# Thread: GCD and least common multiple Problem

1. ## GCD and least common multiple Problem

Please give me help to find out the answer for following question

With the usual notation show that
i. For a > 0, and b > 0,[a,b](a,b) = ab
ii. [a,b, c](ab,bc, ca) = abc if a,b, c are positive integers.

2. (i) let (a,b)=d,then there exist positive integers m and n such that a=md,b=nd,and (m,n)=1.
Thus [a,b]=mnd; ab=mdnd; => [a,b](a,b)=ab. Q.E.D
(ii)By (i)(proved), we have:
$[a,b,c]=[\frac{ab}{(a,b)},c]=\frac{\frac{abc}{(a,b)}}{(\frac{ab}{(a,b)},c)}=\f rac{abc}{(a,b)(\frac{ab}{(a,b)},c)}=\frac{abc}{(ab ,bc,ac)}$
Thus [a,b,c](ab,bc,ac)=abc; Q.E.D

3. Originally Posted by dhammikai
Please give me help to find out the answer for following question

With the usual notation show that
i. For a > 0, and b > 0,[a,b](a,b) = ab
ii. [a,b, c](ab,bc, ca) = abc if a,b, c are positive integers.

These follow directly from unique prime factorization theorem. Do you know that? Can you express (a,b) and [a,b] in terms of prime factors of a,b?

4. You wrote me:
"....These follow directly from unique prime factorization theorem. Do you know that? Can you express (a,b) and [a,b] in terms of prime factors of a,b?"

5. Originally Posted by dhammikai
You wrote me:
"....These follow directly from unique prime factorization theorem. Do you know that? Can you express (a,b) and [a,b] in terms of prime factors of a,b?"

Fundamental theorem of arithmetic - Wikipedia, the free encyclopedia

6. Originally Posted by Shanks
(i) let (a,b)=d,then there exist positive integers m and n such that a=md,b=nd,and (m,n)=1.
Thus [a,b]=mnd; ab=mdnd; => [a,b](a,b)=ab. Q.E.D
(ii)By (i)(proved), we have:
$[a,b,c]=[\frac{ab}{(a,b)},c]=\frac{\frac{abc}{(a,b)}}{(\frac{ab}{(a,b)},c)}=\f rac{abc}{(a,b)(\frac{ab}{(a,b)},c)}=\frac{abc}{(ab ,bc,ac)}$
Thus [a,b,c](ab,bc,ac)=abc; Q.E.D
@dhammikai - Please note this is a nice proof however Shanks has used some intermediate theorems like

1. [a,b,c]=[[a,b],c]
2. (a,b,c)=((a,b),c)
3. (ka,kb)=k(a,b) etc

This I feel is much better than the proof I was proposing