The multiplication of rational Cauchy sequences is cancellative, i.e. for all rational Cauchy sequences \((x_n)_{n\in\mathbb N}\), \((y_n)_{n\in\mathbb N}\) and \((z_n)_{n\in\mathbb N}\) such that \((z_n)_{n\in\mathbb N}\) is not convergent to \(0\), there exists \(N\in\mathbb N\) such that for all sequence members with indices \(n > N\), following laws (both) are fulfilled:
Left cancellation property: If the equation \((z_n)_{n > N}\cdot (x_n)_{n > N}=(z_n)_{n > N} \cdot (y_n)_{n > N}\) holds, then \((x_n)_{n > N}=(y_n)_{n > N}\).
Right cancellation property: If the equation \((x_n)_{n > N}\cdot (z_n)_{n > N}=(x_n)_{n > N} \cdot (z_n)_{n > N}\) holds, then \((x_n)_{n > N}=(y_n)_{n > N}\).
Conversely, for a rational Cauchy sequence \((z_n)_{n\in\mathbb N}\), which is not convergent to \(0\), there exists a natural number \(N\) such that the equation \((x_n)_{n > N}=(y_n)_{n > N}\) implies * \((z_n)_{n > N}\cdot (x_n)_{n > N}=(z_n)_{n > N} \cdot (y_n)_{n > N}\) and * \((x_n)_{n > N}\cdot (z_n)_{n > N}=(x_n)_{n > N} \cdot (z_n)_{n > N}\).
Proofs: 1
