I have to prove that for a fixed positive integer n, the subset A of Q consisting of rationals with denumerator that divide n under addition, forms a group under addition.
I just did that it's not empty ( 0/1 is in A and is its identities ), and also that if we begin choosing q1,q2 with denumator that divides n, we have q1= a/(n/k1) = k1a/n and q2 = b/(n/k2) = k2b/n for a,b integers and k1,k2 divisors of n. Then, its clear that q1 - q2 has a denumator that divides n because k1a/n - k2b/n = (k1a - k2b)/n has a denumator that divides n, and by the subgroup criteria, it forms a subgroup.
So that should prove it ? did i make any mistake ?
I'm thinking that because i've seen two proofs of that same problem but they seem to go a bit more complex , they go :
Being q1= a/c and q2 = b/t where c and t divides n, we can see that if d = gcd(c,t) , then we can write the denumerators c,t of q1 and q2 as c = d.k1 and t d.k2. So, q1 - q2 = a/(dk1) - b/(dk2) = (a.d.k2 - b.k1.d )/ (d²k1k2) = (ak2 - bk1)/(dk1k2) = (ak2 - bk2)/lcm(c,t) and hence stating that if c and t divide n, lcm(c,t) also divides n .
I'm just wondering if i did something wrong, because my proof sounds so more simple.
Also, theres something i'm really suspicious of, why could they replace dk1k2 with lcm(dk1, dk2) = lcm(c,t) ?? I worked out that empirically, and lcm(ka,kb) = k.a.b only when a and b are relatively prime, but generally it should be lcm (ka,kb) = k.lcm(a,b).
Thanks a lot in advance
Thanks a lot