Proving that $\int_a^b\frac1{\lfloor x\rfloor}\;\mathrm{d}x$ is irrational, given $b - a$ is irrational

Question

I believe - am pretty much certain - that $\int_a^b \frac1{\lfloor x\rfloor} dx$ will be irrational if $b - a$ is irrational, due to the discrete nature of the floor function's and its reciprocal's rectangles. However, I do not know how this theory would be expressed proof-wise. I've been trying some funky stuff with moduli, but I still struggling expressing my observation, and I don't even know if modular arithmetic is the way to go. Any ideas?

Answer 1

...(assuming we don't have to worry about $0$ ... i.e. assuming $0 < a < b$ or equivalent) ...

Well

$\int_a^b \frac 1{\lfloor x \rfloor}dx =$

$\int_a^{\lceil a\rceil}\frac 1{\lceil a\rceil-1}dx + \sum_{n=\lceil a\rceil}^{\lfloor b\rfloor -1}\int_n^{n+1}\frac 1ndx + \int_{\lfloor b\rfloor }^b\frac 1{\lfloor b\rfloor }dx=$

$(\lceil a\rceil - a)\frac 1{\lceil a\rceil-1} + \sum_{n=\lceil a\rceil}^{\lfloor b\rfloor -1}\frac 1n+ (b- \lfloor b\rfloor)\frac 1{\lfloor b\rfloor}=$

$\{\frac b{\lfloor b\rfloor} - \frac a{{\lceil a\rceil}-1}\}+\{\sum_{n=\lceil a\rceil}^{\lfloor b\rfloor -1}\frac 1n +\frac {\lceil a\rceil}{\lceil a\rceil-1}- 1\}$ will be rational if and only if

$\{\frac b{\lfloor b\rfloor} - \frac a{{\lceil a\rceil}-1}\}$ is.

and from there it's probably not to difficult to come up with a counter example. The only really requirement is that (as $\lfloor b\rfloor$ and $\lceil a \rceil - 1$ are both integers) is that $a$ and $b$ be in a rational proportion to each other (yet themselves irrational).

My attempt:

If we let $ k-1 < a < k < k+1< b $ we'd have $b - a\cdot(\frac{\lfloor b\rfloor}{{\lceil a\rceil}-1})=$
$b-a\cdot \frac{k+1}k=b -a - \frac 1k a $ which could be rational if, say,
$b-a\cdot \frac{k+1}k = 0$
$b = a\frac {k+1}k$ and $a$ is an arbitrary irrational number and $k$ and arbitrary positive integer. Then $b- a = \frac 1k a$ is irrational.