(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:
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?"
Please let me know about the unique prime factorization theorem please..
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?"
Please let me know about the unique prime factorization theorem please..
(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:
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