Show that $a_0\leq \alpha \le a_0 +1$

Question

It is given that $$\alpha=[a_0;a_1,a_2,...,a_n]$$ where $a_0,...,a_n$ are all positive integers.

We need to show that $$a_0\leq \alpha \le a_0 +1$$

My question is when the equality holds ? I guess the question is wrong ... there can't be the equality sign in the question as these are positive integers.

Answer 1

The question isn't wrong. Note that it is true, for example, that $2\leq 2.5\leq 3$. It is also true that $2<2.5<3$. Just because the strict inequality is true does not mean the inequality with the equal signs is false. In fact, the strict inequality implies the one with equal signs. Indeed, the statement is true because $$ \alpha=a_0+\underbrace{\frac{1}{a_1+\frac{1}{a_2+\frac1{a_3+\frac1\ddots}}}}_{\mathrm{strictly\ between}\ 0\ \mathrm{and}\ 1}. $$