# Proof

According to the definition of negative and positive integers, we can represent an integer $$x$$ by an ordered pair of a natural number $$a$$ and $$0$$ in three ways: $x=\begin{cases}[a,0],~a\neq 0&\Longleftrightarrow \text{ if }x\text{ is a positive integer}\\ [0,0],&\Longleftrightarrow \text{ if }x\text{ equals 0}\\ [0,a],~a\neq 0&\Longleftrightarrow \text{ if }x\text{ is a negative integer}\\ \end{cases}$

According to the definition of multiplying integers, we have for two integers $$x=[a,b]$$ and $$y=[c,d]$$:

$\begin{array}{ccl} x\cdot y:=[a,b] \cdot [c,d] &:=& [a\cdot c + b\cdot d,~ a\cdot d + c\cdot b]=[ac + bd,~ ad + bc], \end{array}$

Because the multiplication of integers is commutative, it is sufficient to prove the following four cases:

### $$(1)$$ A positive integer times a positive integer gives a positive integer

Let $$x=[a,0]$$ and $$y=[c,0]$$. It follows

$[a,0] \cdot [c,0] =[a\cdot c + 0\cdot 0,~ a\cdot 0 + c\cdot 0]=[ac,0],$ which is a positive integer.

### $$(2)$$ A negative integer times a positive integer gives a negative integer.

Let $$x=[0,b]$$ and $$y=[c,0]$$. It follows

$[0,b] \cdot [c,0] =[0\cdot c + b\cdot 0,~ 0\cdot 0 + c\cdot b]=[0,bc],$ which is a negative integer.

### $$(3)$$ A negative integer times a negative integer gives a positive integer.

Let $$x=[0,b]$$ and $$y=[0,d]$$. It follows

$[a,0] \cdot [c,0] =[0\cdot 0 + b\cdot d,~ 0\cdot d + 0\cdot b]=[bd,0],$

which is a positive integer.

### $$(4)$$ Zero times any integer gives zero.

Let $$x=[0,0]$$ and $$y=[c,d]$$. It follows

$[0,0] \cdot [c,d] =[0\cdot c + 0\cdot d,~ 0\cdot d + c\cdot 0]=[0,0],$

which is zero.

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### References

#### Bibliography

1. Kramer Jürg, von Pippich, Anna-Maria: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013