Revision as of 19:01, 3 February 2009 by Kmbrumba (Talk | contribs)

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Problem Statement: Show that U(14) = <3> = <5>. Is U(14) = <11>?

Answer: The long way of doing it (versus the tricks taught on 2/3/09):

U(14) = {1,3,5,9,11,13}

Test using 3 as the generator: 3^1 = 3 3^2 = 9 3^3 = 27 = -1 = 13 3^4 = 13*3 = 39 = 11 3^5 = 11*3 = 33 = 5 3^6 = 5*3 = 15 = 1 (end of cycle and 3 generates all values in U(14), therefore is a generator)

Test using 5 as the generator: 5^1 = 5 5^2 = 25 = -3 = 11 5^3 = -15 = -1 = 13 5^4 = -1*5 = -5 = 9 5^5 = -5*5 = -25 = 3 5^6 = 3*5 = 15 = 1 (end of cycle and 5 generates all values in U(14), therefore is a generator)

To see if U(14) = <11> test the powers of 11 to see if they generate all values in U(14): 11^1 = 11 11^2 = -3*-3 = 9 11^3 = 9*-3 = -27 = 1 (because <11> did not generate all values of U(14), <11> is not a generator)

-K. Brumbaugh, 23:01, 3 February 2009

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