How do you handle the floor and ceiling function in an equation? I tried to do some math in a blog post of mine and came to one with a floor function. I wasn't sure how to deal with it so I just ignored it, and then added the ceiling function in my final equation as that seemed to give me the result I wanted. I'm wondering what is the correct way of handling these functions in equations?
What I did was this:
$$\begin{align}
G(n) &= \left\lfloor n\log{\varphi}-\dfrac{\log{5}}{2}\right\rfloor+1 \\\\
n\log{\varphi} &= G(n)+\dfrac{\log{5}}{2}-1 \\\\
n &= \left\lceil\dfrac{G(n)+\dfrac{\log{5}}{2}-1}{\log\varphi}\right\rceil
\end{align}$$
How should I have done this in a correct way? How do I work with the ceiling and floor functions when I shuffle around with equations?
 A: Observe that
\begin{eqnarray}
G(n) = \left \lfloor n \log \varphi - \log \sqrt{5} \right \rfloor + 1 = \left \lceil n \log \varphi - \log \sqrt{5} \right \rceil
\end{eqnarray}
and write
\begin{eqnarray}
\left \lceil \frac{G(n)}{\log \varphi} + \log_{\varphi} \sqrt{5} \right \rceil & = & \left \lceil \tfrac{1}{\log \varphi} \left \lceil n \log \varphi - \log \sqrt{5}  \right \rceil + \log_{\varphi} \sqrt{5} \right \rceil = n.
\end{eqnarray}
A: You can replace $\lfloor x \rfloor$ with $x - \theta$, where $\theta \in [0,1)$ is some unknown quantity. Similarly, $\lceil x \rceil = x + \theta$ (a different $\theta$ within the same range).
Another helpful identity is $\lfloor x \rfloor + n = \lfloor x + n \rfloor$ for any integer $n$.
A: Your final expression gives you the number you want.  
According to your blog post, you're looking for the smallest integer $n$ (i.e., the "first Fibonacci number with 1000 digits") that satisfies 
$$G(n) = \left\lfloor n \log \varphi - \frac{\log 5}{2} \right\rfloor + 1.$$
There may, of course, be more than one integer $n$ for which this is true.
By definition of the floor function, the values of $n$ that satisfy this are the values that satisfy $$G(n) - 1 \leq n \log \varphi - \frac{\log 5}{2} < G(n),$$
which, since $\log \phi > 0$, are the values that satisfy
$$\frac{G(n) + \frac{\log 5}{2}}{\log \varphi} - \frac{1}{\log \varphi} \leq n  < \frac{G(n) + \frac{\log 5}{2}}{\log \varphi}.$$
Since $\frac{1}{\log \varphi} \approx 4.78$, there are either four or five integers in this interval.  But the smallest one is obtained by taking the ceiling of the lower endpoint of the interval; i.e.,
$$\left\lceil\frac{G(n) + \frac{\log 5}{2} - 1}{\log \varphi}\right\rceil.$$
Incidentally, this argument also apparently shows that there are either four or five Fibonacci numbers that have a given number of digits.  (Except in the single-digit case, where there are six (not counting 0).  But your formula for $G(n)$ doesn't hold when $n=1$, so we shouldn't expect this calculation to be true in the single-digit case anyway.)
