If $\frac{a}{b}\in \left[\frac{p-1}{q},\frac{p}{q}\right]$, is then $b\ge q$?

Question

Let $x=\frac{a}{b}$ be a rational number (in its lowest terms) in $[0,1]$. Let $x\in \left[\frac{p-1}{q},\frac{p}{q}\right]$ for some positive integers $p,q$ with $p\le q$.

Is it true that $b\ge q$ ?

I think the answer is yes. I can prove this if $x\in \left[0,\frac{1}{q}\right]$. But unable to prove this for any $p$ satisfying above. Any help is appreciated.

Answer 1

Counterexample:

If one takes $p = 2$ and $q=3$ then we get $x \in \left[\frac{1}{3}, \frac{2}{3}\right]$. But $x = \frac{1}{2} \in \left[\frac{1}{3}, \frac{2}{3}\right]$ and $2 \not\geq 3$.