Let $x$ be an irrational number. Prove that there exist infinitely many rational numbers $\dfrac pq$ that satisfy the following

$$\bigg|\,x-\dfrac pq\,\bigg|<\dfrac 1{q^2+q}$$

My idea would be to solve the inequality for $\frac pq$ and then somehow use the pigeonhole principle. Is this heading in the right direction? Any hints would be helpful!

Solution 1:

The key here is continued fractions. Look them up.

An irrational number has an infinite number of convergents to its continued fraction. Each of then satisfies $|x-\frac{p_n}{q_n}| < \frac1{q_n q_{n+1}} $. Since $q_{n+1} > q_n $, this proves your statement.

Solution 2:

Here is the standard argument, perhaps what you had in mind.

HINT: Let $Q\ge 1$ integer and consider the $Q$ fractional parts $\{a x\}$ for $1\le a \le Q$. Either one of them is closer to $0$ or $1$ by less than $\frac{1}{Q+1}$, or two of them are closer by less than $\frac{1}{Q+1}$ ( think of the $Q$ points $\frac{1}{Q+1}$, $\frac{2}{Q+1}$, $\ldots$, $\frac{Q}{Q+1}$ in $[0,1]$). From here one concludes that there exists integers $c$ and $1 \le d \le Q$ so that $|d x - c| < \frac{1}{Q+1}$.