# Show that there does not exist positive integers $r,s$ such that $r[a]=s[b]$

Consider group $\Bbb Z_{pq}$. Consider the sets $\{[p],[2p],\cdots ,[(q-1)p]$ and $\{[q],[2q],\cdots ,[(p-1)q]$.

Show that there does not exist positive integers $r,s$ such that $r[a]=s[b]$ where $[a]\in \{[p],[2p],\cdots ,[(q-1)p]\}$ and $[b]\in \{[q],[2q],\cdots ,[(p-1)q]\}$.

Since $[a]\in \{[p],[2p],\cdots ,[(q-1)p]\}$ we have $[a]=[lp]$ where $1\le l\le q-1$ and $[b]=[kq]$ where $1\le k\le p-1$.

Assume that there exist positive integers $r,s$ such that $r[a]=s[b]\implies [rlp]=[skq]$

But I don't understand how to get a contradiction from above?

From $[rlp]=[skq]$, we get that $rlp-skq \equiv 0 \pmod{pq}$. Said differently, we have $rlp-skq=pqt$. This means $skq \equiv 0 \pmod{p}$. Assuming $p \neq q$, we get $s \equiv 0 \pmod{p}$ (because $1 \leq k <p$), likewise $r \equiv 0 \pmod{q}$. But then both $r[a]$ and $s[b]$ are the zero class $[0]_{pq}$.
Note: The way you have stated the question, we can have $r=q$ and $s=p$, in which case there does exist positive integers which will satisfy the given statement. However if we are looking for positive integers in the sense of modulo $pq$, then what I have stated above proves your statement.