(related to Proposition: Existence of Integer One (Neutral Element of Multiplication of Integers))
By the definition of integers, an integer number \(x\in\mathbb Z\) can be represented by a pair of natural numbers \(a,b\in\mathbb N\): \[x:=[a,b].\]
Since \(1_{\in\mathbb N}\), i.e. the natural one exists, and since \(0_{\in\mathbb N}\), i.e. the natural zero exists, it is also true that the integer one \(1_{\in\mathbb Z}\) exists, because it can be represented by a pair of the natural numbers \[1=1_{\in\mathbb Z}:=[h+1_{n\in\mathbb N},h],\quad h\in\mathbb N,\] in particular (for \(h=0\in\mathbb N\)): \[1=1_{\in\mathbb Z}=[1_{\in\mathbb N},0_{\in\mathbb N}].\]
We will prove that \(1\) is neutral with respect to the multiplication of integers by virtue of the following mathematical definitions and concepts: * definition of multiplication of integers "\( \cdot \)", * definition of multiplication of natural numbers, * commutativity law for adding natural numbers, * the natural number \(0_{\in\mathbb N}\) is neutral with respect to the addition of natural numbers, and * the natural number \(1_{n\in\mathbb N}\) is neutral with respect to the multiplication of natural numbers.
\[\begin{array}{rcll} x \cdot 1&=&[a,b]\cdot[1_{n\in\mathbb N},0_{\in\mathbb N}]&\text{by definition of integers}\\ &=&[a\cdot 1_{n\in\mathbb N} + b\cdot 0_{\in\mathbb N},~ a\cdot 0_{\in\mathbb N} + b\cdot 1_{n\in\mathbb N}]&\text{by definition of multiplication of integers}\\ &=&[a\cdot 1_{n\in\mathbb N} + 0_{\in\mathbb N},~ 0_{\in\mathbb N} + b\cdot 1_{n\in\mathbb N}]&\text{by definition of multiplication of natural numbers (in particular the multiplication by }0_{\in\mathbb N}\text{)}\\ &=&[a\cdot 1_{n\in\mathbb N} + 0_{\in\mathbb N},~ b\cdot 1_{n\in\mathbb N}+ 0_{\in\mathbb N}]&\text{by commutativity of addition of natural numbers}\\ &=&[a\cdot 1_{n\in\mathbb N},~ b\cdot 1_{n\in\mathbb N}]&0_{\in\mathbb N}\text{ is neutral with respect to the addition of natural numbers}\\ &=&[a,b]&1_{n\in\mathbb N}\text{ is neutral with respect to the multiplication of natural numbers}\\ &=&x&\text{by definition of integers} \end{array} \] In other words, the integer \(1:=[1,0]\) is neutral with respect to the multiplication of integers.
It remains to be shown that also the equation \(1\cdot x=x\) holds for all \(x\in\mathbb Z\). It follows immediately from the commutativity of multiplying rational numbers.
