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--[[User:Mmohamad|Mmohamad]] 22:07, 14 September 2008 (UTC) | --[[User:Mmohamad|Mmohamad]] 22:07, 14 September 2008 (UTC) | ||
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| + | Yes, that's what the four means. As far as I know, we have to do the calculation until we find a cycle, but as was shown in class, there are shortcuts sometimes. | ||
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| + | --[[User:Dfreidin|Dfreidin]] 22:06, 14 September 2008 (UTC) | ||
Latest revision as of 17:06, 14 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)
One question, how can we know how many of them before the cycle starts again. Do we have to do the calculation until we can figure out it cycles again, or is there any other way that's easier to figure this thing out? And for the question above, is the ord(7 + 15$ : \mathbb{Z} $)=4 means there like 4 numbers before the cycle starts again? I'm a bit confused with this. Just want to make sure about it.
--Mmohamad 22:07, 14 September 2008 (UTC)
Yes, that's what the four means. As far as I know, we have to do the calculation until we find a cycle, but as was shown in class, there are shortcuts sometimes.
--Dfreidin 22:06, 14 September 2008 (UTC)
