Show that if g.c.d.(x,y) = g.c.d.(x,z) = 1 then g.c.d.(x,yz) = 1

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- February 17th 2009, 07:57 AMmaths900[SOLVED] gcd proof 1
Show that if g.c.d.(x,y) = g.c.d.(x,z) = 1 then g.c.d.(x,yz) = 1

- February 17th 2009, 10:02 AMGrandadGCD
- February 17th 2009, 10:05 AMmaths900
Thanks. Is there another way to proof this-without having a contradiction or or something?

- February 17th 2009, 10:23 AMGrandadGCD
- February 17th 2009, 10:25 AMmaths900
- February 17th 2009, 11:29 AMo_O
One important fact:

- If and and , then

Basically, any common divisor of two numbers will divide the gcd of those two numbers.

___________

Let .

Multiply both sides by :

Since and , then . (Why?)

But this means and which implies because of .

But . Can you conclude? - February 17th 2009, 01:23 PMGrandadDivisors and multiples
- February 17th 2009, 01:30 PMo_O
Whoops! Yes, that's what I meant.