Euclidean Algorithm/Examples/9n+8 and 6n+5
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Examples of Use of Euclidean Algorithm
The GCD of $9 n + 8$ and $6 n + 5$ is found to be:
- $\gcd \set {9 n + 8, 6 n + 5} = 1$
Hence:
- $2 \paren {9 n + 8} - 3 \paren {6 n + 5} = 1$
Proof
\(\text {(1)}: \quad\) | \(\ds 9 n + 8\) | \(=\) | \(\ds \paren {6 n + 5} + \paren {3 n + 3}\) | |||||||||||
\(\text {(2)}: \quad\) | \(\ds 6 n + 5\) | \(=\) | \(\ds \paren {3 n + 3} + \paren {3 n + 2}\) | |||||||||||
\(\text {(3)}: \quad\) | \(\ds 3 n + 3\) | \(=\) | \(\ds \paren {3 n + 2} + 1\) |
Thus:
- $\gcd \set {9 n + 8, 6 n + 5} = 1$
Then we have:
\(\ds 1\) | \(=\) | \(\ds \paren {3 n + 3} - \paren {3 n + 2}\) | from $(3)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {3 n + 3} - \paren {\paren {6 n + 5} - \paren {3 n + 3} }\) | from $(2)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 \paren {3 n + 3} - \paren {6 n + 5}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \paren {\paren {9 n + 8} - \paren {6 n + 5} } - \paren {6 n + 5}\) | from $(1)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 \paren {9 n + 8} - 3 \paren {6 n + 5}\) |
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $2$: Some Properties of $\Z$: Exercise $2.3$