(related to Proposition: Multiplication of Rational Cauchy Sequences Is Cancellative)
Because the multiplication of rational Cauchy sequences is commutative, it is without loss of generality sufficient to show the right cancellation property, i.e. \[\exists N\in\mathbb N:~ (x_n)_{n > N} \cdot (z_n)_{n > N}=(y_n)_{n > N} \cdot (z_n)_{n > N}\Longleftrightarrow (x_n)_{n > N}=(y_n)_{n > N},\] for all rational Cauchy sequences \((x_n)_{n\in\mathbb N}, (y_n)_{n\in\mathbb N}, (z_n)_{n\in\mathbb N}\) such that \((z_n)_{n\in\mathbb N}\) is not convergent to \(0\).
Because by hypothesis \((z_n)_{n\in\mathbb N}\) is a rational Cauchy sequence not convergent to \(0\), there exist an index \(N\in\mathbb N\) such that \(|z_n|>0\) for all \(n > N\). For all sequence members with indices \(n > N\), it follows from the cancellation property of multiplication of rational numbers and from the definition of multiplying rational Cauchy sequences that
\[\begin{array}{rcl} (x_n)_{n > N} \cdot (z_n)_{n > N}=(y_n)_{n\in\mathbb N} \cdot (z_n)_{n > N}&\Longleftrightarrow&(x_n\cdot z_n)_{n > N}=(y_n\cdot z_n)_{n > N}\\ &\Longleftrightarrow&(x_n)_{n > N}=(y_n)_{n > N}. \end{array}\]
