November 16th, 2006, 07:12 AM  #1 
Newbie Joined: Nov 2006 Posts: 20 Thanks: 0  GCD question
Find the GCD of 24 and 49 in the integers of Q[sqrt(3)], assuming that the GCD is defined. (Note: you need not decompose 24 or 49 into primes in Q[sqrt(3)]. Please help me! 
November 16th, 2006, 07:35 AM  #2 
Newbie Joined: Nov 2006 Posts: 1 Thanks: 0 
Hint? If u divides x and y then u divides mx+ny for all ordinary integers m and n.

November 16th, 2006, 10:29 AM  #3 
Newbie Joined: Nov 2006 Posts: 20 Thanks: 0 
Hi Larry, Could you please show me how? I got stuck with this question for long. It never state in my book about doing the GCD of a quadratic integer, but it turns out that I have to do this question. Honestly, I don't really get it even with your hint. Please kindly show me the way. Thank you very much. 
November 16th, 2006, 11:47 AM  #4 
Member Joined: Nov 2006 From: Norway Posts: 33 Thanks: 0  GCD
A common divisor of 24 and 49 will also divide 49  24 and 49  2*24 
November 16th, 2006, 12:27 PM  #5 
Newbie Joined: Nov 2006 Posts: 20 Thanks: 0 
The way you showed me is by Euclidean Algorithm. The gcd end up to be 1. But the question is asking to find the GCD OF 24 AND 49 IN THE INTEGERS OF Q[sqrt(3)]. Not only to find the GCD of 24 and 49..... 
November 16th, 2006, 09:59 PM  #6 
Member Joined: Nov 2006 From: Norway Posts: 33 Thanks: 0 
OK, let d be a common divisor of 24 and 49 Then, for some e,f 24 = de, 49 = df Then we get df  2de = 1 giving that d(f2e) = 1 Thus a common divisor of 24 and 49 must be a divisor of 1 A divisor of 1 is called a unit, and when we search for a GCD, two common divisors are considered equal if their quotient is a unit. So the GCD is 1. Two things should be noted here: 1) Although there might not be a euclidean algorithm in a ring, it is always allowed to try, and if the algorithm ends after finitely many steps, you may use the result. 2) There ARE nontrivial units in Z(sqrt(3)), all the numbers + (2+sqrt(3))^n, n any (positive or negative) integer. Einar 

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