text: prove lcm(m,n) = m*n/gcd(m,n)?   its equivalent to proving m*n = (lcm(m,n) ) * ( gcd(m,n) )  now, i'll have to use basic number theory. the idea is to show that the number of times a prime factor of m*n occurs on the lhs is equal to the number of times a prime factor of (lcm(m,n) ) * ( gcd(m,n) ) occurs on the rhs. note : lhs = left hand side        rhs = right hand side  for any prime factor p of m*n, call it p. now say p occurs to the power of h in m and power of k in n. that is to say m "contains" p^h while n contains p^k. note that h,k are nonnegative integers.   now, in lcm(m,n), we would choose p^ [ min(h,k) ] while in gcd(m,n) we would choose  p^ [ max(h,k) ]    note that min (h,k) means minimum of h and k, similarly  max (h,k)means maximum of h and k . for example, if h=2 and k=3, then min (2,3) = 2. while max(2,3) = 3. if say h = k, then obviously in both lcm and gcd of m*n we choose p^h to be in its product. this is true for every prime factor of m*n.  so, consider the product (lcm(m,n) ) * ( gcd(m,n) ). for each prime factor p of m*n, we have shown above that it occurs as { p^ [ min(h,k) ] } * { p^ [ max(h,k) ]   }, that is our rhs product contains    p^ [ min(h,k) +  max(h,k) ] = p^(h+k) now, in our lhs, p occurs as p^h * p^k = p^(h+k) too!!! thus, we have shown for each prime factor, p of m*n, it occurs as the same product in both the lhs and rhs. and thus we have lhs = rhs. done.  one question may circle in your mind, that is how i know that [ min(h,k) +  max(h,k) ] = (h+k) well, this is a fact.  further, if you want to show it, consider 3 cases.  case 1. ( h = k) here min(h,k) = min(h,h) = h and  max(h,k) = max(h,h) = h so, [ min(h,k) +  max(h,k) ] = h + h = 2h. and (h+k)= h + h = 2h.  case 2. ( h  k ) here min(h,k) =  k, because k is smaller. and  max(h,k) = h, because h is bigger. so, [ min(h,k) +  max(h,k) ] = k + h