Show that $\mathbb{Z} + \mathbb{Z} =\mathbb{Z} $. We wil use defitinion.
Definition. $\mathbb{Z} +\mathbb{Z} =${$z_{1}+z_{2}$: $z_{1},z_{2}$ $\in \mathbb{Z}$}. So, how can I prove? Can you give hint me?
 A: 1) Let $z \in \mathbb{Z}$. Then $z = z + 0$. So $z \in \mathbb{Z} + \mathbb{Z}$.
This means that $\mathbb{Z} \subset \mathbb{Z} + \mathbb{Z}$
2) Let $z \in \mathbb{Z} + \mathbb{Z}$. Then there are $z_1 \in \mathbb{Z}$ and $z_2 \in \mathbb{Z}$ such that $z= z_1 + z_2$. But the sum of integers is integer. So $z \in \mathbb{Z}$. 
This means that $\mathbb{Z} + \mathbb{Z} \subset \mathbb{Z}$. 
From 1) and 2) you get that $\mathbb{Z} + \mathbb{Z} = \mathbb{Z}$
A: The main problem of the OP seems to be to see that the sum of two integers is an integers. While this may be called obvious, it perhaps isn't. It is never a bad idea to question notions that are - at first glance - obvious.
So we need to make ourselves clear: What is $\Bbb Z$? and what is $+$?
To answer this formally, takes one deeper into the bowels of mathematical foundation than one might eoriginally expect.
Set theory gives us the notion of ordinals (sets that are transitive and that are well-ordered by $\in$). For these we have the principles of transfinite induction and recursion.
We can use this to define addition of ordinals: If $a$ is an ordinal, we define $a+0=a$ (here we use the usual identification $0=\emptyset$). If $a$ is an ordinal ad $b$ a non-zero ordinal, we define $a+b$ as the least ordinal that is greater than all $a+c$ with $c<b$.
In particular, if $b$ is the successor ordinal of $b'$ then $a+b$ is the successor ordinal of $a+b'$.
Another consequence of the axioms of set theory is the existence of infinite ordinals, in particular of a smallest infinite ordinal $\omega$.
Then $\omega$ is closed under taking the successor and all non-zero elements of $\omega$ are successor ordinals.
Lemma. If $a,b\in \omega$ then $a+b\in\omega$.
Proof. Assume otherwise and let $(a,b)$ be a counterexample (i.e., $a,b\in\omega, a+b\notin\omega$) with minimal $b$. Then Certainly $b\ne 0$ as otherwise $a+b=a+0=a\in\omega$. Then $b$ is a successor ordinal, say of $b'$. By minimalitly of $b$, $a+b'\in\omega$, hence also the successor $a+b$ is $\in\omega$. $\square$
One further verifies that on $\omega$ (this restriction is important!) the operation $+$ is associative and commutative .
Remark: We may as well write $\Bbb N$ instead of $\omega$.
On $\omega\times \omega$ we can define the relation $(a,b)\sim (a',b')\iff a+b'=a'+b$.
As $+$ is associative and commutative, $\sim$ is an equivalence relation.
Definition. We let $\Bbb Z=\omega\times\omega/\sim$.
We check that $(a+c,b+d)\sim(a'+c',b'+d')$ if $(a,b)\sim(a',b')$ and $(c,d)\sim(c',d')$. Therefore the it addition is well-defined by $[(a,b)]+[(c,d)]:=[(a+c,b+d)]$. This is of course a map $\Bbb Z\times \Bbb Z\to \Bbb Z$ - by definition.
A: An integer is defined as an orderd pair (a,b) where a,b are natural numbers. Two integer (a,b), (c,d) are considered to be equal iff a+d = c+b. The sum of two integers is defined as (a,b)+(c,d) := (a+c, b+d) where a,b,c,d are all natural numbers.
Since sum of any two natural numbers is again a natural number (prove by induction: keeping second number fixed, induct on first number), so (a+c, b+d) is again a integer.
