Thread: 2 Questions
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May31-09, 08:25 PM
P: 92
Would this be sufficient for the first one:

Assume Nn|Nm, then (t)Nn=Nm, t≥1. For all m,n≥1, there exists q,r such that m=qn + r, 0≤r<n.
Next assume m does not divide n, therefore 1≤r<n.
Thus Nm=Nqn+r=10r(s)Nn+Nr
Then, (t)Nn=Nqn+r=10r(s)Nn+Nr
Thus, Nr=Nn(t-s10r)=Nn(d), d≥1, which implies r≥n, which contradicts our hypothesis of r<n.
Thus Nn|Nm iff n|m.