Question

Show that if a and b are positive integers, then there are divisors c of a and d of b with (c,d): i and cd=[a,b]

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

Let a,b∈N such that a≠b. Consider p,q∈N such that a=(a,b)p and b=(a,b)q. Then (p,q)=1 and (p,(a,b))=1. Hence (p,q(a,b))=1 If we define c=p and d=q(a,b) we have what we wanted.

CORRECTION: Let a,b∈N such that a≠b. Consider p,q∈N such that a=(a,b)p and b=(a,b)q. Then (p,q)=1 and [a,b]=(a,b)pq. Now we want to factorize (a,b) in a suitable way, say (a,b)=xy, in order to get c=xp and d=yq coprime.

Let s∈N such that (a,b)=(p,(a,b))⋅(q,(a,b))⋅s. Notice that (p,q)=1 implies that (p,(a,b)) and (q,(a,b)) are coprime. Notice also that s is coprime to p and q.

Now consider c=p⋅(p,(a,b))⋅s and d=q⋅(q,(a,b)). We have that cd=[a,b], c|a and d|b. The remaining part to show is that c and d are coprime.

Since p and q are coprime and (q,(a,b)) divides q, we conclude that p is prime to (q,(a,b)) and then it is prime to d as well.

Now, since (p,(a,b)) divides p, and (p,d)=1 then (p,(a,b)) is prime to d.

Similarly, since s and q are coprime, s and (q,(a,b)) are coprime. Thus s is prime to d.

Therefore c=p⋅(p,(a,b))⋅s is prime to d.