Let $A=\{\frac{\lfloor nb\rfloor }{n}\},b>0$ prove that $supA=b$

Question

Let $A=\{\frac{\lfloor nb\rfloor }{n}|n\in \mathbb{N}\},b>0$ prove that $supA=b$

My attempted :

for $0<b\in \mathbb{N}$ we get $A=\{b\} \rightarrow supA=b$

for $0<b\in \mathbb{Q}$ and I stuck here any hint how to continue

thanks

Answer 1

Hint

$$nb-1<\lfloor nb\rfloor \le nb$$