(related to Proposition: Existence of Inverse Rational Numbers With Respect to Addition)
By definition, any rational number \(x\) can be represented by two integers \(a,b\in\mathbb Z\) with \(x=\frac ab\) and \(b\neq 0_{\in\mathbb Z}\), where the symbol \(0_{\in\mathbb Z}\) denotes the integer zero. By definition of multiplying integers, \(ab\) denotes the integer resulting by multiplying the integer \(a\) by the integer \(b\). Note that there exists an inverse integer \(-ab\), such that \(ab + (-ab)=0_{\in\mathbb Z}\). By definition of adding rational numbers, we get
\[\frac ab+\frac{-a}b=\frac{ab+(-ab)}{b^2}=\frac {0_{\in\mathbb Z}}{b^2}.\]
Since the rational zero \(0_{\in\mathbb Q}\) can be represented by the (integer) zero \(0_{\in\mathbb Z}\) and an arbitrary integer \(d\neq 0\): \[0=0_{\in\mathbb Q}:=\frac {0_{\in\mathbb Z}}{d},\quad d\in\mathbb Z\setminus\{0\},\] we can take \(d:=b^2\), which completes the proof.
