$\{\lfloor na\rfloor\mid n\in\mathbb N\}\cup\{\lfloor nb\rfloor\mid n\in\mathbb N\}$ is a permutation of $\mathbb N$ iff $ab=a+b$ 
Given $a,b\in\mathbb R^+$, show that if the set $S=\{\lfloor na\rfloor\mid n\in\mathbb N\}\cup\{\lfloor nb\rfloor\mid n\in\mathbb N\}$ contains every natural number exactly once, then $a,b$ are irrational, and satisfy $ab=a+b$.

I've been able to solve the reverse implication. Indeed, if $N\not\in S$, then we can find $k,\ell$ such that $$\frac{k}{N}<\frac{1}{a}<\frac{k+1}{N+1}\quad\text{and}\quad\frac{\ell}{N}<\frac{1}{b}<\frac{\ell+1}{N+1}.$$ Adding these gives $k+\ell<N<k+\ell+1$, contradiction.
Secondly, if $N=\lfloor ka\rfloor=\lfloor\ell b\rfloor$, then $$\frac{N}{a}<k<\frac{N+1}{a}\quad\text{and}\quad\frac{N}{b}<\ell<\frac{N+1}{b}.$$ Adding gives $N<k+\ell<N+1$, contradiction again.
However, I'm having trouble doing the problem as stated above.
 A: Write $S_a=\{\lfloor an\rfloor:n\in\mathbb N\},S_b=\{\lfloor bn\rfloor:n\in\mathbb N\}$. Our assumption is that $S_a\cap S_b=\varnothing, S_a\cup S_b=\mathbb N$.
First we will show $a,b$ can't both be rational. If they were rational, say $a=\frac{k}{m},b=\frac{l}{m}$ (we can always write them with common denominator), then $\lfloor a\cdot lm\rfloor=kl=\lfloor b\cdot km\rfloor$, so $S_a\cap S_b\neq\varnothing$. Hence at least one of them is irrational. Later we will deduce they both are irrational.
Denote by $[N]$ the set $\{1,2,\dots,N\}$. We shall consider sets $S_a\cap [N],S_b\cap [N]$. By our assumption, their intersection is empty and their union is $[N]$. Hence the sum of their sizes is precisely $N$.
What are their sizes? Well, $\lfloor an\rfloor\in S_a\cap[N]$ iff $an<N+1$ iff $n<\frac{N+1}{a}$. This inequality holds precisely for $n=1,2,\dots,\lfloor\frac{N+1}{a}\rfloor$, unless this last number is an integer, in which case it holds for $n=1,2,\dots,\frac{N+1}{a}-1$. In either case, this gives $S_a\cap[N]$ is between $\frac{N+1}{a}-1$ and $\frac{N+1}{a}$. Therefore
$$\frac{N+1-a}{N}\frac{1}{a}\leq\frac{S_a\cap[N]}{N}\leq\frac{N+1}{N}\frac{1}{a}.$$
If $N$ is very large, then both $\frac{N+1-a}{N}=1+\frac{1-a}{N}$ and $\frac{N+1}{N}=1+\frac{1}{N}$ are very close to $1$, therefore both the upper and lower bounds for $\frac{S_a\cap[N]}{N}$ are very close to $\frac{1}{a}$, hence so is $\frac{S_a\cap[N]}{N}$ itself. Since you've said you are not familiar with analysis, I am trying to state this in somewhat informal terms, but I can assure you this can be formalized using the notion of a limit (so called squeeze theorem is of use here). Similarly, $\frac{S_b\cap[N]}{N}$ will be arbitrarily close to $\frac{1}{b}$.
But now note that $\frac{S_a\cap[N]}{N}+\frac{S_b\cap[N]}{N}=\frac{N}{N}=1$. Hence, by the above discussion, we will have $\frac{1}{a}+\frac{1}{b}=1$. From here two things are now clear:


*

*since one of $a,b$ is irrational, so is the other, and

*multiplying both sides by $ab$, we get $a+b=ab$.

