Monday, February 06, 2006

Set of Integers

The set of integers can be constructed from the Natural Numbers.

We can define the set of integers as the set NxN where "x" is the Cartesian Product.

In set theory, the Cartesian Product is the combination of all the elements of one set with all the elements of the other set forming pair.

So, if N = { 1, 2, 3 ... }, then NxN = { (1,1), (1,2), (1,3), ..., (2,1),(2,2), ..., }

Here is the definition for integers:

Definition 1: Set of Integers: Z is the set NxN where (a,b) is the same number as (c,d) if and only if a + d = b + c.

NOTE: Z stands for Zahlen which is German for number.

From this perspective, each number corresponds to the set of all values (a,b) where a-b is equal to the number.

So 0 for example corresponds to { (1,1), (2,2), .... }

And -1 corresponds to { (1,2), (2,3), ... }

We can use the definition to show that (1,1) ~ (2,2) in that 1 + 2 = 1 + 2.

From this, we can now define addition, multiplication, and subtraction.

Definition 2: (a,b) + (c,d) = (a + c, b + d)

Example:

-1 + -3 = (2,3) + (5,8) = (2+5,3+8) = (7,11) = -4

Definition 3: (a,b) * (c,d) = (ac + bd, ad+bc)

Example:

-1 * -1 = (1,2)*(1,2) = (1*1 + 2*2,1*2 + 2*1) = (1 + 4,2 + 2) = (5,4) = 1

-1 * 3 = (1,2)*(4,1) = (1*4 + 2*1,1*1 + 2*4) = (4 + 2,1 + 8) = (6,9) = -3

Definition 4: (a,b) - (c,d) = (a + d, b + c)

Example:

4 - 3 = (5,1) - (4,1) = (5 + 1, 1 + 4) = (6, 5) = 1

(-3) - (-5) = (2,5) - (1,6) = (2 + 6, 5 + 1) = (8, 6) = 2

3 - 4 = (4,1) - (5,1) = (4 + 1, 1 + 5) = (5,6) = -1

Definition 5: (a,b) is less than (c,d) if and only if a + d is less than b + c

Example:

-3 is less than 4 since (2,5) is less than (5,1) since 2+1=3 is less than 5 + 5=10.

Lemma 1: Z is closed under addition.

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

(2) a + c is a natural number and b + d is a natural number (since addition is closed for natural numbers, see here)

QED

Lemma 2: Z is closed under multiplication

(1) (a,b) * (c,d) = (ac + bd, ad + bc)

(2) ac, bd, ad, bc are all natural numbers since N is closed under multiplication (see here)

(3) ac + bd, ad + bc are natural numbers since N is closed under addition (see here)

QED

Lemma 3: Z is closed under subtraction

(1) (a,b) - (c,d) = (a + d, b + c)

(2) a + d, b + c are both natural numbers since N is closed under addition (see here)

QED

Lemma 4: Z has 0 as the identity element for addition

n + 0 = (n+1,1) + (1,1) = (n + 2,2) = n

QED

Lemma 5: Z has 1 as the identity element for multiplication

n * 1 = (n + 1, 1)*(2,1) = (2*(n+1) + 1*1, (n+1)*1 + 2) = (2n + 3,n + 3) = n

QED

Lemma 6: Each element of Z has an inverse element for addition that is determined by reversing the pairs (b,a) is the inverse element for (a,b)

n + (-n) = (n+1,1) + (1,n+1) = (n + 1 + 1, 1 + n + 1) = (n+2,n+2) = 0

QED

Lemma 7: Z is commutative on addition: a + b = b + a

a + b = (a+1,1) + (b+1,1) = (a + b + 2,2)
b + a = (b+1,1) + (a+1,1) = (b + a + 2,2) = (a + b + 2,2) [Since natural numbers are commutative on addition, see here]

QED

Lemma 8: Z is commutative on multiplication: a*b=b*a

a*b = (a+1,1)*(b+1,1) = ((a+1)(b+1) + 1*1,(a+1)*1 + 1*(b+1)) =
= (ab + a + b + 1 + 1, a + 1 + b + 1) =
= (ab + a + b + 2,a + b + 2) = ab

b*a = (b+1,1)*(a+1,1) = ((b+1)(a+1) + 1*1,(b+1)*1 + 1*(a+1)) =
= (ba + a + b + 2, a + b + 2) =
= (ab + a + b + 2, a + b + 2) = ab
[Since natural numbers are commutative on multiplication, see here]

QED

Lemma 9: Z is associative on addition

(a+b) + c = [(a+1,1) + (b+1,1)] + (c+1,1) = (a+b+2,2) + (c+1,1) = (a+b+c+3,3)=
= a+b+c


a + (b+c) = (a+1,1) + [(b+1,1) + (c+1,1)] = (a+1,1) + (b+c+2,2) = (a+b+c+3,3) =
= a+b+c


QED

Lemma 10: Z is associative on multiplication

(a*b)*c = [(a+1,1)*(b+1,1)]*(c+1,1)=
= [(a+1)(b+1) + (1)(1),(a+1)(1) + 1(b+1)](c+1,1) =

= (ab + a + b + 2, a+1+b+1)(c+1,1) = (ab+1,1)(c+1,1) =
= ((ab+1)(c+1) + (1)(1), (ab+1)(1)+(1)(c+1)) =
= (abc + ab + c + 1 + 1, ab + 1 + c + 1) =

= (abc + ab + c + 2, ab + c + 2) = abc

a*(b*c) = (a+1,1)[(b+1,1)(c+1,1)] =
= (a+1,1)((b+1)(c+1) + (1)(1),(b+1)(1)+(1)(c+1)) =

= (a+1,1)(bc + b + c + 1 + 1,b+1+c+1) =
= (a+1,1)(bc+b+c+2,b+c+2)=(a+1,1)(bc+1,1)=

= ((a+1)(bc+1) + (1)(1),(a+1)(1) + (1)(bc+1)) =
= (abc + a + bc + 1 + 1,a + 1 + bc + 1) =

= (abc + 1, 1) = abc

QED

Lemma 11: Z is distributive: (a+b)c =ac + bc.

(a+b)c = [(a+1,1)+(b+1,1)](c+1,1) = [(a + b + 2,1+1)](c+1,1) = (a+b+1,1)(c+1,1) =
= ((a+b+1)(c+1) + (1)(1),(a+b+1)(1) + (1)(c+1)) =
= (ac + bc +c + a + b + 1 + 1,a + b + 1 + c + 1) =
= (ac + bc + 1, 1) = ac + bc.

QED

References

4 comments:

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Kimpapop Mat said...

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Kimpapop Mat said...
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