# Proof

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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