## Sunday, February 05, 2006

### Natural Numbers

The natural numbers are the numbers that are most familiar to us; they are the counting numbers that usually begin at 1 (some mathematicians hold that the natural numbers include 0). This also matches with the history of mathematics where the use of 0 was a major innovation.

The mathematician Kronecker once said:
"God made the natural numbers. Everything else is the work of man." (Quoted from here)

At the same time, this leads us to an interesting question. Is it possible to define the natural numbers in terms of a set of axioms similiar to what Euclid did for geometry?

The successful definition of natural numbers was done by two men: Richard Dedekind in his book What is a Number and the logician Guiseppe Peano. Today, these axioms are known as Peano's Axioms.

Here are a set of postulates based on Peano's Axioms:

I. The natural numbers is a set of numbers starting with 1 where each number has a unique successor which we will characterize by S(a) where a is a natural number.

II. All numbers but 1 have are themselves successors to a unique number.

III. a = b if and only if S(a) = S(b)

IV. If a property is possessed by 1 and possession by a value a implies it is also true of S(a), then we say it is true of all numbers. (Axiom of Induction)

Definition 1: m + n

m + 1 = S(m)
m + S(n) = S(m + n)

Example:

1 + 2 = S(1 + 1) = S(S(1)) = S(2) = 3

Definition 2: m * n

m * 1 = m
m * S(n) = m + (m * n)

Example:

2 * 2 = 2 + (2*1) = 2 + 2 = 4

Definition 3: m is less than n

m is less than n if there exists a natural number d such that m + d = n.

Definition 4: m - n

S(m) - 1 = m
S(m) - S(n) = m - n

If m is less than or equal to n, then it is undefined.

Lemma 1: Natural numbers are closed under addition.

(1) m+1 = S(m)
(2) Assume that there exists a value n such that m + n is a natural number.
(3) m + S(n) = S(m+n)
(4) Which must also be a natural number by Axiom I.

QED

Lemma 2: Natural numbers are closed under multiplication

(1) m*1 = m
(2) Assume that there exists a value n such that m * n is a natural number
(3) m*S(n) = m + (m*n)
(4) Which is a natural number by Lemma 1 above.

QED

Lemma 3: Natural numbers are not closed under subtraction.

m - n is undefined for the case where m ≤ n.

QED

Lemma 4: 1 + m = S(m)

(1) Case n=1:

1 + 1 = S(1)

(2) Assume that it is true up to n.

(3) 1 + (n+1) = 1 + S(n) = S(n+1)

(4) So, it is true by Axion IV.

QED

Lemma 5: Natural numbers are commutative under addition: m + n = n + m

(1) For case n=1:

m + 1 = S(m)

(2) From Lemma 4, 1 + m = S(m)

(3) Assume it is true up to n so that m+n=n+m

(4) m + (n+1) = m + S(n) = S(m+n)

(5) (n+1) + m = S(n) + m = S(n+m) = S(m+n)

(6) So, we apply Axiom IV and we are done.

QED

Lemma 6: 1*m = m*1

(1) 1*1 = 1*1

(2) S(m)*1 = S(m) [By Definition 2 above]

(3) 1*S(m) = 1 + m = m + 1 = S(m)

(4) We are done since all numbers are either 1 or a successor to another number. (by Axion II)

QED

Lemma 7: Distributive Law: (a + b)m = am + bm

(1) Case m = 1:

(a + b)1 = a + b [By Definition 2]

a(1) + b(1) = a + b [By Definition 2]

(2) So we assume that it is true up to n so that (a+b)n = an + bn.

(3) (a+b)S(n) = (a + b) + (a+b)n = a + b + an + bn = a + an + b + bn

(4) a(S(n)) + b(S(n)) = a + an + b + bn

(5) So by the Axiom of Induction we are done (see Axiom IV)

QED

Lemma 8: Natural numbers are commutative under multiplication: m * n = n * m

(1) Case n=1:

m * 1 = 1 * m [From Lemma 6 above]

(2) Assume that it is true up to n so that we have m*n=n*m

(3) m*(n+1) = m + mn [By definition 2]

(4) (n+1)*m = nm + m = m + nm = m + mn

(5) So we are done by the Axion of Induction (Axiom IV above)

QED

Lemma 9: Natural numbers are associative under addition:
(a + b) + n = a + (b + n)

(1) Case n = 1

(a + b) + 1 = S(a + b)

a + (b + 1) = a + S(b) = S(a + b)

(2) (a + b) + S(n) = S(a + b + n)

(3) a + (b + S(n)) = a + S(b + n) = S(a + b + n)

(4) It is true now in all cases since all numbers are either 1 or are a successor (by Axiom II above)

QED

Lemma 10: Natural numbers are associative under multiplication:
(a*b)*n = a*(b*n)

(1) Case n = 1

(a*b)*1 = a*b

a*(b*1) = a*b

(2) (ab)*S(n) = S(n)(ab) = (n+1)(ab) = abn + ab

(4) a*(b*S(n)) = a*(b*(n+1)) = a*((n+1)*b) = a*(bn+b) = (bn+b)*a = abn+ab

QED

References