Peano Axioms | Number System | Discrete Mathematics

Introduction :
The set of natural numbers is axiomatically defined below. G. Peano, an Italian mathematician, and J. W. R. Dedekind, a German mathematician, are credited with these axioms. The purpose of these axioms is to prove the existence of one natural number before defining a function to create the remaining natural numbers, known as the successor function.

Peano Axioms :
A premise or starting point for further reasoning and argumentation is an axiom, postulate, or assumption, which is a statement that is assumed to be true. The axioms developed by G.Peano are –

1. P1. 0 ∈ N ; 0 is a natural number –
   Axiom 5 actually replaces 0 with 1 in different versions of the Peano axioms. This yields a nearly identical set of natural numbers, known as “positive whole numbers” . The context determines whether or not a mathematician includes 0 in the natural numbers. We follow the standard practice of including 0 as a natural number.
   Only the existence of a single natural number, 0, is guaranteed at this point. The successor function is used in the next axiom to construct other natural numbers. The successor function is a function S with the domain N, as its name implies. The co-domain of S is also N, according to the next axiom.
   The next three axioms describe the equality relation.
2. ∀x ∈ N ⇒ x=x ; 
   Reflexive equality. The fourth axiom, known as the closure of equality axiom, states that if “anything” is equal to a natural number, then that “anything” must also be a natural number.
3. ∀ x, y ∈ N ; and if x=y ⇒  y=x ; 
   Symmetric equality.If one natural number equals another, the second number should be equal to the first. This is known as the axiom of symmetry.
4. ∀ x, y, z ∈ N ; and if x=y & y = z  ⇒ x=z; 
   Transitive equality.The next property states that if one natural number equals a second, and that the second natural number equals a third, then the first and third are equal. The transitivity axiom is what it’s called.
5. ∀ a, b ; if a ∈ N and a=b   ⇒ b is also a natural number.
6. P2. If x ∈ N  ⇒ S(x) ∈ N.
   In Peano’s initial axioms, 1 was employed instead of 0 as the “first” natural number. 
   Most recent formulations of the Peano axioms begin with 0. This is because 0 is the additive identity in arithmetic. 
   The successor of x is also a natural number, if x is a natural number. 
   S(x) will be referred to as the successor of x, as the axiom indicates. 
   Intuitively, S(x) should be interpreted as x+1.
    We are still a long way from having the natural numbers as we know them at this point. Axioms 1 and 6 define a unary representation of natural numbers: 
   S(0) = 0 +1 = 1
   S(S(0)) = S(1) = 1+1 = 2 .
   The attributes of this representation are defined by the next two axioms.
7. If n ∈ N ; S(n) ≠ 0 . 
   If  n ∈ N, then the successor of n cannot be  0.
8. ∀ a, b ∈ N;  if S(a) = S(b) ⇒  a = b.
   S is an injection (one -one mapping, i.e, successor of every number is unique)
   The preceding axiom has some significant implications. From axiom 1, it rules out the option of defining N as simply 0 and 1. 
   To understand why, consider that S(0) = 1 already exists and that S(1) = 1 is not possible due to injection mapping.
   The possibility of S(1) = 0 is ruled out by Axiom 6. As a result, S(1) must be another natural number, which we call 2. 
   Thus : 2 = S(1) .
   S(2) cannot be 0, 1, or 2 according to a similar reasoning. 
   As a result, it must be a different natural number, which we refer to as 3. Following this pattern, we can deduce that N must contain all of the natural numbers we are aware of. At this point, we know that N must contain 0 and that its successor 1 = S(0), its successor 2 = S(1), and so on. Thus, every number has a unique successor. 
   So, if we say that the 2 successors  are the same, then it implies that they are the successors of the same number.
9. If V is an inductive set; i.e. ; 0 ∈ V and every natural number n ∈ V, then S(n) ∈ V  ⇒  N ⊂ V
   As previously established, the first eight axioms guarantee that { 0, 1, 2, 3,… } ∈ N.  We know that the set  { 0, 1, 2, 3,… } is an inductive set. As a result of Axiom 9,  N ⊂ { 0, 1, 2, 3,…}  must be true. As a result, we got the set equality we were looking for: N = { 0, 1, 2, 3,…}

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