# Proposition: Multiplication of Rational Cauchy Sequences Is Cancellative

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

Github: ### References

#### Bibliography

1. Kramer Jürg, von Pippich, Anna-Maria: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013