Proving that Spec(α) and Spec(ß) partition positive integers iff α and ß are irrational and 1/α + 1/ß = 1 From Concrete Math, problem 3.13 asks:
"Let α and ß be positive real numbers. Prove that Spec(α) and Spec(ß) partition positive integers if and only if α and ß are irrational and 1/α + 1/ß = 1"
The solution claims:
"If they form a partition, the text's formula for N(α,n) implies that 1/α + 1/ß = 1, because the coefficients of n in the equation N(α,n) + N(ß,n) = n must agree if the equation is to hold for large n" (it goes on to the next part of the proof but I only care about this part)
In this chapter, they define Spec(α) to mean an infinite multiset of integers: $\{\lfloorα\rfloor,\lfloor2α\rfloor, ...\}$, and define N(α,n) to be the number of elements in Spec(α) that are $\le$ n. They show that N(α,n) = $\lceil(n+1)/α\rceil - 1$. They also show that a necessary condition for Spec(α) and Spec(ß) to partition the positive integers is N(α,n) + N(ß,n) = n  
I can write out the equation N(α,n) + N(ß,n) = n by substituting that equation:
$$\lceil(n+1)/ß\rceil + \lceil(n+1)/α\rceil - 2 = n$$
then converting to floor
$$\lfloor (n+1)/ß\rfloor + \lfloor (n+1)/α\rfloor = n$$
then splitting the floor into the fractional and actual part of its arguments
$$n(1/ß + 1/α) + 1/α + 1/ß - \{(n+1)/ß\} - \{(n+1)/α\} = n$$
...but, at that point, I don't see how I can conclude that 1/α + 1/ß = 1. I see that 1/α + 1/ß appears as a coeffecient of n, but there's the problem of the fractional stuff on the right. For their claim to be correct, the fractional parts would have to add up to 1 to cancel out the 1 from  1/α + 1/ß. I know that the fractional parts both have values less than one (since they are fractional parts of real numbers), but I don't see how to conclude that they sum to 1. Can I just claim that, since n appears on the right hand side with a coefficient of 1, that the coefficient for n on the left must be 1?

The hint is in the phrase "for large $n$".
As you proved, we must have that for all $n$, 
$$n(1/\beta + 1/\alpha) + 1/\alpha + 1/\beta - \{(n+1)/\beta\} - \{(n+1)/\alpha\} = n.$$
Let us denote $1/\beta + 1/\alpha$ by $c$, to rewrite this as
$$nc + c - \{(n+1)/\beta\} - \{(n+1)/\alpha\} = n.$$
As $0 \le \{x\} < 1$ for all $x$ and here we have $\{(n+1)/\beta\} - \{(n+1)/\alpha\} = nc + c - n$, this means that
$$0 \le nc + c -n < 2.$$
Now consider what happens for very large $n$. If $c < 1$, then $nc + c - n = c - n(1-c)$ eventually becomes negative (in fact, it becomes negative for $n > \frac{c}{1-c}$), violating the "$0 \le$" inequality. And if $c > 1$, then for very large $n$, we'll have $nc + c - n > 2$: specifically, for $n > 2/(c-1)$, we'll have $nc + c - n > nc - n > 2$. So we must have $c = 1$.
Another way of writing the same thing is to divide $0 \le nc + c - n < 2$ by $n$ and say that $$0 \le c + \frac{c}{n} - 1 < \frac{2}{n},$$
and taking the limit as $n \to \infty$ gives $c - 1 = 0$, as it is sandwiched between the left- and right-limits, both equal to $0$.
