Rational Multiplication is Closed
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Theorem
The operation of multiplication on the set of rational numbers $\Q$ is well-defined and closed:
- $\forall x, y \in \Q: x \times y \in \Q$
Proof 1
Follows directly from the definition of rational numbers as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
So $\struct {\Q, +, \times}$ is a field, and therefore a fortiori $\times$ is well-defined and closed on $\Q$.
$\blacksquare$
Proof 2
From the definition of rational numbers, there exists four integers $p$, $q$, $r$, $s$, where:
- $q \ne 0$
- $s \ne 0$
- $\dfrac p q = x$
- $\dfrac r s = y$
We have that:
- $p \times r \in \Z$
- $q \times s \in \Z$
Since $q \ne 0$ and $s \ne 0$, we have that:
- $q \times s \ne 0$
Therefore, by the definition of rational numbers:
- $x \times y = \dfrac {p \times r} {q \times s} \in \Q$
Hence the result.
$\blacksquare$
Sources
- 1964: Walter Rudin: Principles of Mathematical Analysis (2nd ed.) ... (previous) ... (next): Chapter $1$: The Real and Complex Number Systems: Introduction
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $1$: Integral Domains: $\S 2$. Operations: Example $1$
- 1973: G. Stephenson: Mathematical Methods for Science Students (2nd ed.) ... (previous) ... (next): Chapter $1$: Real Numbers and Functions of a Real Variable: $1.1$ Real Numbers
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: The Real Number System: $3$