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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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It should probably be added that $m$, $n$, and $k$ are positive. If you already have and can use the Unique Factorization Theorem (aka the Fundamental Theorem of Arithmetic), then it comes down to comparing exponents of the primes. If this comes before Unique Factorization, more detailed work is involved. –  André Nicolas Jan 16 '12 at 0:21
    
What is your definition of $\operatorname{gcd}$? The first one is certainly true—can you try to prove it? –  Dylan Moreland Jan 16 '12 at 0:22
    
@Dylan I have it on good authority that his definition of gcd is taking the minimum exponent of every prime ;). –  aelguindy Jan 16 '12 at 9:04
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1 Answer

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.

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