Let p and q be coprime. There exists $X_0>0$ such that if $X>X_0$ then $\exists p^aq^b$ in $(\frac{3}{4}X, X)$. I was reading this article and the author states the following lemma as trivial.

In the lemma, $a$ and $b$ are positive integers and $p$ and $q$ are coprime positive integers, both greater than 1 (actually, for this lemma, I think it's enough that they are not multiples, but let's go with coprime for simplicity).
Is there a simple way to prove it?
A friend and I came up with a proof by using that there exists $x$ and $y$ positive integers such that $\frac{3}{4}<\frac{q^x}{p^y}<1$. But, as the author said it is trivial, I guess there might be a nicer way of proving it. It would be nice to have a nice heuristic to this problem too. Any ideas?
 A: Lemma: If $\alpha>0$ is irrational and $\epsilon>0$ then for some $Y_0$ we have, for all $Y>Y_0,$ that there is a pair of non-negative integers $a,b$ such that $Y<a+b\alpha<Y+\epsilon.$
Proof: Later.
Main answer: Let $\alpha=\log_pq.$ Since $p,q$ coprime, $\alpha$ is irrational.
Let $\epsilon=\log_p(4/3).$ Then there is some $Y_0$ which satisfies the theorem. Now, if $X>\frac{4}3p^{Y_0},$ then $\log_p(3X/4)>Y_0,$ and we know there are $a,b$ such that $$\log_p(3X/4)<a+b\alpha <\log_p(3X/4)+\log_p(4/3)=\log_p(X).$$
But $a+b\alpha =\log_p(p^aq^b),$ so we are done.

Sublemma: If $\mu>0$ and $(a,b)$ is an interval with $a>0$ and $\mu<b-a,$ then there is a positive integer $m$ such that $m\mu\in(a,b).$
Proof: Left to reader. Hint: Let $m$ be the smallest integer such that $m\mu>a.$

Proof of Lemma: Let $\frac{P_n}{Q_n}$ be the convergents for the continued fraction for $\alpha.$ Then you have, for $n$ even:
$$0<P_n-Q_n\alpha<\frac{1}{Q_n}.$$ Choose $n$ even so that $\frac1{Q_n}<{\epsilon}$
Letting $\mu=P_n-Q_n\alpha,$ then the sublemma means, given $K>0,$ there is an $m$ so that $$m(P_n-Q_n\alpha)\in (K,K+\epsilon)$$
If $K+\epsilon\leq 1,$ then $m<\frac{1}{P_n-Q_n\alpha}.$
Let $Y_0=\frac{P_n}{P_n-Q_n\alpha}.$ Given $Y>Y_0,$ let $a_1$ be the smallest integer greater than $Y$. If $a_1<Y+\epsilon$ then we are done - we pick $a=a_1,b=0.$
Otherwise, let $K=a_1-Y-\epsilon.$ Find $m$ so that $$m(P_n-Q_n\alpha)\in(a_1-Y-\epsilon,a_1-Y).$$ Then:
$$a_1-m(P_n-Q_n\alpha)\in(Y,Y+\epsilon).$$
But that means we can set $a=a_1-mP_n$ and $b=mQ_n.$ We know $a>0$ since $K+\epsilon\leq 1,$ we know $$mP_n<\frac{P_n}{P_n-Q_n\alpha}=Y_0<Y<a_1$$

If you absolutely must have $b>0$ let $Y_0’=Y_0+\alpha.$ Then if $Y>Y_0’,$ $Y-\alpha>Y_0$ so there must be $a>0,b\ge 0$ such that $a+b\alpha\in(Y-\alpha,Y-\alpha+\epsilon)$ or $a+(b+1)\alpha\in(Y,Y+\epsilon).$

For example, if $p=2,q=3$, $0<8-5\log_23<\log_2(4/3),$ and we can take $Y_0=107.$ Then $X_0=2^{107}\approx 1.6\cdot 10^{32}.$ That gives you a sense of how big $X_0$ needs to be.
