Peano Axioms | Number System | Discrete Mathematics

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. These axioms aim 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,...}

Solved Examples on Peano Axioms

Example 1: Prove that 1 is a natural number.

Proof:

By axiom 1, 0 is a natural number.

By axiom 2, S(0) exists and is a natural number.

We define 1 as S(0).

Therefore, 1 is a natural number.

Example 2: Prove that 2 is not equal to 0.

Proof:

Assume, for the sake of contradiction, that 2 = 0.

We know that 2 = S(1) = S(S(0)).

If 2 = 0, then S(S(0)) = 0.

But this contradicts axiom 3, which states that 0 is not the successor of any natural number.

Therefore, our assumption must be false, and 2 ≠ 0.

Example 3: Prove that for any natural number n, n + 1 ≠ n.

Proof:

Let n be any natural number.

n + 1 is defined as S(n).

Assume, for the sake of contradiction, that n + 1 = n.

This would mean S(n) = n.

But this contradicts axiom 4, which states that if S(m) = S(n), then m = n.

(In this case, S(n) = n would imply n = n-1, which is impossible for natural numbers.)

Therefore, our assumption must be false, and n + 1 ≠ n for any natural number n.

Example 4: Use mathematical induction to prove that for all natural numbers n, n + 1 > n.

Proof:

Base case: When n = 0

0 + 1 = 1, and 1 > 0 (by the definition of the successor function and the ordering of natural numbers).

Inductive step: Assume the statement is true for some natural number k, i.e., k + 1 > k.

We need to prove that (k + 1) + 1 > (k + 1).

(k + 1) + 1 = S(k + 1) (by the definition of addition)

= S(S(k)) (by the inductive hypothesis)

Example 6 : Prove that addition is associative for natural numbers.

Proof:

We need to show that (a + b) + c = a + (b + c) for all natural numbers a, b, and c.

We can use induction on c.

Base case: c = 0

(a + b) + 0 = a + b (by definition of addition)

a + (b + 0) = a + b (by definition of addition)

So, (a + b) + 0 = a + (b + 0)

Inductive step:

Assume (a + b) + k = a + (b + k) for some natural number k.

We need to prove (a + b) + S(k) = a + (b + S(k))

(a + b) + S(k) = S((a + b) + k) (by definition of addition)

= S(a + (b + k)) (by inductive hypothesis)

= a + S(b + k) (by definition of addition)

= a + (b + S(k)) (by definition of addition)

By induction, (a + b) + c = a + (b + c) for all natural numbers a, b, and c.

Example 5 : Prove that there is no largest natural number.

Proof by contradiction:

Assume there exists a largest natural number n.

Consider S(n), the successor of n.

By axiom 2, S(n) is a natural number.

By the definition of the successor function, S(n) > n.

This contradicts our assumption that n is the largest natural number.

Therefore, there is no largest natural number.

Example 6 : Prove that for any natural number n, n + 0 = n.

Proof:

Proof by induction:

Base case: n = 0

0 + 0 = 0 (by definition of addition)

Inductive step:

Assume k + 0 = k for some natural number k.

We need to prove S(k) + 0 = S(k)

S(k) + 0 = S(k + 0) (by definition of addition)

= S(k) (by inductive hypothesis)

By induction, n + 0 = n for all natural numbers n.

Example 7 : Prove that multiplication distributes over addition for natural numbers.

Solution:

We need to show that a * (b + c) = (a * b) + (a * c) for all natural numbers a, b, and c.

We can use induction on c.

Base case: c = 0

a * (b + 0) = a * b (by properties of addition and multiplication)

(a * b) + (a * 0) = (a * b) + 0 = a * b

Inductive step:

Assume a * (b + k) = (a * b) + (a * k) for some natural number k.

We need to prove a * (b + S(k)) = (a * b) + (a * S(k))

a * (b + S(k)) = a * S(b + k) (by definition of addition)

= (a * (b + k)) + a (by definition of multiplication)

= ((a * b) + (a * k)) + a (by inductive hypothesis)

= (a * b) + ((a * k) + a) (by associativity of addition)

= (a * b) + (a * S(k)) (by definition of multiplication)

By induction, a * (b + c) = (a * b) + (a * c) for all natural numbers a, b, and c.

Example 9 : Prove that for any natural numbers a and b, if a + b = a, then b = 0.

Proof:

Assume a + b = a for some natural numbers a and b.

By the cancellation property of addition (which can be proved using Peano axioms), we can subtract a from both sides:

(a + b) - a = a - a

The left side simplifies to b, and the right side to 0:

b = 0

Therefore, if a + b = a, then b must be 0.

Example 10 : Prove that for any natural number n, 1 * n = n.

Proof by induction:

Base case: n = 0

1 * 0 = 0 (by definition of multiplication)

Inductive step:

Assume 1 * k = k for some natural number k.

We need to prove 1 * S(k) = S(k)

1 * S(k) = (1 * k) + 1 (by definition of multiplication)

= k + 1 (by inductive hypothesis)

= S(k) (by definition of successor function)

By induction, 1 * n = n for all natural numbers n.

Practice Problems on Peano Axioms

1. Prove that for any natural number n, n + 1 ≠ n.

2. Show that addition is commutative for natural numbers: a + b = b + a for all natural numbers a and b.

3. Prove that multiplication is associative: (a * b) * c = a * (b * c) for all natural numbers a, b, and c.

4. Demonstrate that 0 is the additive identity: n + 0 = n for any natural number n.

5. Prove the principle of mathematical induction as derived from the Peano axioms.

6. Show that for any natural numbers a and b, if a + b = a, then b must be 0.

7. Prove that there is no largest natural number.

8. Demonstrate that multiplication distributes over addition: a * (b + c) = (a * b) + (a * c) for all natural numbers a, b, and c.

9. Prove that for any natural number n, 1 * n = n.

10. Show that the successor function is injective: if S(a) = S(b), then a = b for any natural numbers a and b.

Summary

The Peano axioms are a set of fundamental rules that define the natural numbers and form the basis for arithmetic. They consist of five key principles: 0 is a natural number, every natural number has a unique successor, 0 is not the successor of any natural number, different natural numbers have different successors, and the principle of mathematical induction holds. These axioms allow us to rigorously define and prove basic properties of natural numbers, such as the commutativity and associativity of addition and multiplication, the existence of additive and multiplicative identities, and the lack of a largest natural number. They also enable us to establish more complex concepts like the well-ordering principle and the division algorithm. By providing a formal foundation for arithmetic, the Peano axioms play a crucial role in mathematics, underpinning our understanding of numbers and serving as a starting point for more advanced mathematical theories.