Prove there exists n where na < b and (n+1)a > b if a does not divide b

[Ceci020]

** 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 = q₁ * a + r₁ for r₁ is the first remainder, and r₁ < 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.

 

[Dick]

If you have the Euclidean algorithm, then use it! Pick n=q1. n*a<b right? Why is (n+1)*a>b?