Integer Multiplication is Commutative

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Theorem

The operation of multiplication on the set of integers $\Z$ is commutative:

$\forall x, y \in \Z: x \times y = y \times x$


Proof

From the formal definition of integers, $\left[\!\left[{a, b}\right]\!\right]$ is an equivalence class of ordered pairs of natural numbers.


Let $x = \left[\!\left[{a, b}\right]\!\right]$ and $y = \left[\!\left[{c, d}\right]\!\right]$ for some $x, y \in \Z$.


Then:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle x \times y\) \(=\) \(\displaystyle \left[\!\left[{a, b}\right]\!\right] \times \left[\!\left[{c, d}\right]\!\right]\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \left[\!\left[{ac + bd, ad + bc}\right]\!\right]\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          by definition of integer multiplication          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \left[\!\left[{ca + db, da + cb}\right]\!\right]\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          Natural Number Multiplication is Commutative          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \left[\!\left[{ca + db, cb + da}\right]\!\right]\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          Natural Number Addition is Commutative          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \left[\!\left[{c, d}\right]\!\right] \times \left[\!\left[{a, b}\right]\!\right]\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          by definition of integer multiplication          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle y \times x\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    

$\blacksquare$


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