[proofplan] We use the continued-fraction convergents of $\theta$. For each sufficiently large index, two adjacent convergents and their mediant have three explicit approximation constants. Borel's algebraic lemma says that at least one of the reciprocal constants is strictly larger than $\sqrt{5}$, because the only equality case would force a rational quotient of consecutive denominators to equal the irrational golden ratio. This gives infinitely many Hurwitz approximations. For sharpness, we test the golden ratio $\varphi=[1;1,1,\dots]$: its convergents have limiting constant $1/\sqrt{5}$, and Legendre's criterion forces every stronger-than-$1/2$ approximation to be a convergent. [/proofplan]

[step:Write the three local approximation constants] Let \begin{align*} \theta=[a_0;a_1,a_2,\dots] \end{align*} be the simple continued fraction expansion of $\theta$. Let the convergents be denoted by \begin{align*} \frac{p_n}{q_n}, \qquad q_n>0. \end{align*} For $n\geq 2$, define the complete quotient \begin{align*} \alpha_{n+1}:=[a_{n+1};a_{n+2},a_{n+3},\dots] \end{align*} and define \begin{align*} y_n:=\frac{q_n}{q_{n-1}}. \end{align*} Then $\alpha_{n+1}>1$ and $y_n>1$. The complete-quotient formula gives \begin{align*} \theta=\frac{\alpha_{n+1}p_n+p_{n-1}}{\alpha_{n+1}q_n+q_{n-1}}. \end{align*} Using the determinant identity \begin{align*} |p_nq_{n-1}-p_{n-1}q_n|=1, \end{align*} we subtract the convergent \begin{align*} \frac{p_n}{q_n} \end{align*} from the complete-quotient expression and obtain \begin{align*} \left|\theta-\frac{p_n}{q_n}\right| =\frac{1}{q_n(\alpha_{n+1}q_n+q_{n-1})}. \end{align*} Multiplying by $q_n^2$ gives \begin{align*} q_n^2\left|\theta-\frac{p_n}{q_n}\right| = \frac{1}{\alpha_{n+1}+1/y_n}. \end{align*} The same calculation with the preceding convergent gives \begin{align*} q_{n-1}^2\left|\theta-\frac{p_{n-1}}{q_{n-1}}\right| = \frac{1}{y_n+1/\alpha_{n+1}}. \end{align*} Now consider the mediant \begin{align*} \frac{p_n+p_{n-1}}{q_n+q_{n-1}}. \end{align*} It is reduced: any common divisor of $p_n+p_{n-1}$ and $q_n+q_{n-1}$ divides \begin{align*} (p_n+p_{n-1})q_n-p_n(q_n+q_{n-1}) =p_{n-1}q_n-p_nq_{n-1}, \end{align*} and hence divides $1$. Subtracting the mediant from the complete-quotient expression gives \begin{align*} \left|\theta-\frac{p_n+p_{n-1}}{q_n+q_{n-1}}\right| = \frac{\alpha_{n+1}-1}{(\alpha_{n+1}q_n+q_{n-1})(q_n+q_{n-1})}. \end{align*} Therefore \begin{align*} (q_n+q_{n-1})^2 \left|\theta-\frac{p_n+p_{n-1}}{q_n+q_{n-1}}\right| = \frac{(\alpha_{n+1}-1)(y_n+1)}{\alpha_{n+1}y_n+1}. \end{align*} [/step]

[step:Prove Borel's algebraic lemma] We claim that if $x>1$ and $y>1$, then at least one of \begin{align*} x+\frac{1}{y}, \qquad y+\frac{1}{x}, \qquad \frac{xy+1}{(x-1)(y+1)} \end{align*} is at least $\sqrt{5}$. Moreover, if the maximum of these three quantities is exactly $\sqrt{5}$, then \begin{align*} x=y=\varphi:=\frac{1+\sqrt{5}}{2}. \end{align*} Let $r:=\sqrt{5}$, and suppose the first two displayed quantities are at most $r$. Then \begin{align*} x\leq r-\frac{1}{y} \end{align*} and \begin{align*} y+\frac{1}{y}\leq r. \end{align*} Since $y>1$, the last inequality is equivalent to $1<y\leq\varphi$. Define the map \begin{align*} C:(1,\infty)\times(1,\infty)\to\mathbb R \end{align*} by \begin{align*} C(x,y):=\frac{xy+1}{(x-1)(y+1)}. \end{align*} For fixed $y>1$, this map is strictly decreasing in $x$, because \begin{align*} \frac{\partial C}{\partial x} = -\frac{(y+1)^2}{(x-1)^2(y+1)^2} <0. \end{align*} Thus, under the constraint $x\leq r-1/y$, \begin{align*} C(x,y)\geq C\left(r-\frac{1}{y},y\right). \end{align*} A direct simplification gives \begin{align*} C\left(r-\frac{1}{y},y\right)\geq r \end{align*} if and only if \begin{align*} (r-2)(\varphi-y)(y+\varphi^2)\geq 0. \end{align*} This holds for $1<y\leq\varphi$. Therefore, if the first two quantities are at most $r$, then the third quantity is at least $r$. The equality case in this argument forces $y=\varphi$ and then $x=r-1/y=\varphi$. Conversely, when $x=y=\varphi$, all three displayed quantities equal $\sqrt{5}$. This proves the lemma. [/step]

[step:Apply Borel's lemma to produce infinitely many approximations] Apply the lemma with \begin{align*} x=\alpha_{n+1}, \qquad y=y_n. \end{align*} Since $y_n$ is a quotient of positive integers, it is rational, while $\varphi$ is irrational. Hence the equality case $x=y=\varphi$ is impossible, so at least one of the three displayed quantities in Borel's lemma is strictly larger than $\sqrt{5}$. If \begin{align*} \alpha_{n+1}+\frac{1}{y_n}>\sqrt{5}, \end{align*} then the formula for the convergent \begin{align*} \frac{p_n}{q_n} \end{align*} gives \begin{align*} \left|\theta-\frac{p_n}{q_n}\right| < \frac{1}{\sqrt{5}q_n^2}. \end{align*} If \begin{align*} y_n+\frac{1}{\alpha_{n+1}}>\sqrt{5}, \end{align*} then the formula for the preceding convergent \begin{align*} \frac{p_{n-1}}{q_{n-1}} \end{align*} gives \begin{align*} \left|\theta-\frac{p_{n-1}}{q_{n-1}}\right| < \frac{1}{\sqrt{5}q_{n-1}^2}. \end{align*} Finally, if \begin{align*} \frac{\alpha_{n+1}y_n+1}{(\alpha_{n+1}-1)(y_n+1)}>\sqrt{5}, \end{align*} then the mediant formula gives \begin{align*} \left|\theta-\frac{p_n+p_{n-1}}{q_n+q_{n-1}}\right| < \frac{1}{\sqrt{5}(q_n+q_{n-1})^2}. \end{align*} In every case, we have found a reduced rational satisfying the Hurwitz inequality. As $n\to\infty$, the smallest denominator among \begin{align*} q_{n-1}, \qquad q_n, \qquad q_n+q_{n-1} \end{align*} tends to infinity. Therefore the construction gives infinitely many distinct rational numbers satisfying \begin{align*} \left|\theta-\frac{p}{q}\right|<\frac{1}{\sqrt{5}q^2}. \end{align*} [/step]

[step:Show that the constant is sharp] Let \begin{align*} \varphi:=\frac{1+\sqrt{5}}{2}=[1;1,1,1,\dots]. \end{align*} Let the convergents of $\varphi$ be denoted by \begin{align*} \frac{P_n}{Q_n}. \end{align*} Since every complete quotient of $\varphi$ is again $\varphi$, the exact convergent error formula from the first step gives \begin{align*} Q_n^2\left|\varphi-\frac{P_n}{Q_n}\right| = \frac{1}{\varphi+\frac{Q_{n-1}}{Q_n}}. \end{align*} The denominator ratios \begin{align*} \frac{Q_n}{Q_{n-1}} \end{align*} converge to $\varphi$, because they satisfy the recurrence \begin{align*} \frac{Q_n}{Q_{n-1}}=1+\frac{Q_{n-2}}{Q_{n-1}} \end{align*} and the positive fixed point of $t\mapsto 1+1/t$ is $\varphi$. Hence \begin{align*} Q_n^2\left|\varphi-\frac{P_n}{Q_n}\right| \longrightarrow \frac{1}{\varphi+1/\varphi} = \frac{1}{\sqrt{5}}. \end{align*} Let $C<1/\sqrt{5}$. By the convergence just proved, there is an integer $N$ such that for every $n\geq N$, \begin{align*} \left|\varphi-\frac{P_n}{Q_n}\right| \geq \frac{C}{Q_n^2}. \end{align*} Because $C<1/\sqrt{5}<1/2$, any reduced rational number satisfying \begin{align*} \left|\varphi-\frac{p}{q}\right|<\frac{C}{q^2} \end{align*} also satisfies the hypothesis of Legendre's continued-fraction criterion, \begin{align*} \left|\varphi-\frac{p}{q}\right|<\frac{1}{2q^2}. \end{align*} Therefore the rational number must be a convergent of $\varphi$. The preceding lower bound excludes every convergent with index at least $N$, so only the finitely many convergents with index less than $N$ can satisfy the strict inequality with the constant $C$. Thus, for the irrational number $\varphi$, every constant $C<1/\sqrt{5}$ gives only finitely many rational approximations satisfying \begin{align*} \left|\varphi-\frac{p}{q}\right|<\frac{C}{q^2}. \end{align*} Consequently no smaller universal constant can replace $1/\sqrt{5}$ in [Hurwitz's theorem](/theorems/3358). [/step]

[step:Give the guided proof]The following guided version repeats the argument in a single continuous pass.[/step]

[guided]Let \begin{align*} \theta=[a_0;a_1,a_2,\dots] \end{align*} be the simple continued fraction expansion of the irrational number $\theta$, and let its convergents be \begin{align*} \frac{p_n}{q_n}, \qquad q_n>0. \end{align*} For every $n\geq2$, define \begin{align*} \alpha_{n+1}:=[a_{n+1};a_{n+2},a_{n+3},\dots], \qquad y_n:=\frac{q_n}{q_{n-1}}. \end{align*} Then \begin{align*} \alpha_{n+1}>1, \qquad y_n>1. \end{align*} The complete-quotient formula and the determinant identity give \begin{align*} \theta=\frac{\alpha_{n+1}p_n+p_{n-1}}{\alpha_{n+1}q_n+q_{n-1}}, \qquad |p_nq_{n-1}-p_{n-1}q_n|=1. \end{align*} Subtracting the two adjacent convergents gives \begin{align*} q_n^2\left|\theta-\frac{p_n}{q_n}\right| = \frac{1}{\alpha_{n+1}+1/y_n} \end{align*} and \begin{align*} q_{n-1}^2\left|\theta-\frac{p_{n-1}}{q_{n-1}}\right| = \frac{1}{y_n+1/\alpha_{n+1}}. \end{align*} The mediant \begin{align*} \frac{p_n+p_{n-1}}{q_n+q_{n-1}} \end{align*} is reduced because any common divisor of its numerator and denominator divides the determinant \begin{align*} p_{n-1}q_n-p_nq_{n-1}. \end{align*} Subtracting this mediant from the complete-quotient formula gives \begin{align*} (q_n+q_{n-1})^2 \left|\theta-\frac{p_n+p_{n-1}}{q_n+q_{n-1}}\right| = \frac{(\alpha_{n+1}-1)(y_n+1)}{\alpha_{n+1}y_n+1}. \end{align*} We now use Borel's algebraic lemma. For $x>1$ and $y>1$, at least one of \begin{align*} x+\frac{1}{y}, \qquad y+\frac{1}{x}, \qquad \frac{xy+1}{(x-1)(y+1)} \end{align*} is at least $\sqrt{5}$, and the case where the maximum is exactly $\sqrt{5}$ forces \begin{align*} x=y=\varphi:=\frac{1+\sqrt{5}}{2}. \end{align*} To prove the lemma, put \begin{align*} r:=\sqrt{5}. \end{align*} If the first two quantities are at most $r$, then \begin{align*} x\leq r-\frac{1}{y}, \qquad y+\frac{1}{y}\leq r. \end{align*} For $y>1$, this gives \begin{align*} 1<y\leq\varphi. \end{align*} The map \begin{align*} C:(1,\infty)\times(1,\infty)\to\mathbb R, \qquad C(x,y):=\frac{xy+1}{(x-1)(y+1)} \end{align*} is decreasing in $x$, since \begin{align*} \frac{\partial C}{\partial x}<0. \end{align*} Hence the third Borel quantity is minimized at \begin{align*} x=r-\frac{1}{y}. \end{align*} At that value, the inequality \begin{align*} C\left(r-\frac{1}{y},y\right)\geq r \end{align*} is equivalent to \begin{align*} (r-2)(\varphi-y)(y+\varphi^2)\geq0, \end{align*} which holds on $1<y\leq\varphi$. Equality forces $y=\varphi$ and then $x=\varphi$. Apply the lemma with \begin{align*} x=\alpha_{n+1}, \qquad y=y_n. \end{align*} The equality case is impossible because $y_n$ is rational and $\varphi$ is irrational. Therefore one of the three reciprocal constants is strictly larger than $\sqrt{5}$. The corresponding one of the three reduced rational numbers \begin{align*} \frac{p_n}{q_n}, \qquad \frac{p_{n-1}}{q_{n-1}}, \qquad \frac{p_n+p_{n-1}}{q_n+q_{n-1}} \end{align*} satisfies \begin{align*} \left|\theta-\frac{p}{q}\right| < \frac{1}{\sqrt{5}q^2}. \end{align*} As $n$ tends to infinity, all possible denominators in this list tend to infinity, so the construction supplies infinitely many distinct rational approximations. For sharpness, take \begin{align*} \varphi=\frac{1+\sqrt{5}}{2}=[1;1,1,1,\dots]. \end{align*} Let its convergents be \begin{align*} \frac{P_n}{Q_n}. \end{align*} The exact error formula gives \begin{align*} Q_n^2\left|\varphi-\frac{P_n}{Q_n}\right| = \frac{1}{\varphi+\frac{Q_{n-1}}{Q_n}}. \end{align*} The denominator ratios converge as \begin{align*} \frac{Q_n}{Q_{n-1}}\longrightarrow\varphi, \end{align*} so \begin{align*} Q_n^2\left|\varphi-\frac{P_n}{Q_n}\right| \longrightarrow \frac{1}{\varphi+1/\varphi} = \frac{1}{\sqrt{5}}. \end{align*} Let \begin{align*} C<\frac{1}{\sqrt{5}}. \end{align*} Then all sufficiently late convergents satisfy \begin{align*} \left|\varphi-\frac{P_n}{Q_n}\right| \geq \frac{C}{Q_n^2}. \end{align*} If a reduced rational number satisfied \begin{align*} \left|\varphi-\frac{p}{q}\right| < \frac{C}{q^2}, \end{align*} then, since \begin{align*} C<\frac{1}{2}, \end{align*} Legendre's continued-fraction criterion would imply that this rational number is a convergent of $\varphi$. Only finitely many early convergents are not excluded by the preceding lower bound. Thus every constant smaller than $1/\sqrt{5}$ gives only finitely many approximations to $\varphi$, so no smaller universal constant can replace the Hurwitz constant.[/guided]