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After constant long division we get to the base equation where there is still a remainder of 1. | After constant long division we get to the base equation where there is still a remainder of 1. | ||
Therefore <math> 5*n + 3 </math> and <math> 7*n + 4 </math> are relatively prime. | Therefore <math> 5*n + 3 </math> and <math> 7*n + 4 </math> are relatively prime. | ||
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| + | == Even easier would be to use the Euclidean Algorithm == | ||
| + | <math>gcd(5*n + 3, 7*n + 4) = gcd(5*n + 3, 2*n + 1)</math> | ||
| + | <math> = gcd(3*n + 2, 2*n + 1)</math> | ||
| + | <math> = gcd(n + 1, 2*n + 1)</math> | ||
| + | <math> = gcd(n + 1, n)</math> | ||
| + | <math> = gcd(1, n)</math> | ||
| + | <math> = 1</math> | ||
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| + | Since the GCD of the two expressions is one for all n, they are relatively prime. | ||
Revision as of 20:43, 6 September 2008
Show that $ 5*n + 3 $ and $ 7*n + 4 $ are relatively prime.
$ 7*n + 4 = 5*n + 3 + 2*n + 1 $
$ 5*n + 3 = 2*(2*n + 1) + n + 1 $
$ 2*n + 1 = 1*(n + 1) + n $
$ n + 1 = 1*n + 1 $
After constant long division we get to the base equation where there is still a remainder of 1.
Therefore $ 5*n + 3 $ and $ 7*n + 4 $ are relatively prime.
Even easier would be to use the Euclidean Algorithm
$ gcd(5*n + 3, 7*n + 4) = gcd(5*n + 3, 2*n + 1) $ $ = gcd(3*n + 2, 2*n + 1) $ $ = gcd(n + 1, 2*n + 1) $ $ = gcd(n + 1, n) $ $ = gcd(1, n) $ $ = 1 $
Since the GCD of the two expressions is one for all n, they are relatively prime.
