Quote:

Proof:

1) LCM(a,b)·GCD(a,b)=ab

2) LCM(a,b,c)=LCM(LCM(a,b),c) or see below

3) Let H=gcd(a,b), then

4) LCM(a,b)=ab/H

4) LCM(a,b,c)=LCM(ab/H,c)

5) LCM(a,b,c)=[abc/H]/GCD(ab/H,c)

6) LCM(a,b,c)·H·GCD(ab/H,c)=abc,

EXAMPLE

LCM(4,6,15)x2xGCD(2x6,15)=4x6x15

LCM(4,6,15)x2x3=4x6x15

LCM(4,6,15)=60

You get same answer for any order of a,b,c

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LCM(a,b,c)=LCM(LCM(a,b),c), by prime factorization which is easy to see for specific cases but awkard to type in the general case, or:

7) Let Q=LCM(LCM(a,b),c)

8) a|LCM(a,b), LCM(a,b)|Q, → a|Q

9) b|LCM(a,b), LCM(a,b)|Q, → b|Q

10) c|Q

So Q is a common multiple of (a,b,c). Now you have to show Q divides every common multiple of (a,b,c). Then Q =LCM(a,b,c)

CM(a,b,c) =ra=sb=tc

ra=sb =CM(a,b)

CM(a,b,c)=CM(a,b)=tc

Q|CM(a,b), Q|c → Q|CM(a,b,c)

The beauty of the proof is its easy to type.