HCF×LCM 原來等於... 

我也中一，看傻了

DSE 考試前最後衝刺

好事嚟

我自學Number theory，我覺得 ab = (a,b)[a,b]我可以嘗試Prove

First, Consider gcd(a,b) = d, then there exist some relatively prime integer c,f where cd = a, fd = b Once we can prove that cd*fd = c[a,b], then the prove finish. [a,b] is the lcm of a and b, The beautiful of Mathematics is in here ! the lcm is actually dcf, But it is not enough, dcf is just the common multiple, we have to prove dcf is the least common multiple ! Let m is a common multiple of a and b, we have m = ka = cdk for some k, But because m is the common multiple of b, then b divides m, so df divides kdc, so f divides kc, Since c and f are relatively prime, so f divides k. Proof Finish, QED