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Show that <math> 5*n + 3 </math> and <math> 7*n + 4 </math> are relatively prime. | Show that <math> 5*n + 3 </math> and <math> 7*n + 4 </math> are relatively prime. | ||
| − | <math> | + | |
| − | 7*n + 4 = 5*n + 3 + 2*n + 1< | + | <math>7*n + 4 = 5*n + 3 + 2*n + 1</math> |
| − | 5*n + 3 = 2*(2*n + 1) + n + 1 | + | |
| − | 2*n + 1 = 1*(n + 1) + n | + | <math>5*n + 3 = 2*(2*n + 1) + n + 1</math> |
| − | n + 1 = 1*n + 1 | + | |
| − | </math> | + | <math>2*n + 1 = 1*(n + 1) + n</math> |
| + | |||
| + | <math>n + 1 = 1*n + 1</math> | ||
| + | |||
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. | ||
Revision as of 20:14, 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.
