View source for Chapter 3: Problem 8 (6th ed.) - Rhea

[[Category:MA453Spring2009Walther]]

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. --[[User:Podarcze|Podarcze]] 13:05, 4 February 2009 (UTC)


[[Category:MA453Spring2009Walther]] 

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