_n-q_n+1)∙q’_n+1

– p_n∙(p’_n+1-p’_n) – (p_n+1-p_n)∙p’_n+1

All violet terms are positive or 0, the second and fourth violet term cannot be both zero at the same time, and the same can be stated about terms three and five. From this it follows that:

q_n+1∙q’_n+1 – p_n+1∙p’_n+1 < q_n∙q’_n – p_n∙p’_n

This is the second CI property so {p_n∙p’_n, q_n∙q’_n} is a CI series that must be convergent to r∙r’ so the multiplication can be defined as:

{p_n,q_n}_CI ∙ {p’_n,q’_n}_CI = {p_n∙p’_n, q_n∙q’_n}_CI → r∙r’

Note that due to the CI properties, it can be demonstrated that any other two CI sequences convergent to r and r’ respectively, would converge to the same r∙r’ when multiplied according to the recipe formulated above.

Distributive property of multiplication over addition and subtraction

The following equation:

(r+r’)∙r”=r∙r”+r’∙r”

is true for all number types we introduced thus far. Is it true for real: r, r’, r” as well? To start with , let’s grab three CI sequences convergent to real numbers r, r’, r”>0.

{p_n,q_n}_CI → r
{p’_n,q’_n}_CI → r’
{p”_n,q”_n}_CI → r”

We have:
({p_n,q_n}_CI+{p’_n,q’_n}_CI)∙{p”_n,q”_n}_CI → (r+r’)∙r”

that can be transformed into:

({p_n,q_n}_CI+{p’_n,q’_n}_CI)∙{p”_n,q”_n}_CI = {(p_n+p’_n)∙p”_n, (q_n+q’_n)∙q”_n}_CI =

= {p_n∙p”_n + p’_n∙p”_n, q_n∙q”_n + q’_n∙q”_n}_CI =

= {p_n∙p”_n, q_n∙q”_n}_CI + {p’_n∙p”_n, q’_n∙q”_n}_CI =

= {p_n, q_n}_CI∙{p”_n, q”_n}_CI + {p’_n, q’_n}_CI∙{p”_n, q”_n}_CI → r∙r”+r’∙r”

thus for real r, r’, r”>0 it must be true that:

(r+r’)∙r”=r∙r”+r’∙r”

Does it hold for negative real numbers? Let’s prove this one first:

(r-r’)∙r”=r∙r”-r’∙r”

for r>r’>0 and r”>0. The left hand side is:

({p_n, q_n}_CI+(-{p’_n,q’_n}_CI))∙{p”_n, q”_n}_CI=

= {p_n∙p”_n – q’_n∙p”_n, q_n∙q”_n – p’_n∙q”_n}_CI → (r-r’)∙r”

and for the right side (r∙r”-r’∙r”):

{p_n, q_n}_CI∙{p”_n, q”_n}_CI – {p’_n,q’_n}_CI∙{p”_n, q”_n}_CI=

={p_n∙p”_n – q’_n∙q”_n, q_n∙q”_n – p’_n∙p”_n}_CI → r∙r”-r’∙r”

Note that because of: p”_n<q”_n, the following must be true:

p_n∙p”_n – q’_n∙q”_n < p_n∙p”_n – q’_n∙p”_n

q_n∙q”_n – p’_n∙q”_n < q_n∙q”_n – p’_n∙p”_n

so altogether for every index n:

p_n∙p”_n – q’_n∙q”_n<p_n∙p”_n – q’_n∙p”_n< (r-r’)∙r” <q_n∙q”_n – p’_n∙q”_n<q_n∙q”_n – p’_n∙p”_n

and this means that:

{p_n∙p”_n – q’_n∙q”_n, q_n∙q”_n – p’_n∙p”_n}_CI → (r-r’)∙r”

thus:

(r-r’)∙r”=r∙r”-r’∙r”

for r>r’>0 and r”>0.

Multiplication and negative real numbers

It is obvious that for real positive numbers it must be true that:

0 = (r-r)∙r’ = r∙r’ – r∙r’ = r∙r’ + (-(r∙r’))

We would like the following to be true as well:

0 = (r+(-r))∙r’ = r∙r’ + (-r)∙r’

so to make it happen, we need to adopt:

-(r∙r’) = (-r)∙r’

Also, we can write:

0 = r∙(r’-r’) = r∙r’-r∙r’ = r∙r’ + (-(r∙r’))

and we would like the following to hold too:

0 = r∙(r’+(-r’)) = r∙r’ + r∙(-r’)

thus we will adopt:

-(r∙r’) = r∙(-r’)

Finally, it’s worth to remember that (-r)∙(-r’) can be transform using the properties above into:

(-r)∙(-r’)=-(r∙(-r’))=-(-(r∙r’))=r∙r’

All of that leads to the conclusion that we don’t have to multiply negative numbers. All we need to do is to pull the negative sign in front of the multiplication.

Properties of real numbers multiplication

The commutative:

r∙r’ = r’∙r

and associative property:

(r∙r’)∙r” = r∙(r’∙r”)

of multiplication for positive real numbers and zero are a direct consequence of their definitions in terms of contracting interval sequences (they simply inherit the properties of rational numbers). Using the property proven in the previous section:

-(r∙r’) = (-r)∙r’ = r∙(-r’)

it is possible to show that the commutative and associative properties hold for negative numbers too. We have done it for whole numbers and the procedure is exactly the same so there is no point to repeat it here.

In the chapter about whole numbers it was also demonstrated that the following properties:

(r+r’)∙r”=r∙r”+r’∙r’
for r,r’,r”≥0

(r-r’)∙r”=r∙r”-r’∙r”
for r>r’≥0, r”≥0

and:
-(r∙r’) = (-r)∙r’ = r∙(-r’)

translate into the distributive property of multiplication over addition:

(r+r’)∙r”=r∙r”+r’∙r’

for numbers that are positive, negative and zero. We have shown that the same properties hold also for real numbers so the conclusion for them must be the same i.e.:

(r+r’)∙r”=r∙r”+r’∙r’

holds for real numbers, positive, negative and zero alike.

Division of real numbers

In the chapter about rational numbers we learnt that division of some number r’ by a number r is equivalent to multiplying r’ times 1/r where 1/r is inverse to r:

1=(1/r)∙r=r∙(1/r)

Let’s adopt this for real numbers as well. The question is how to find a number inverse to given. We will start with:

{p_n,q_n}_CI → r > 0

We can’t choose {1/p_n,1/q_n} because 1/p_n>1/q_n due to p_n<q_n. How about {1/q_n,1/p_n}? The first CI property of {p_n,q_n}_CI is:

p_n ≤ p_n+1 < q_n+1 ≤ q_n

so:

1/p_n ≥ 1/p_n+1 > 1/q_n+1 ≥ 1/q_n

thus {1/q_n,1/p_n} has the first CI property as well. What about the second?
1/q_n+1 ≥ 1/q_n multiplied by -1 produces:

-1/q_n+1 ≤ -1/q_n

that can be combined with:

1/p_n+1≤1/p_n

into:

1/p_n+1 – 1/q_n+1 < 1/p_n – 1/q_n

In the inequality above there is “<” instead of “≤” because according to the second CI property of {p_n,q_n}_CI the possibility that both: -1/q_n+1 = -1/q_n and 1/p_n+1 = 1/p_n are true at the same time, is excluded. The inequality above constitutes the second CI property for {1/q_n,1/p_n} so it must converge to some real number. If we could prove that

{p_n,q_n}_CI∙{1/q_n,1/p_n}_CI={p_n/q_n, q_n/p_n}_CI → 1

then we would know:

{1/q_n,1/p_n}_CI→ 1/r

Let’s take a closer look at p_n/q_n. We can transform it as follows:

p₀<q_n for every n so it must be true that:

In a similar way for q_n/p_n it can be shown that:

After putting it altogether we get:

With rising n the leftmost and rightmost side both get closer and closer to 1 so:

{p_n/q_n, q_n/p_n}_CI → 1

thus

{1/q_n,1/p_n}_CI→ 1/r

If we search for the inverse of -r<0, then we find 1/r first and the inverse of -r is
-(1/r) because:

Preservation of equalities and inequalities

The following two properties:

n=p n+w=p+w

n=p n∙w=p∙w

for real: n, p, and w are a trivial consequence of the definitions adopted above.

Regarding inequalities, they can be introduced because for every two segments they are either equal or one of them must be longer.

Negative length makes no sense but we have introduced the concept of movement direction which is forward (positive) and backward (negative). This in turn can be used to introduce the concept of greater/smaller for negative numbers too. It was already done for rational numbers so I won’t repeat this here. To make a long story short, it simply boils down to the equivalence:

n>p    n–p>0

and real number n-p is greater than zero if it is possible to insert between 0 and n-p a positive rational number.

The equivalence above can be used to prove the following:

n>p n+w>p+w
n≥p n+w≥p+w

For w>0:
n>p w∙n>w∙p

For w<0:
n>p w∙n<w∙p

For w≥0:
n≥p w∙n≥w∙p

For w≤0:
n≥p w∙n≤w∙p

The procedure is exactly the same as for rational numbers.

More than infinity? 😮

Irrational numbers turned out to be very elusive. We can get as close to them as we wish but can’t quite put a finger on any of them. Maybe we could at least name them all. In the positional system with the base equal to 26 we can use letters as digits so names can be replaced with numbers that can be put on a list in ascending order with 0 representing a no-name.



Whole numbers go from minus to plus infinity. Can we name them all i.e. place them on a list from 0 to infinity? Let’s define a variable w_i:

w₀ = 0

w_2∙n-1 = n

w_2∙n = -n

It’s clear that the list w₀, w₁, w₂, w₃, w₄ etc. contains every whole number.

What about rational numbers? This time we will introduce a variable with two indices: r_i,j:

r_0,m = 0/m

r_2∙n-1,m = n/m

r_2∙n,m = -n/m

Now every r_{i≥0, j≥1} represents a ratio and every ratio is represented by some
r_{i≥0, j≥1}. Next, we can put all r_i,j on the list in such a way that the first one will have the sum of indices i+j=1, then there will be the ones with i+j=2, then i+j=3 i.e.:

i+j=1: r_0,1 = 0

i+j=2: r_0,2 = 0, r_1,1 = 1

i+j=3: r_0,3 = 0, r_1,2 = 1/2, r_2,1 = -1

etc.

Every rational number is on the list, many of them more then once e.g. 0, 1/2, 2/4, 4/8 etc. but it doesn’t really matter. If they can be placed on the list with duplications, all the more so without them.

What about real numbers? Surely we can name them as well. After all we have an infinite number of names available. Let’s focus on numbers greater than 0 and smaller than 1. All rational numbers in this range can be put on the list. This time r_i,j will be defined as:

r_i,j=i/j

and the list will contain only r_i,j for which 0<i<j: r_1,2, r_1,3, r_2,3, r_1,4, r_2,4, r_3,4 etc. Next, let’s get rid of duplications and write down our list using a positional notation:

0.d_1,1 d_1,2 d_1,3 d_1,4 d_1,5 …

0.d_2,1 d_2,2 d_2,3 d_2,4 d_2,5 …

0.d_3,1 d_3,2 d_3,3 d_3,4 d_3,5 …

…

where d_m,n is the n-th digit after a decimal dot of the m-th number on the list. Now imagine a number x:

x = 0.x₁ x₂ x₃ x₄ x₅ …

with its digits x_i meeting the requirement:

0 < x_i ≠ d_i,i < 9

From this it follows that x must be greater than 0 and smaller than 1. Also, it is different that the first number on our list because it differs on first fractional position: x₁≠ d_1,1, different than the second number on the list because its second fractional position is not the same: x₂≠ d_2,2. The same can be told about third and every other number on the list. In other words, x is not on the list. Every rational number between 0 and 1 is on the list so x must be irrational.

Note that it is not possible to put together a list like that for all real numbers (between 0 and 1) because number x cannot be on the list even when it contains irrational numbers too. In other words, there are so many irrational numbers that it is not possible to name them all despite the fact that we have an infinite number of names! How bizarre🤔

Approximating roots with rational numbers

We have found irrational numbers by showing that a square root of 2 i.e. 2^1/2 cannot be rational. Then we demonstrated that although irrational numbers are extremely elusive, it is possible to approximate them with CI sequences. Let’s approximate a square root (n^1/2) of an arbitrary rational number n with rational numbers d and g:
g > n^1/2> d > 0

and with a predefined accuracy Δ:
g – d < Δ

A random rational number greater than 0 is either greater or smaller than n^1/2. Let’s assume it is greater:
x₊>n^1/2>0.

After multiplying both sides of this equation by n^1/2 we get:
n^1/2 ∙ x₊>n

so:
n^1/2>n/x₊

thus altogether we can write:
x₊>n^1/2>n/x₊>0

In a similar way we can prove that for some n^1/2>x_– it must be true that:
n/x_–>n^1/2>x_–>0

In other words, for any x>0, n^1/2 must be between x and n/x. From this it follows that the number exactly between x and n/x i.e.
x + (n/x-x)/2=(x+n/x)/2

must be closer to n^1/2 than x and n/x:

Let’s start with some random number x₀>0. n^1/2 must be between x₀ and n/x₀. That is our first approximation.

We know that x₁ = (x₀+n/x₀)/2 is closer to n^1/2 than both x₀ and n/x₀ so x₁ is a better approximation than x₀.

n^1/2 must be between x₁ and n/x₁ so x₂ = (x₁+n/x₁)/2 is even better approximation.

We can continue this process by evaluating a formula:
x_i+1 = (x_i+n/x_i)/2

The question is how much closer to n^1/2 we get after every step of the iteration.

To begin with, let’s define a relative difference between x_i and n^1/2 as:
ε_i = (x_i-n^1/2) / n^1/2

If we could calculate how much smaller ε_i+1 is in comparison to ε_i, that would give us a rate of convergence.

We will start with ε_i+1:
ε_i+1 = (x_i+1-n^1/2) / n^1/2

and transform it in such a way that we can see how ε_i+1 depends on ε_i:

x₀>0 so ε₀>-1 and from the above equation it follows that ε_i>0 as long as i>0.

If ε_i>0 then:
1>ε_i/(ε_i+1)

so after multiplying both sides times ε_i/2 we get:

meaning: after every iteration the relative distance between x_i and n^1/2 is slashed at least in half.

If we are interested in absolute distance between x_i and n^1/2 we can easily obtain it from the relative. For i>0:

After multiplying both sides times 2∙n^1/2:

so:

and x_i – n^1/2>0 for i>0 thus:

Also, because n^1/2 is between x_i+1 and n/x_i+1 we must have:

Let’s estimate 2^1/2 starting with x₀=1.

x₁=(1 + 2)/2=3/2

x₂=(3/2 + 2∙2/3)/2=17/12

x₃=(17/12 + 2∙12/17)/2=577/408

etc.

If we want to reach some predefined accuracy Δ, we simply iterate until:

Δ > x_i+1 – 2/x_i+1 > x_i+1 – 2^1/2

What about cube root n^1/3. For them it can easily be shown that:

x₊>n^1/3>n/x₊²>0

n/x_–²>n^1/3>x_–>0

and the procedure is similar. Of course, we can also calculate roots of some irrational number n. In this case it would be more complicated though because n/x is not rational and we would have to approximate it as well. These days smart phones do this on our behalf. That’s a lot of work done in split second.

Rational powers

The last thing we will cover in this chapter are rational powers. We won’t have a clear aim here. We will just sail into the unknown to see what happens. Later on, the discoveries made will turn out to be useful.

The 1/m-th power of number r is defined as such a number r^1/m that when we multiply it m times by itself, it produces r:

Of course we can multiply r^1/m by itself n times instead. Would the following be true:

Let’s multiply the left side by itself m times:

and this in turn can be rewritten as:

so indeed:

and to shorten the notation we can define r^n/m as:

This in turn leads to the question whether for whole a, b, c≠0, d≠0 we could write:

Using what we have established thus far, it is easy to show that both sides raised to power b∙d yield r^a∙c so indeed:

and also:

All of this leads to the conclusion that we can write:

Note that the following must be true either:

Just raise both sides to power b to find out.

Let’s continue our exploration further. For whole powers a and b we have:

r^a+b = r^a ∙ r^b

Would that be true for rational powers as well? Let’s find out:

Indeed, this holds. Additionally, for rational powers f, g, h we have:

r^f+g = r^f ∙ r^g = r^g ∙ r^f = r^g+f

r^(f+g)+h = (r^f ∙ r^g) ∙ r^h = r^f ∙ (r^g ∙ r^h) = r^f+(b+h)

so all is good. Our definition of rational power is consistent with commutative and associative properties of addition and multiplication. There is one hurdle in all of this though! We have:

and we expect:

The problem is that (-1)^1/6 doesn’t exist! Also, as we know from the section about whole powers that:

and 1/0 doesn’t exist either. In other words, the rational power defined as:

make sense only for r>0.

In closing, let’s examine whether powers preserve the inequalities.
For real: 0<x<y we have a relation:

0<x<y 0<x^m<y^m

for whole m>1 so we must have:

0<x^1/m<y^1/m 0<x<y

Let’s assume that the following implication is false:

x<y x^1/m<y^1/m

This would be equivalent to x<y and x^1/m≥y^1/m being both true. If that’s the case then x≥y and we have a contradiction: x<y and x≥y true at the same time. The implication above must be true then and as a result this one must be true as well:

x^m<y^m x<y

Altogether we have established the equivalences:

x<y x^1/m<y^1/m

x<y x^m<y^m

that together lead to the conclusion:

x<y x^m/n<y^m/n

for whole m,n>1. In a similar way it can be shown that:

x≤y x^m/n≤y^m/n

Summary

In this chapter we finally managed to find numbers adequate to describe lengths of segments. Additionally, we have described their properties in the way independent from the concept of segments so we can use them in any area where it is possible to introduce a divisible unit.

We also have proved that real numbers fundamentally have the same properties as rational numbers, however, at the same time we have made a shocking discovery that there is more of them than infinity! Well, kind of, anyway.

Ability to measure distances between points in space suggests that we should be able to describe location of any object with numbers. We will go deep into the weeds of this problem in the next chapter.