Prove that, if $x$ is any irrational number, and if $p/q$ is a rational fraction in lowest terms, with $q\geq 1$, such that $$\Big|x-\frac{p}{q}\Big|<\frac{1}{2q^2}$$ then it can be proved that $p/q$ is necessarily one of the convergents of the simple continued fraction expansion of $x$.
Note: This theorem is important in quantum computing in the sense that the final step of Shor's order finding algorithm is based on this.
If $x$ is an irrational number and $[x]=a_1$ be the greatest integer less than $x$. Then $$x=a_1+\dfrac{1}{x_2}\;,\; 0<\dfrac{1}{x_2}<1$$ where $x_2=\dfrac{1}{x-a_1}$ is irrational since if an integer is subtracted from an irrational number, the result and the reciprocal of the result are irrational. This can be repeated to obtain $$ x=a_1+\cfrac{1}{a_2+\cfrac{1}{a_3+\cfrac{1}{a_4+\cdots}}} $$ the infinite simple continued fraction. The convergents $c_n=p_n/q_n$ satisfies $$ p_n=a_np_{n-1}+p_{n-2}\\ q_n=a_nq_{n-1}+q_{n-2} $$
These properties are the same for both rational as well as irrational numbers.
What I can prove
Let $x$ be any real number and $n$ be a positive integer. Define $x_k$ so that $kx=x_k(\mod 1)$, i.e., $kx=m_k+x_k\forall k=0,1,\cdots,n$ such that $m_k\in\mathbb{Z}$ and $x_k\in[0,1)$. Let's divide the internal of size $1/n$, so we have $n+1$ number of $x_k's$ and $n$ intervals. By Pigeonhole principle, at least two of $x_k's$ are in the same interval, say $x_i$ and $x_j$ with $i<j$.
i.e., there are distinct $i$ and $j$ in the range $0,1,\cdots,n$ such that $|x_j-x_i|<1/n$.
$|(j-i)x-(m_j-m_i)|=|jx-mj-(ix-mi)|=|x_j-x_i|<1/n\\\implies \Big|x-\frac{m_j-m_i}{j-i}\Big|\leq \frac{1}{(j-i)n}\leq \frac{1}{(j-i)^2}$
This implies that, for any real number $x$ and natural number $n$, there exists integers $p$ and $q$ such that $1\leq q\leq n$ with $\Big|x-\frac{p}{q}\Big|<\frac{1}{nq}\leq\frac{1}{q^2}$.
Proof from Wiki : link
A possible proof by contradiction can be found in the link given. i.e., assuming that $|x-a/b|<1/2q^2$ but $a/b$ is not one of the convergence of $x$.
It s also taken that $q_r\leq b\leq q_{r+1}$ for a unique integer $r$, but how do we take such an assumption ?