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Mathematics 14 Online
OpenStudy (zzr0ck3r):

number theory: given theorem:if (a,b)=1 ^ (a,c) = 1 -> (a,bc) = 1 for a,b,c in Z nts: (a,b) = 1 -> (a^2,b^2) = 1 this is how I went about it, but other people in my class did it in a much longer fashion and are sceptical of my proof pf: (a,b) = 1 ^ (a,b) = 1 -> (a*a,b) = (a^2,b) = 1 then (a^2,b) = 1 ^ (a^2,b) = 1 -> (a^2,b^2) = 1Does this look ok?

OpenStudy (anonymous):

This is correct as long as you give a valid proof for why:\[\gcd (a,b)=1\Longrightarrow \gcd (a^2,b)=1\]and:\[\gcd (a^2,b)=1\Longrightarrow \gcd (a^2,b^2)=1\]

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