Integer Addition is Closed

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Theorem

The set of integers is closed under addition:

$\forall a, b \in \Z: a + b \in \Z$


Proof

Let us define $\left[\!\left[{\left({a, b}\right)}\right]\!\right]_\boxminus$ as in the formal definition of integers.

That is, $\left[\!\left[{\left({a, b}\right)}\right]\!\right]_\boxminus$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxminus$.

$\boxminus$ is the congruence relation defined on $\N \times \N$ by $\left({x_1, y_1}\right) \boxminus \left({x_2, y_2}\right) \iff x_1 + y_2 = x_2 + y_1$.


In order to streamline the notation, we will use $\left[\!\left[{a, b}\right]\!\right]$ to mean $\left[\!\left[{\left({a, b}\right)}\right]\!\right]_\boxminus$, as suggested.


Integer addition is defined as:

$\forall a, b, c, d \in \N: \left[\!\left[{a, b}\right]\!\right] + \left[\!\left[{c, d}\right]\!\right] = \left[\!\left[{a + c, b + d}\right]\!\right]$


We have that:

$\forall a, b, c, d \in \N: \left[\!\left[{a, b}\right]\!\right] \in \Z, \left[\!\left[{c, d}\right]\!\right] \in \Z$

Also:

$\forall a, b, c, d \in \N: \left[\!\left[{a, b}\right]\!\right] + \left[\!\left[{c, d}\right]\!\right] = \left[\!\left[{a + c, b + d}\right]\!\right]$

But:

$a + c \in \N, b + d \in \N$

So:

$\forall a, b, c, d \in \N: \left[\!\left[{a + c, b + d}\right]\!\right] \in \Z$


Therefore integer addition is closed.

$\blacksquare$


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