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7 was the number that we started with, so it was kind of arbitrary, and g=7. 4, 13, and 1 are <math>7^{2}, 7^{3}, 7^{4}</math> respectively, all taken mod 15. We know it's a cycle because obviously, <math>7^{0}=1</math>, so if we take <math>(7^{4}*7) + n{Z}</math> (the next value if we were to continue), this will just be equal to 7 again, and so if we were to continue in this manner, it would continue in a cycle. | 7 was the number that we started with, so it was kind of arbitrary, and g=7. 4, 13, and 1 are <math>7^{2}, 7^{3}, 7^{4}</math> respectively, all taken mod 15. We know it's a cycle because obviously, <math>7^{0}=1</math>, so if we take <math>(7^{4}*7) + n{Z}</math> (the next value if we were to continue), this will just be equal to 7 again, and so if we were to continue in this manner, it would continue in a cycle. | ||
| − | [[User:Dfreidin|Dfreidin]] 20:39, 10 September 2008 (UTC) | + | |
| + | --[[User:Dfreidin|Dfreidin]] 20:39, 10 September 2008 (UTC) | ||
Revision as of 15:39, 10 September 2008
I am confused on this example that was done in class.
ord(7 + 15$ : \mathbb{Z} $)=4
g$ \in $ 7+15$ : \mathbb{Z} $
$ g^{0} =1, g^{1} =7, g^{2}= 4, g^{3}=13, g^{4}=1 $
How did we get 7,4,13, and 1 and how do we know that these are cycles?
--Akcooper 11:23, 10 September 2008 (UTC)
7 was the number that we started with, so it was kind of arbitrary, and g=7. 4, 13, and 1 are $ 7^{2}, 7^{3}, 7^{4} $ respectively, all taken mod 15. We know it's a cycle because obviously, $ 7^{0}=1 $, so if we take $ (7^{4}*7) + n{Z} $ (the next value if we were to continue), this will just be equal to 7 again, and so if we were to continue in this manner, it would continue in a cycle.
--Dfreidin 20:39, 10 September 2008 (UTC)
