Product of Coprime Factors
From ProofWiki
Theorem
Let $a, b, c \in \Z$ such that $a$ and $b$ are coprime.
Let both $a$ and $b$ be divisors of $c$.
Then $ab$ is also a divisor of $c$.
That is:
- $a \perp b \land a \backslash c \land b \backslash c \implies a b \backslash c$
Proof
We have:
- $a \backslash c \implies \exists r \in \Z: c = a r$
- $b \backslash c \implies \exists s \in \Z: c = b s$
So:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle a\) | \(\perp\) | \(\displaystyle b\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle \exists m, n \in \Z: m a + n b\) | \(=\) | \(\displaystyle 1\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Integer Combination of Coprime Integers | ||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle c m a + c n b\) | \(=\) | \(\displaystyle c\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle b s m a + a r n b\) | \(=\) | \(\displaystyle c\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle a b \left({s m + r n}\right)\) | \(=\) | \(\displaystyle c\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle a b\) | \(\backslash\) | \(\displaystyle c\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
$\blacksquare$
Sources
- Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): Exercise $2.6$