We will prove that for all times, there always exist carpet with side length $p,q$ such that $(p-1)(q-1)\geq 0$ First step is obvious since $a,b>1$, now suppose that after $k$ steps, there not exist such carpet So in $(k+1)^{th}$ step, such carpet from $k^{th}$ step must be changed, let that carpet have side length $u,v$ If $u,v\geq 1$, then we have $1)$ $(\frac{1}{u}-1)(\frac{1}{v}-1)=\frac{(u-1)(v-1)}{uv}\geq 0$ and $2)$ If $c\geq 1$, then $(c-1)(v-1)\geq 0$ and if $c<1$ then $(\frac{u}{c}-1)(v-1)>0$ Similarly, if $u,v\leq 1$ we easily get that we can't get all carpet with $(p-1)(q-1)<0$ So we get contradiction and our claim is true, then it's immediately finish the problem

We show that this is not possible. Proceeding by induction, we start by seeing that the base case is obvious. Now assume that after $n$ moves he has $k$ carpets all of whose dimensions are either both $>1$ or $<1$, and choose any one carpet with dimension $a \cdot b$. WLOG assume $a,b>1$. Then there are two(three in fact) transformations: $\bullet$ $(a,b) \mapsto \left( \tfrac{1}{a},\tfrac{1}{b} \right)$, but then $\tfrac{1}{a}<1$ and $\tfrac{1}{b}<1$. $\bullet$ $(a,b) \mapsto \left(c \cdot b, \tfrac{a}{c} \cdot b \right)$. Since $b>1$, we must have $c<1$. But then $b,\tfrac{a}{c}>1$. $\bullet$ $(a,b) \mapsto \left(c \cdot a, \tfrac{b}{c} \cdot a \right)$. Since $a>1$, we must have $c<1$. But then $a,\tfrac{b}{c}>1$. Hence this is not possible. $\square$