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I did it the same way, but once I got to 3^5=5 and 5^5=3 they were equal.  It that right to assume. --[[User:Podarcze|Podarcze]] 13:05, 4 February 2009 (UTC)
 
I did it the same way, but once I got to 3^5=5 and 5^5=3 they were equal.  It that right to assume. --[[User:Podarcze|Podarcze]] 13:05, 4 February 2009 (UTC)
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[[Category:MA453Spring2009Walther]]
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I did it the same way as well.  What did you assume from taking 3^5 and 5^5?

Revision as of 07:17, 5 February 2009


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

I did it the same way, but once I got to 3^5=5 and 5^5=3 they were equal. It that right to assume. --Podarcze 13:05, 4 February 2009 (UTC)

I did it the same way as well. What did you assume from taking 3^5 and 5^5?

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva