Exercise 6.2.4 (Condition for $\beta$ to be irrational)

Proof. Assume, for the sake of contradiction, that $\beta$ is a rational number. Then $\beta =a∕b$, for some integers $a,b$ such that $b>0$ and $a\wedge b=1$. By hypothesis, the inequality

has infinitely many rational solutions $h∕k$. So there are infinitely many rational solutions $h∕k\ne a∕b$.

A fortiori, there are infinitely many pairs $(h,k)\in \mathbb{Z}\times {\mathbb{N}}^{\text{∗}}$ such that $h∕k\ne a∕b$ which satisfy the inequality (1). Since $h∕k\ne a∕b$, then $\mathit{ak}-\mathit{bh}\ne 0$. Multiplying (1) by $\mathit{bk}$, we obtain

This implies $k^{\alpha -1}\le b$, thus $k\le k_0=\lfloor \exp <!--\; FUNCTION\; APPLICATION\; -->\left(\frac{1}{\alpha -1}\log <!--\; FUNCTION\; APPLICATION\; -->b\right)\rfloor$ (we used here $\alpha >1$). So there are infinitely many pairs $(h,k)\in \mathbb{Z}\times {\mathbb{N}}^{\text{∗}}$ such that $0<k\le k_0$ and $1\le |\mathit{ak}-\mathit{bh}|\le \frac{b}{k^{\alpha -1}}.$

But then for every $k\in [[1,k_0]]$, $|\mathit{bh}|-|\mathit{ak}|\le |\mathit{ak}-\mathit{bh}|\le b∕k^{\alpha -1}$, thus

Therefore, for every $k\in [[1,k_0]]$, there are only finitely many values of $h$ satisfying (2), so (2) has only finitely many solutions $(h,k)\in \mathbb{Z}\times {\mathbb{N}}^{\text{∗}}$. This contradiction shows that $\beta$ is irrational. □