Inclusion of the natural numbers in the integers

We're going to see the operations of integers

\mathbb{N}_0 includes in \mathbb{Z} associating to each a\in\mathbb{N}_0 the equivalence class of (a, 0)

The natural number 0 in \mathbb{Z} is (0, 0), which is the kind of equivalence given by \{(a, a) | a \in\mathbb{N}_0\}, which is the element of the sum in \mathbb{Z}

The natural 1 in \mathbb{Z} is (1, 0), and its kind of equivalence, which is the neutral element of the product in \mathbb{Z}

Unlike in \mathbb{N}_0, in \mathbb{Z}, given a number, you can always find another one that adds to the first give zero (the neutral element of the sum). If we have (a, b), just take (b, a) and it is fulfilled:

(a, b)+(b, a)=(a + b, a + b)\sim (0, 0)

This number (b, a), which is the opposite (a, b), and we denote it as -(a, b). Which allows us to define the subtraction:

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

Or equivalently:

(a, b) - (c, d)=(r + s) \Longleftrightarrow (a, b)=(r, s) + (c, d)

Operation sum

Given that (a, b) it is the way that we have in \mathbb{Z} to indicate the subtraction a-b and in the same way, (c, d) represents c-d, there's nothing more to think about (a-b)+(c-d)=(a+c)-(b+d) to give a proper definition of the sum:

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

Operation product

Similarly, to define the product we have to (a-b)\cdot(c-d)=(a\cdot c + b\cdot d)-(a\cdot d+b\cdot c), therefore the appropriate definition for the product:

(a,b)\cdot (c,d)=(a\cdot c+b\cdot d, a\cdot d+ b\cdot c)

Relationship of equivalence

For the definitions in \mathbb{N}_0\times\mathbb{N}_0 given above are valid in \mathbb{Z}=\mathbb{N}_0\times\mathbb{N}_0/\sim, we need to prove that they are compatible with the relationship of equivalence, that is, if we have (a_1,b_1)\sim(a_2, b_2)\text{ y }(c_1,d_1)\sim(c_2, d_2) it is fulfilled that:

\begin{cases} (a_1,b_1)+(c_1,d_1)\sim(a_2,b_2)+(c_2,d_2) \\ (a_1,a_1)\cdot(c_1,d_1)\sim(a_2,b_2)\cdot(c_2,d_2) \end{cases}

Order relationship

There is also to define the order relationship in \mathbb{Z}. To define when (a,b)\leq(c,d), let's think once again about (a,b) as in \displaystyle a-b and in (c,d) as c-d. Then just look at that a-b\leq c-d is equivalent to a+d\leq b+c to realize that the definition that we seek has to be (a,b)\leq (c,d)\Leftrightarrow a+d\leq b+c

As in the sum and the product, we must check the compatibility of that definition with the equivalence ratio, that is, if we have (a_1,b_1)\sim(a_2,b_2)\text{ y }(c_1,d_1)\sim(c_2,d_2), is fulfilled that:

(a_1,b_1)+(c_1,d_1)\sim(a_2,b_2)+(c_2,d_2)