(related to Proposition: Existence of Rational One (Neutral Element of Multiplication of Rational Numbers))

By definition of rational numbers, the rational number \(x\in\mathbb Q\) can be represented by a pair of integers \(a,b\in\mathbb Z\) with \(b\neq 0\). \[x:=\frac ab.\]

Since \(1\in \mathbb Z\), i.e. the integer one exists, the (rational) one \(1\in\mathbb Q\) also exists, because it can be represented by any pair of the integers \(h\in\mathbb Z\), \(h\neq 0\), in particular by \(h=1\in\mathbb Z\): \[1_{\in\mathbb Q}:=\frac hh=\frac {1_{\in\mathbb Z}}{1_{\in\mathbb Z}}\]

We will prove that \(1\) is neutral with respect to the multiplication of rational numbers by virtue of the following mathematical definitions and concepts: * definition of multiplication of rational numbers "\( \cdot \)", and * \(1\in\mathbb Z\) is neutral with respect to the multiplication of integers:

\[\begin{array}{rcll} x \cdot 1&=&\frac ab\cdot\frac 11&\text{by definition of rational numbers}\\ &=&\frac {a1}{b1}&\text{by definition of multiplication of rational numbers}\\ &=&\frac ab&\text{because }1\text{ is neutral with respect to multiplication of integers}\\ &=&x&\text{by definition of rational numbers} \end{array} \]

In other words, the rational number \(1\) is neutral with respect to the multiplication of rational numbers.

It remains to be shown that also the equation \(1\cdot x=x\) holds for all \(x\in\mathbb Q\). It follows immediately from the commutativity of multiplying rational numbers.

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