Deriving Numbers from First Principles

In the late 19th century, mathematician Giuseppe Peano worked out Peano's axioms as a starting point for defining numbers.

The equality axioms for members of some set S are:

1. For x in S, x = x -- reflexive
   
2. For x and y in S, x = y implies y = z -- symmetric
   
3. For x, y, and z in S, x = y and y = z implies x = z -- transitive
   
4. for x in S, x = y implies y in S -- closure
   

Peano's axioms depend on the notion of successor S(x); they are:

1. 0 is a Peano number (PN)
   
2. For every PN x, S(x) is also a PN
   
3. For every PN x, S(x) != 0
   
4. For all PN's x and y, S(x) = S(y) implies x = y
   
5. If a property P is true of 0, and P being true for PN x implies its truth for S(x), then P is true of all PN's -- mathematical induction
   

Addition one defines with

1. for every PN x, x + 0 = x
   
2. For all PN's x and y, x + S(y) = S(x+y)
   

Multiplication one defines with

1. for every PN x, x*0 = 0
   
2. For all PN's x and y, x*S(y) = (x*y) + x
   

I will now get into commutative, associative, distributive, identity, and zero properties of addition and multiplication.

First, addition.

Additive Identity = 0
0 + x = x + 0 = x

Right identity?
A1: x + 0 = x

Left identity?
Mathematical induction:
0 + 0 = 0 + 0 = 0
0 + x = x + 0 implies 0 + S(x) = S(x) + 0

First:
A1: 0 + 0 = 0
Second LHS:
A2: 0 + S(x) = S(0 + x)
Premise: S(0 + x) = S(x)
Second RHS:
A1: S(x) + 0 = S(x)
LHS = RHS

Success: both left and right

Addition is Associative
Mathematical induction:
x + (y + 0) = (x + y) + 0
x + (y + z) = (x + y) + z implies x + (y + S(z)) = (x + y) + S(z)
First LHS:
A1: x + (y + 0) = x + y
First RHS:
A1: (x + y) + 0 = x + y
LHS = RHS
Second LHS:
A2: x + (y + S(z)) = x +S(y + z) = S(x + (y + z))
Second RHS:
A2: (x + y) + S(z) = S((x + y) + z)
Premise: S((x + y) + z) = S(x + (y + z))
LHS = RHS

Success

Addition is Commutative
Addition with 0: additive-identity property
0 + x = x + 0 = x

Definition of 1: S(0)
Successor property in terms of adding 1:
S(x) = x + 1
Identity: S(x) = S(x + 0)
A2: S(x + 0) = x + S(0) = x + 1

Addition with 1: mathematical induction
1 + 0 = 0 + 1
1 + x = x + 1 implies 1 + S(x) = S(x) + 1
First: additive-identity property
Second LHS:
Succ.: 1 + S(x) = 1 + (x + 1)
Assoc: 1 + (x + 1) = (1 + x) + 1
Second RHS:
Succ.: S(x) + 1 = (x + 1) + 1
Premise: (x + 1) + 1 = (1 + x) + 1
RHS = LHS

Addition in general: mathematical induction
x + 0 = 0 + x
x + y = y + x implies x + S(y) = S(y) + x
First: additive-identity property
Second LHS:
Succ.: x + S(y) = x + (y + 1)
Assoc. x + (y + 1) = (x + y) + 1
Second RHS:
Succ.: S(y) + x = (y + 1) + x
Assoc.: (y + 1) + x = y + (1 + x)
Commut. 1: y + (1 + x) = y + (x + 1)
Assoc. y + (x + 1) = (y + x) + 1
Premise: (y + x) + 1 = (x + y) + 1
LHS = RHS

Success

Now, multiplication.

Multiplicative Zero = 0
0*x = x*0 = 0

Right zero?
M1: x*0 = 0

Left zero?
mathematical induction:
0*0 = 0
0*S(x) = 0*x
First:
M1: 0*0 = 0
Second LHS:
M2: 0*S(x) = 0*x + 0 = 0*x
LHS = RHS

Success: both left and right

Multiplicative Identity = 1
1*x = x*1 = x

Right identity?
M2: x*1 = x*S(0) = x*0 + x
M1: x*0 + x = 0 + x = x

Left identity?
Mathematical induction
1*0 = 0
1*x = x implies 1*S(x) = S(x)
First:
M1: 1*0 = 0
Second LHS:
1*S(x) = 1*x + 1
Premise: 1*x + 1 = x + 1
Second RHS:
Succ.: S(x) = x + 1
LHS = RHS

Success: both left and right

Multplication is Distributive over Addition
Left distributive?
x*(y+z) = (x*y) + (x*z)

Mathematical induction
x*(y+0) = (x*y) + (x*0)
x*(y+z) = (x*y) + (x*z) implies x*(y+S(z)) = (x*y) + (x*S(z))
First LHS:
x*(y+0) = x*y
First RHS:
M1: (x*y) + (x*0) = (x*y) + 0 = x*y
LHS = RHS
Second LHS:
M2: x*(y+S(z)) = x*(S(y+z)) = x*(y+z) + x
Second RHS:
M2: (x*y) + (x*S(z)) = (x*y) + (x*z + x) = ((x*y) + (x*z)) + x
Premise: ((x*y) + (x*z)) + x = x*(y+z) + x
LHS = RHS

Right distributive?
(x+y)*z = (x*z) + (y*z)

Mathematical induction:
(x+y)*0 = (x*0) + (y*0)
(x+y)*z = (x*z) + (y*z) implies (x+y)*S(z) = (x*S(z)) + (y*S(z))
First LHS:
M1: (x+y)*0 = 0
First RHS:
M1: (x*0) + (y*0) = 0 + 0 = 0
Second LHS:
M2: (x+y)*S(z) = (x+y)*z + (x+y)
Second RHS:
M2: (x*S(z)) + (y*S(z)) = (x*z + x) + (y*z + y)
Comm.,Assoc.: (x*z + x) + (y*z + y) = (x*z + y*z) + (x + y)
Premise: (x*z + y*z) + (x + y) = (x+y)*z + (x + y)
LHS = RHS

Success: both left and right

Multiplication is Associative
Mathematical induction:
x*(y*0) = (x*y)*0
x*(y*z) = (x*y)*z implies x*(y*S(z)) = (x*y)*S(z)
First LHS:
M1: x*(y*0) = x*0 = 0
First RHS:
M1: (x*y)*0 = 0
Second LHS:
M2: x*(y*S(z)) = x*(y*z + y)
Distrib.: x*(y*z + y) = x*(y*z) + x*y
Second RHS:
M2: (x*y)*S(z) = (x*y)*z + x*y
Premise: (x*y)*z + x*y = x*(y*z) + x*y
LHS = RHS

Success

Multiplication is Commutative
Mathematical induction:
x*0 = 0*x
x*y = y*x implies x*S(y) = S(y)*x
First:
zero property
Second LHS:
x*S(y) = x*y + x
Second RHS:
Succ.: S(y)*x = (y + 1)*x
Distrib.: (y + 1)*x = y*x + x
Premise: y*x + x = x*y + x
LHS = RHS

Success

My previous post may have seemed hopelessly long-winded, but it should give an idea of how one can prove some familiar properties of numbers from first principles.

We can go on from there to consider a simple equation in Peano numbers. Is there a Peano number x such that

x + a = 0

where a is a nonzero Peano number? Since a is nonzero, a = S(a') where a' is another Peano number, and

x + a = x + S(a') = S(x + a) != 0

from the definition of addition and Peano axiom #3. But there is a way to get around that, and that involves ordered pairs (a,b) that satisfy this equality relation:

(a,b) = (c,d) if a+d = b+c

This satisfies the axioms of equality:

Reflexivity: (a,b) = (a,b) if a+b = a+b

Symmetry: (a,b) = (c,d) implies (c,d) = (a,b) means
a+d = b+c implies c+b = d+a

Transitivity: (a,b) = (c,d) and (c,d) = (e,f) implies (a,b) = (e,f) means
LHS 1: a+d = b+c
LHS 2: c+f = d+e
Sum: a+c+d+f = b+c+d+e
RHS: a+f = b+e
Add c+d: a+c+d+f = b+c+d+e
Success

To relate these new numbers to Peano numbers, we have
(a,b) = c if a = b+c

We can move backward with the successor relation. If a and b are nonzero, then a = S(a') and b = S(b'). Is
(a,b) = (a',b')

If so, then a+b' = a'+b
which is S(a') + b' = a' + S(b')
and both sides equal a' + b' + 1

Thus, one can move backward until one arrives at these types of numbers: (a,0), (0,0), and (0,a) where a is nonzero. The first two equal Peano numbers a and 0, while the third one equals no Peano number.


Now for addition. Define (a,b) + (c,d) = (a+c,b+d)

It is easy to see that it is commutative and associative.
Also, (a,a) = 0 is the additive identity: x + 0 = x

We also notice that (a,b) and (b,a) look similar. Let us now add them:
(a,b) + (b,a) = (a+b,a+b) = 0

Thus, (b,a) is the additive inverse of (a,b), and more specifically, (0,a) is the additive inverse of (a,0) or minus (a,0). The Peano numbers and minus the Peano numbers thus cover all these ordered-pair numbers. These numbers are also known as the integers, and the definition of Peano numbers here is of nonnegative integers. A nonzero Peano integer is a positive integer, and minus it is a negative integer.

Turning to multiplication, its definition is a bit more elaborate:
(a,b) * (c,d) = (a*c+b*d, a*d+b*c)

But one can show that it is commutative and associative, that it is distributive over addition, and that its identity is 1 = (1,0) -- 1*x = x for all x.

Thus, integer multiplication has all the properties expected from Peano-number multiplication.

One can even define negation with the help of -1: -x = (-1)*x, from (b,a) = (0,1)*(a,b)

Now a note on "positive" vs. "negative" numbers. The terms date from half a millennium ago, when European mathematicians found it hard to believe that negative numbers could possibly be reasonable numbers.

But some centuries before, mathematicians in India like Brahmagupta had no trouble understanding negative numbers, which they called debt numbers as opposed to positive numbers being asset numbers. It is unfortunate that this nice analogy for negative numbers did not catch on in Europe.


Now that we have defined integers, we can consider solutions of equations involving multiplications: b*x = a

This will have integer solutions x only for some sets of integers a and b, but we can use the ordered-pair approach as before to define new kinds of numbers.

Equality: (a,b) = (c,d) if a*d = b*c

Equality to an Integer: (a,b) = c if a = b*c

It satisfies the axioms of equality stated earlier.

Addition: (a,b) + (c,d) = (a*d + b*c, b*d)
Multiplication: (a,b)*(c,d) = (a*c,b*d)

As with integers, these new numbers have the familiar commutative, associative, and distributive properties. These numbers not only have additive inverses, (a,b) -> (-a,b), but also multiplicative inverses, (a,b) -> (b,a). The latter are also known as reciprocals, and these pairs of integers represent rational numbers.

We can also define an ordering for Peano numbers:

a <= b is equivalent to there being some Peano number x such that a + x = b, or else that one can get b from a by applying the successor operation x times to a.

with similar definitions for a < b, a >= b, and a > b.

Ordering is antisymmetric:
a <= b and b <= a imply a <= b

Proof:
a <= b implies a + x = b
b <= a implies b + y = a

Adding y to both sides of the first one gives
a + x + y = a

Applying Peano axiom 4 a times to the above expression, we find x + y = 0, which by axiom 3 implies that x = y = 0.

Ordering is transitive:
a <= b and b <= c imply a <= c

Proof:
x <= b implies a + x = b
b <= c implies b + y = c

Adding y to both sides of the first equation yields
a + (x + y) = c
and (x + y) is, of course, a Peano number.

Extending to integers, we can define an ordering with:
a <= b is equivalent to a - b being a nonnegative integer, that is, equivalent to a Peano number.

One can demonstrate the above axioms in the integer case with analogies with the above proofs, and this integer ordering includes nonnegative-integer ordering as a special case. Addition to both cases preserves the relations, as does multiplication by a positive integer. Multiplication by a negative integer, however, reverses their direction.


For rational numbers, it gets a bit trickier.
(a,b) <= (c,d) if a*d - b*c is a nonnegative integer, where we have set b and d to be positive integers.

The familiar axioms hold, and rational numbers equivalent to integers are placed in the right order.

I'll move on to real numbers, but first a proof that rational-number inequality is transitive and then one that integers have the right order.

Transitivity:
(a,b) <= (c,d) if a*d - b*c is nonnegative
(c,d) <= (e,f) if c*f - d*e is nonnegative
Multiply the first definition by f and the second one by b, then add. The result is
d*(a*f - b*e), which implies
(a,b) <= (e,f)

Integers:
(a,1) <= (b,1) implies a - b is nonnegative, as expected for the corresponding integers.


There are various ways of getting real numbers from rational numbers. The most useful IMO is with Cauchy sequences of rational numbers. These are infinite sequences that satisfy a convergence criterion independent of their limits, since their limits need not be in the set that the members are defined in. For a series a(n), and for every e > 0, there will be a N such that

|a(n) - a(m)| < e for all m,n >= N

Using more and more digits of a decimal or other place-system representation of a number is constructing a Cauchy sequence of rational numbers, and various approximation algorithms also generate Cauchy sequences. A simple one is Newton's method for square roots:

To find sqrt(a), take an initial value x(1) and find the next ones using

x(n+1) = (1/2)*(x(n) + a/x(n))

If a and x(1) are rational, then all the x(n) will be rational, even though the square root of a may be irrational.


If the Cauchy sequence has a limit a(lim) defined for some set of numbers that includes its members, then for every e > 0, there will be a N such that

|a(n) - a(lim)| < e for all m,n >= N

One defines addition and multiplication in term-by-term fashion:

(a + b)(n) = a(n) + b(n)
(a * b)(n) = a(n) * b(n)

Two Cauchy sequences have equal values if their difference has a limiting value of 0. For every e > 0, there will be a N such that

|a(n) - b(n)| < e for all n >= N

and they can be shown to produce Cauchy sequences:
|a(m) + b(m) - a(n) - b(n)| <= |a(m) - a(n)| + |b(m) - b(n)|
|a(m)*b(m) - a(n)*b(n)| <= |b(m)|*|a(m) - a(n)| + |a(n)|*|b(m) - b(n)|

These operations can be shown to have the expected commutative, associative, distributive, identity, and inverse properties. When the series have limits defined in some set of numbers, those limits have the expected behavior under those operations.

Ordering is most easily defined in the absence of the equality case:

a < b if and b are unequal and if there is some N such that

a(n) < b(n) for all n > N

or else there is some positive x and some N such that

a(x) + x <= b(x) for all n > N

One includes the equality case to get the a <= b condition.

This ordering definition can be shown to have the expected properties, and to produce the right order of rational numbers. Thus, for transitivity:

a(n) + x <= b(n) for all n > N1
b(n) + y <= c(n) for all n > N2
a(n) + (x + y) <= c(n) for all n > max(N1,N2)

Having created a construction for real numbers, can we conclude that they are all rational numbers?

The first counterevidence was discovered about 2500 years ago by some follower of Pythagoras. The Pythagoreans had a strong belief in the mystical powers of numbers, and they ended up doing a lot of mathematics. But one of them discovered some Satanic Verses: he showed that the square root of 2 is an irrational number. The proof is simple:

If sqrt(2) is a rational number, then it can be expressed as a/b, where a and b are two relatively-prime positive integers. One finds from it that

a^2 = 2*b^2

meaning that a must be divisible by 2. Let a = 2*c, where c is a positive integer. Then

(2*c)^2 = 4*c^2 = 2*b^2

Dividing both sides by 2 yields

2*c^2 = b^2

meaning that b is also divisible by 2, which contradicts the rational-number hypothesis.

This proof can easily be extended to the square root of any other prime number, and from there, to the square root of any integer that has an odd power of any of its prime factors.


Having taken on sqrt(2) I will now take on sqrt(-1). Every real number has a nonnegative square, so sqrt(-1) must be a new kind of number, an "imaginary number". That name dates back to when some mathematicians considered it an illegitimate sort of number, but they are as "real" as any other kind of number. As the great mathematician Karl Friedrich Gauss had noted two centuries ago:

That this subject [imaginary numbers] has hitherto been surrounded by mysterious obscurity, is to be attributed largely to an ill adapted notation. If, for example, +1, -1, and the square root of -1 had been called direct, inverse and lateral units, instead of positive, negative and imaginary (or even impossible), such an obscurity would have been out of the question.

To construct complex numbers, we take some numbers and create ordered pairs with these equality, addition and multiplication operations:
(a,b) = (c,d) for a=c, b=d (trivial equality)
(a,b) + (c,d) = (a+c, b+d)
(a,b) * (c,d) = (a*c - b*d, a*d + b*c)

The originals a are equivalent to (a,0). The imaginary unit i is equivalent to (0,1).

Additive identity: 0 = (0,0)
Additive inverse: - (a,b) = (-a,-b)
Multiplicative identity: 1 = (1,0)
Multiplicative inverse: 1/(a,b) = (a/(a^2+b^2), -b/(a^2+b^2))

The operations also have the expected commutative, associative, and distributive properties.

They cannot be completely ordered; whatever ordering scheme one uses will make different complex numbers equal.

Thus, one can define complex integers and complex rational numbers as well as the complex extension of real numbers.

Some of these extensions of numbers have been done by taking various insoluble equations and treating their solutions as additional kinds of numbers.

Thus, Peano numbers -> integers -> rational numbers

We can extend the rational numbers with every solution of a polynomial equation with rational coefficients:

a(n)*x^n + a(n-1)*x^(n-1) + ... + a(1)*x + a(0) = 0

The result is the algebraic numbers. In general, they will be complex, with their real and imaginary parts both being real algebraic numbers. A real number that is not an algebraic number is called a "transcendental number", and some of them are e (base of natural logarithms) and pi (circle constant).

Rational numbers are all algebraic, of course, but some algebraic numbers are not rational numbers, like sqrt(2) and i.

Addition and multiplication of algebraic numbers can be defined by composing defining polynomials for them that are combinations of the defining polynomials of the input numbers. They have all the expected properties, and a polynomial with algebraic-number coefficients will also have algebraic-number solutions.

This polynomial-solution closure property is also shared by the most general complex numbers, those whose real and imaginary parts are both real numbers. Their polynomial-equation solutions are also general complex numbers.

One can compose polynomials for the sums and products of roots of polynomials in terms of the coefficients of those polynomials without having to solve them. The mathematics for doing so is rather involved, but one algorithm for doing so constructs sums of powers of a polynomial's roots along the way, using the Newton-Girard formulas. It's very easy to find the root-power-sum sequences for sums and products of roots in terms of their polynomials' root-power-sum sequences.

A linear polynomial will have a single root, r, and its root-power-sum sequence will have the form

{1, r, r^2, r^3, r^4, ...}

Likewise a quadratic polynomial will have two roots, r1 and r2, with root-power-sum sequence

{2, r1+r2, r1^2+r2^2, r1^3+r2^3, r1^4+r2^4...}

The polynomial x^2 + a*x + b = 0 has solutions

x = (-a + sqrt(a^2-4b))/2 and x = (-a - sqrt(a^2-4b))/2

and the series

{2, -a, a^2 - 2b, -a*(a^2 - 3b), a^4 - 4a^2*b + 2b^2, ...)

The addition formula:

rps(a+b,n) = sum for k = 0 to n for n!/(k!(n-k)!)*rps(a,n-k)*rps(b,k)

The multiplication formula:

rps(a*b,n) = rps(a,n)*rps(b,n)

They can be shown to have many of the expected properties: commutative, associative, distributive, and identities. However, neither inverses nor ordering can easily be defined by this method.