# 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

Thank you to the contributors under CC BY-SA 4.0!

Github:

### References

#### Bibliography

1. Kramer Jürg, von Pippich, Anna-Maria: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}