Can a fraction of irrational numbers be an integer (specific case)?

Question

Can the fraction $\frac{\pi b}{a - e b}$ where $e$ is the usual Euler constant, and $a$,$b$ are integers, be equal to an non-zero integer ?

Answer 1

I want to show that the following elementary conclusion :

Let $\thinspace (a,b)\in\mathbb Z^{2}\setminus \{0\}\thinspace $, then $\thinspace\frac{\pi b}{a-eb}\thinspace$ might be equal to some integer, only if when $\thinspace\frac {e}{\pi}\thinspace$ is irrational .

Let $\thinspace m\in\mathbb Z\thinspace $ and $\thinspace\frac ab =r\in\mathbb Q\thinspace$ with $\thinspace ab≠0\thinspace $, then we have :

$$ \begin{align}\frac{\pi b}{a-e b}&=\frac{\pi}{r-e }=m \end{align} $$

This leads to :

$$ \begin{align}&r=e+\frac {\pi}{m}\\ \implies &\frac {r}{\pi}=\frac {e}{\pi}+\frac 1m \end{align} $$

Since $\thinspace \frac{r}{\pi}\thinspace$ is irrational and $\thinspace\frac 1m\thinspace$ is rational, this implies that $\thinspace\frac {e}{\pi}\thinspace$ is irrational, but this is an open problem , in mathematics .

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Then, I want to show that another elementary conclusion :

If there exists non-zero integer pair $\thinspace(a,b)\thinspace$ such that, $\thinspace m=\frac{\pi b}{a-eb}\thinspace$ is an integer, then there exists at most one integer $\thinspace m\thinspace .$

Indeed, we have :

$$ \begin{align}&\begin{cases}\frac {r_1}{\pi}=\frac {e}{\pi}+\frac {1}{m_1} \\ \frac {r_2}{\pi}=\frac {e}{\pi}+\frac {1}{m_2}\end{cases}\\\\ \implies &\frac {r_1-r_2}{\pi}=\frac {m_2-m_1}{m_1m_2}\end{align} $$

Finally, we observe that if $\thinspace r_1≠r_2\thinspace $, then $\thinspace \frac {m_2-m_1}{m_1m_2}\not\in\mathbb Q\thinspace .$

A contradiction .