Prove that the two numbers $\lfloor ax \rfloor$, $\lfloor bx \rfloor$ for $x=1,2,3,\dots$ comprise of all integers $1,2,3,\dots$, without repetition if $a$ and $b$ are positive irrational numbers such that $\dfrac{1}{a}+\dfrac{1}{b}=1$.
I am unable to solve this problem.
Maybe this would help,
Assume $[ax]=[by]=n$ for some positive integer n. Note that since a and b are irrational, we cannot have $ax=n$ or $by=n$. Hence, we have the strict inequalities:
$$n < ax < n+1 $$ and $$ n < by < n+1 $$
By dividing the first inequality by a, the second by b, then adding them and using the relation $\dfrac{1}{a} + \dfrac{1}{b} =1$ to get:
$$n < x+y < n+1$$
But $x+y$ is an integer. Hence this is a contradiction and we cannot have repetition.
Now let us assume that there exists an n such that neither $[ax]$ nor $[bx]$ equals n for any x. Let p and q be the largest integers such that $ap, bq<n$. Then we have the following inequalities:
$$ ap < n < n+1 < a(p+1)$$ And, $$ bq < n < n+1 < b(q+1)$$
Again dividing the first by a and second by b, we get,
$$ p+q < n < n+1 < p+q+2$$
But clearly this is not possible. Hence, the proof is done.