(Abstract Algebra 1) Least Common Multiple

[@dragonslayer98767]

This channel is literally the only reason I'm passing Number Theory. I know these videos are old, but thank you for explaining your logic using actual words instead of a mess of symbolic logic.

[@ratneshchandak9494]

In the ring of even integers , do 4 and 6 have a lcm and gcd?

[@saumyojitdas4212]

let S = {x ∈ N : a|x and b|x}. Since a | |ab| and b | |ab| then |ab| ∈ S and therefore S not Equal to ∅. By well order principal, S has a smallest element m. Thus, a|m and b|m. This proves (1). To prove (2), we assume that c is an integer such that a|c and b|c. By the Division Algorithm there exist unique integers q and r such that c = mq + r, 0 ≤ r < m. Since a|c then c = aq1 for some q1 ∈ Z. Since a|m then m = aq2 for some q2 ∈ Z. Thus, aq1 = aqq2 + r or a(q1 −qq2) = r. This implies that a|r as q1-qq2 =r/a As q1,q2,q are all integers so integers closed under subtraction therefore "r is divided by a" **IS IT CORRECT UPTO THIS**??? A similar argument with b replacing a we find that b|r. If r > 0 then r ∈ S and this contradicts the definition of m as r will be the least no which gets divided by a and b . So we must have r = 0. Now one might say 0 is also a **Common** Multiple of a and b then why doesn't it gets included in S . That's why S 'set' contains only Natural nos .0 is not natural. So s contains least element 'm' Therefore, c = mq and m|c Is it correct