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**** Homework Statement**

Question is to prove:

If 1 < a < b and if a does not divide b

then there exists n in N such that na < b and (n+1)a > b

**** My thoughts:**

Since a does not divide b for 1 < a < b, I want to refer to the Euclidean algorithm, so:

b = q

_{1}* a + r

_{1}for r

_{1}is the first remainder, and r

_{1}< a

But then I get lost on where to go from here. Actually I thought of using induction for my proof, where I define a set S = {n in N such that na < b and (n+1)a > b}. But when I do the 2nd claim: if n in S, then n+1 in S, I think I find a contradiction because when replace n+1 to n for the first part of the def. of S, (n+1)a < b is false.

So could someone please give me some hints on how should I proceed?

Thank you very much for your help.