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how to prove or disprove the following :-

$\gcd (kn,km) = k\gcd(n,m).$

$\operatorname{lcm}(n,m)\gcd(n,m)=mn.$

$\operatorname{lcm}(kn,km)=k\operatorname{lcm}(n,m).$

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    @Dylan I have it on good authority that his definition of gcd is taking the minimum exponent of every prime ;).2012-01-16

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HINT $\rm\ \ d\ |\ (kn,km)/k \iff dk\ |\ (kn,km) \iff dk\ |\ kn,km \iff d\ |\ n,m \iff d\ |\ (n,m)$

Dually $\rm\ \ \ \ [kn,km]/k\ |\ d \iff [kn,km]\ |\ kd \iff kn,km\ |\ kd \iff n,m\ |\ d \iff [n,m]\ |\ d $

THEOREM $\rm\;\; (a,b)\ =\ ab/[a,b] \;\;$ if $\;\rm\ [a,b] \;$ exists.

Proof $\rm\ \ d\ |\ a,b \;\iff\; a,b\ |\ ab/d \;\iff\; [a,b]\ |\ ab/d \;\iff\; d\ |\ ab/[a,b]\ \ $ QED

These proofs use the universal definitions of GCD and LCM.