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=[[HW1_MA453Fall2008walther|HW1]], Chapter 0, Problem 14, [[MA453]], Fall 2008, [[user:walther|Prof. Walther]]=
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==Problem Statement==
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Show that <math> 5*n + 3 </math> and <math> 7*n + 4 </math> are relatively prime.
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==Discussion==
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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.
  
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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.
 
 
  
 
== Even easier would be to use the Euclidean Algorithm ==
 
== Even easier would be to use the Euclidean Algorithm ==
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Since the GCD of the two expressions is one for all n, they are relatively prime.
 
Since the GCD of the two expressions is one for all n, they are relatively prime.
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[[HW1_MA453Fall2008walther|Back to HW1]]
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[[Main_Page_MA453Fall2008walther|Back to MA453 Fall 2008 Prof. Walther]]

Latest revision as of 15:49, 22 October 2010

HW1, Chapter 0, Problem 14, MA453, Fall 2008, Prof. Walther

Problem Statement

Show that $ 5*n + 3 $ and $ 7*n + 4 $ are relatively prime.


Discussion

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.


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