Continued fraction convergent

Truncating a simple continued fraction gives a convergent; sufficiently accurate rational approximations with bounded denominator occur among these convergents.

Between consecutive continued fraction convergents, intermediate convergents have the form $(t{p}_{k-1}+p_{k-2})/(t{q}_{k-1}+q_{k-2})$ for $1\le t\le a_k$. The final value $t=a_k$ is the next convergent.

For nonsquare $N>1$, write the complete quotients after the initial term as $(\sqrt{N}+m)/d$. They exceed one, while their algebraic conjugates lie between minus one and zero. Thus $d<\sqrt{N}+m<2\sqrt{N}$. Comparing coefficients in $\sqrt{N}=(p_k{\alpha }_{k+1}+p_{k-1})/(q_k{\alpha }_{k+1}+q_{k-1})$ gives $p_k^2-N{q}_k^2=d_{k+1}(p_k{q}_{k-1}-q_k{p}_{k-1})$, and the last determinant has absolute value one. These small residues are useful candidates for smooth numbers in congruence-of-squares factorization.

A rational $p/q$ is a best approximation of the second kind to $\alpha$ when $|q\alpha -p|$ is no larger than $|b\alpha -a|$ for every competing rational $a/b$ with $0<b\le q$. Every simple continued-fraction convergent has this property.

With $M(a)=\left(\begin{array}{cc}a& 1\\ 1& 0\end{array}\right)$,

If $p_n/q_n=[a_0,\dots ,a_n]$, then appending a positive tail $x$ gives

Because each matrix $M(a)=\left(\begin{array}{cc}a& 1\\ 1& 0\end{array}\right)$ is symmetric, reversing the partial quotients transposes their product. If the convergents of $[a_0,\dots ,a_n]$ are $p_j/q_j$, then the last two convergents of $[a_n,\dots ,a_0]$ are