Continued fraction expansion of irrational numbers

In this section we use continued fractions for expansion of irrational numbers.

Theorem 1. Let $\displaystyle (x_n)_{n\geq 0}$ be a sequence of intergers with $\displaystyle x_i>0$ for every $\displaystyle i>0$ . Then the sequence $\displaystyle (p_n/q_n)_{n\geq 0}$ is a convergent sequence, and the its limit is an irrational number. We denote this limit by $\displaystyle [x_0;x_1,x_2,\ldots]$ .

Proof. From [1] we have $\displaystyle q_1\geq q_0=1>0$ and for all $\displaystyle n>1$ , $\displaystyle q_n=x_nq_{n-1}+q_{n-2},$ hence by induction on $\displaystyle n$ , $\displaystyle q_{n+1}>q_n$ for every $\displaystyle n\geq 1$ . Therefore $\displaystyle\lim_{n\to \infty}q_n=\infty$ .

By the Proposition 4 in [1], for all $\displaystyle n\geq 0$ ,

$\displaystyle \frac{p_n}{q_n}-\frac{p_{n+1}}{q_{n+1}}=\frac{(-1)^{n-1}}{q_nq_{n+1}},\quad\quad (1)$

hence $\displaystyle \frac{p_n}{q_n}-\frac{p_{n+2}}{q_{n+2}}=\frac{(-1)^{n-1}(q_{n+2}-q_n)}{q_nq_{n+1}q_{n+2}},\quad \forall n\geq 0.$ Therefore

$\displaystyle \frac{p_1}{q_1}>\frac{p_3}{q_3}>\frac{p_5}{q_5}>\ldots>\frac{p_0}{q_0}$

and

$\displaystyle \frac{p_0}{q_0}<\frac{p_2}{q_2}<\frac{p_4}{q_4}<\ldots<\frac{p_1}{q_1},$

hence $\displaystyle (p_{2n}/q_{2n})_{n\geq 0}$ and $\displaystyle (p_{2n+1}/q_{2n+1})_{n\geq 0}$ are convergent sequences. By (1) and $\displaystyle q_n\to\infty$ we have

$\displaystyle\lim_{n\to\infty}\frac{p_{2n}}{q_{2n}}=\lim_{n\to\infty}\frac{p_{2n+1}}{q_{2n+1}},$ so $\displaystyle (p_n/q_n)_{n\geq 0}$ is a convergent sequence.

Now we prove $\displaystyle \displaystyle \alpha:=\lim_{n\to\infty}\frac{p_n}{q_n}$ is an irrational number. We have

$\displaystyle \frac{p_{2m}}{q_{2m}}<\alpha<\frac{p_{2n+1}}{q_{2n+1}},\quad\forall m,n\geq 0.$

Thus, by (1),

$\displaystyle\left|\alpha-\frac{p_{2n}}{q_{2n}}\right|\leq \frac{1}{q_{2n}q_{2n+1}}<\frac{1}{q_{2n}^2},\quad\forall n\geq 1.$

By the Proposition 2 in [1], $\displaystyle p_{2n}$ and $\displaystyle q_{2n}$ are coprime integers for every $\displaystyle n\geq 1$ , hence there are infinite rational numbers $\displaystyle r/s$ , with $\displaystyle s>0$ and $\displaystyle (r,s)=1$ , such that

$\displaystyle \left|\alpha-\frac{r}{s}\right| <\frac{1}{s^2}.\quad\quad (2)$

Assume that $\displaystyle \alpha$ is rational and write $\displaystyle \alpha=p/q$ , where $\displaystyle p$ and $\displaystyle q>0$ are coprime integers. For all positive integers $\displaystyle s$ , at most two integers $\displaystyle r$ satisfy the equation (2), hence there are coprime integers $\displaystyle r_0$ and $\displaystyle s_0>q$ such that

$\displaystyle\left|\frac{p}{q}-\frac{r_0}{s_0}\right| <\frac{1}{s_0^2}.$

From the inequality we have $\displaystyle \mid ps_0-qr_0\mid <1$ , hence $\displaystyle ps_0=qr_0$ , a contradiction. Therefore $\displaystyle \alpha$ is an irrational number. $\Box$

Theorem 2. Let $\displaystyle \alpha$ be an irrational number. Then there is a unique sequence of integers $\displaystyle (a_n)_{n\geq 0}$ such that

(1) $\displaystyle a_i>0$ for every $\displaystyle i>0$ .

(2) $\displaystyle \alpha =[a_0;a_1,a_2,\ldots]$ .

Proof. In this proof, $\displaystyle [x]$ is the integer part of $\displaystyle x$ . Because $\displaystyle \alpha$ is an irrational number, we have $\displaystyle [\alpha]<\alpha<[\alpha]+1$ , hence there is a real number $\displaystyle u_1>1$ such that

$\displaystyle \alpha=[\alpha]+\frac{1}{u_1}.$

Because $\displaystyle \alpha$ is an irrational and $\displaystyle [\alpha]$ is an integer, $\displaystyle u_1$ is an irrational number. Hence there is an irrational number $\displaystyle u_2>1$ such that

$\displaystyle u_1=[u_1]+\frac{1}{u_2},$

and so on. Therefore we have real numbers $\displaystyle u_0:=\alpha$ , $u_1>1$ , $\displaystyle u_2>1$ , $\displaystyle \ldots$ such that $\displaystyle u_i$ is irrationals for every $\displaystyle i>0$ and

$\displaystyle u_k=[u_k]+\frac{1}{u_{k+1}},\quad\forall k\geq 0.$

We claim that $\displaystyle \alpha=[[u_0];[u_1],[u_2],\ldots]$ . Fix a $\displaystyle k>2$ . We have

$\displaystyle \alpha=[[u_0];[u_1],\ldots, [u_k],u_{k+1}].$

Hence, by Proposition 4 in [1],

$\displaystyle \left|\alpha-\frac{p_k}{q_k}\right|=\frac{1}{q_k(u_{k+1}q_{k}+q_{k-1})}<\frac{1}{q_k^2},$

so $\displaystyle \lim_{n\to\infty}\frac{p_n}{q_n}=\alpha$ . Now assume that

$\displaystyle \alpha =[a_0;a_1,a_2,\ldots]=[b_0;b_1,b_2,\ldots],$

where $\displaystyle (a_n)_{n\geq 0}$ and $\displaystyle (b_n)_{n\geq 0}$ are two sequences of integers such that $\displaystyle a_i>0$ and $\displaystyle b_i>0$ for every $\displaystyle i>0$ .

Because

$\displaystyle [a_0;a_1,a_2,\ldots,a_{n}]=a_0+\frac{1}{[a_1;a_2,\ldots,a_n]},\quad\forall n\geq 0,$

we have

$\displaystyle [a_0;a_1,a_2,\ldots]=a_0+\frac{1}{[a_1;a_2,\ldots]}.$

Hence $\displaystyle a_0=b_0=[\alpha]$ and $\displaystyle [a_1;a_2,a_3,\ldots] = [b_1;b_2,b_3,\ldots]$ . Similarly, $\displaystyle a_1=b_1$ and

$\displaystyle [a_2;a_3,a_4,\ldots] = [b_2;b_3,b_4,\ldots],$

and so on. Therefore $\displaystyle a_i=b_i$ for every $i$ . $\Box$

The equality in the theorem is called an expansion of $\displaystyle \alpha$ into a infinite continued fraction. In that expansion we will call $\displaystyle [a_0;a_1,a_2,\ldots,a_i]$ is the $\displaystyle i-$ th convergent of the continued fraction, or $\displaystyle i-$ th convergent of $\displaystyle \alpha$ . The theorem says that for every irrational number has an expansion into a infinite continued fraction, and this expansion is unique.

Example 1. $\displaystyle \sqrt{2}=[1;2,2,\ldots]$ .

Example 2. The golden ratio $\displaystyle\varphi:=\frac{1+\sqrt{5}}{2}=[1;1,1,\ldots]$ .

Example 3. $\displaystyle e=[2;1,2,1,1,4,1,1,6,1,1,8,\ldots]$ .

A sequence $\displaystyle (a_n)_{n\geq 0}$ is called eventually periodic if $\displaystyle a_{n+T}=a_n$ for some positive integer $\displaystyle T$ and sufficiently large $\displaystyle n$ . A real number is called quadratic irrational number, if there is a polynomial $\displaystyle P(x)$ is of degree two with rational coefficients such that $\displaystyle P(x)$ is an irreducible polynomial (see [3]) over the rational numbers and $\displaystyle \alpha$ is a root of $\displaystyle P(x)$ .

Theorem 3. Let $\displaystyle \alpha$ be an irrational number and $\displaystyle \alpha =[a_0;a_1,a_2,\ldots]$ is the expansion of $\displaystyle\alpha$ into a infinite continued fraction. Then $\displaystyle (a_n)_{n\geq 1}$ is eventually periodic if and only if $\displaystyle \alpha$ is a quadratic irrational.

References

[1] https://nttuan.org/2008/10/12/continued-fractions-the-basics/

[2] https://nttuan.org/2008/11/14/continued-fraction-expansion-of-rational-numbers/

[3] https://nttuan.org/2009/01/11/poly02/

Continued fraction expansion of rational numbers

In this section we use continued fractions ([2]) for expansion of rational numbers. If $\displaystyle x_0$ , $\displaystyle x_1$ , $\displaystyle \ldots$ , are integer nunbers with $\displaystyle x_i>0$ for every $\displaystyle i>0$ then $\displaystyle [x_0;x_1,x_2,\ldots,x_n]\in\mathbb{Q},\quad\forall n\geq 0.$

Conversly, we have the theorem

Theorem 1. Let $\displaystyle r$ and $\displaystyle s$ be coprime integers with $\displaystyle s>0$ . Then there are non negative integer $\displaystyle n$ and integers $\displaystyle a_0$ , $\displaystyle a_1$ , $\displaystyle \ldots$ , $\displaystyle a_n$ such that

(1) $\displaystyle a_i>0$ for every $\displaystyle i=1,2,\ldots,n$ .

(2) $\displaystyle r/s=[a_0;a_1,a_2,\ldots,a_n]$ .

Proof. Let us proceed by induction on $\displaystyle s$ . The case $\displaystyle s=1$ is trivial. Now suppose that the assertion is true for all positive integers up to $\displaystyle s-1$ ( $\displaystyle s>1$ ). Because $\displaystyle (r,s)=1$ and $\displaystyle s>1$ , we have $\displaystyle s\nmid r$ . Hence by the Division Algorithm ([1]), there are integers $\displaystyle a$ and $\displaystyle b$ such that

$\displaystyle r=sa+b,\quad 1\leq b<s.\quad\quad (1)$

By the hypothesis of the induction, there are integers $\displaystyle m>0$ , $\displaystyle a_1$ , $\displaystyle a_2>0$ , $\displaystyle \ldots$ , $\displaystyle a_m>0$ such that

$\displaystyle \frac{s}{b}=[a_1;a_2,a_3,\ldots,a_m].\quad\quad (2)$

Because $\displaystyle s>b$ , we have $\displaystyle a_1>0$ . From (1) and (2) we have

$\displaystyle \frac{r}{s}=a+\frac{1}{s/b}=a+\frac{1}{[a_1;a_2,a_3,\ldots,a_m]}=[a;a_1,a_2,\ldots,a_m],$ completing the induction step. $\Box$

The equality in the theorem is called an expansion of $\displaystyle r/s$ into a finite continued fraction. In that expansion we will call $\displaystyle [a_0;a_1,a_2,\ldots,a_i]$ is the $i-$ th convergent of the continued fraction, or $i-$ th convergent of $\displaystyle r/s$ .

Example 1. Find an expansion of $\displaystyle 43/5$ into a finite continued fraction.

Solution. By the Division Algorithm, we have

$\displaystyle 43=5\cdot 8+3\quad\quad\frac{43}{5}=8+\frac{3}{5}=8+\frac{1}{5/3},$ and

$\displaystyle 5= 3\cdot 1 +2\quad\quad\frac{5}{3}=1+\frac{2}{3}=1+\frac{1}{3/2},$ and $\displaystyle \frac{3}{2}=1+\frac{1}{2}$ . Therefore $\displaystyle 43/5=[8;1,1,2]$ . $\Box$

The theorem says that for every rational number has an expansion into a finite continued fraction. But this expansion is not unique.

Example 2. $\displaystyle 13/5=[2;1,1,2]=[2;1,1,1,1]$ . $\Box$

Theorem 2. Let $\displaystyle \alpha$ be an integer number. Then $\displaystyle \alpha$ has exactly two expansions into a finite continued fraction.

Proof. By the theorem 1, we can write

$\displaystyle \alpha=[a_0;a_1,a_2,\ldots,a_n],$

where $a_0$ , $a_1$ , $\ldots$ , $a_n$ are integers such that $a_i>0$ for every $i=1,2,\ldots,n$ . If $\displaystyle n=0$ then $\displaystyle \alpha=a_0$ and $\displaystyle \alpha=[\alpha]$ is an expansion of $\displaystyle \alpha$ . If $\displaystyle n=1$ then $\displaystyle \alpha=a_0+\frac{1}{a_1}$ , hence $\displaystyle \frac{1}{a_1}$ is an integer, so $\displaystyle a_1=1$ . Therefore $\displaystyle \alpha=[\alpha-1;1]$ is an expansion of $\displaystyle \alpha$ .

Now assume that $\displaystyle n\geq 2$ . We have

$\displaystyle \alpha-a_0=\frac{1}{[a_1;a_2,\ldots,a_n]}$

is an integer number and $\displaystyle [a_1;a_2,\ldots,a_n]>0$ , hence $\displaystyle [a_1;a_2,\ldots,a_n]\leq 1$ . This claim is false because $\displaystyle n\geq 2$ and $\displaystyle a_i\geq 1$ for every $\displaystyle i=1,2,\ldots,n$ . $\Box$

Theorem 3. Let $\displaystyle \alpha$ be a rational number but not an integer. Then $\displaystyle \alpha$ has exactly two expansions into a finite continued fraction.

Proof. Assume that $\displaystyle \alpha=\alpha=r/s$ , where $\displaystyle r$ and $\displaystyle s>1$ are coprime integers. We prove by induction on $\displaystyle s$ that $\displaystyle \alpha$ has exactly two expansions into a finite continued fraction

$\displaystyle \alpha=[a_0;a_1,\ldots,a_n]=[a_0;a_1,\ldots,a_n-1,1],$

where $\displaystyle a_n>1$ . If $\displaystyle s=2$ , because $\displaystyle (r,s)=1$ there is an integer $\displaystyle k$ such that $\displaystyle r=2k+1$ . By the theorem 1, we can write

$\displaystyle \alpha=k+\frac{1}{2}=[a_0;a_1,a_2,\ldots,a_n],$

where $a_0$ , $a_1$ , $\ldots$ , $a_n$ are integers such that $a_i>0$ for every $i=1,2,\ldots,n$ . We have $n>0$ and $a_0=k$ , hence $\displaystyle 2=[a_1;a_2,\ldots,a_n]$ . By the theorem 2, we have $2$ has exactly two expansions into a finite continued fraction, those are $2=[2]$ and $2=[1;1]$ , therefore $\alpha$ has exactly two expansions $\displaystyle \alpha=[k;2]=[k;1,1],$ hence the claim is true for $\displaystyle s=2$ . Now suppose that the claim is true for $\displaystyle 2$ , $\displaystyle 3$ , $\displaystyle \ldots$ , $\displaystyle s-1$ ( $\displaystyle s>2$ ). By the theorem 1, we can write

$\displaystyle \alpha=[a_0;a_1,a_2,\ldots,a_n],$

where $a_0$ , $a_1$ , $\ldots$ , $a_n$ are integers such that $a_i>0$ for every $i=1,2,\ldots,n$ . We have $n>0$ and $a_0=[\alpha]$ (integer part of $\alpha$ ), hence $\displaystyle \frac{1}{\alpha-[\alpha]}=[a_1;a_2,\ldots,a_n].$ By the Division Algorithm, there is an integer $\displaystyle a$ such that $\displaystyle r=s[\alpha]+a$ and $1\leq a<s$ , then

$\displaystyle \frac{s}{a}=[a_1;a_2,\ldots,a_n].$

If $a=1$ then $a_1>1$ and by the theorem 2, we have $s/a$ has exactly two expansions are $s/a=[a_1]$ and $\displaystyle s/a=[a_1-1;1]$ . If $\displaystyle a>1$ then by the hypothesis of the induction (note that $\displaystyle s$ and $\displaystyle a$ are coprime integers), $\displaystyle s/a$ has exactly two expansions are

$\displaystyle \frac{s}{a}=[a_1;b_2,\ldots,b_n]=[a_1;b_2,\ldots,b_n-1,1],$

where $\displaystyle b_n>1$ . Therefore $\displaystyle \alpha$ has exactly two expansions, and the claim is true for $\displaystyle s$ . $\Box$

References

[1] https://nttuan.org/2020/01/14/divisibility/

[2] https://nttuan.org/2008/11/14/continued-fraction-expansion-of-rational-numbers/

Continued fractions: The basics

Let $x_0$ , $\displaystyle x_1$ , $\displaystyle x_2$ , $\displaystyle \ldots$ be variables. We define two sequences of polynomials with complex coefficients $\displaystyle \{p_n\}_{n\geq 0}$ and $\displaystyle \{q_n\}_{n\geq 0}$ by conditions following:

(1) For all non negative integer $\displaystyle n$ , $\displaystyle p_n$ and $\displaystyle q_n$ are polynomials in $\displaystyle x_0,x_1,\ldots,x_n$ .

(2) $\displaystyle p_0=x_0$ and $\displaystyle q_0=1$ .

(3) For all positive integer $\displaystyle n$ , $\displaystyle p_n=x_0p_{n-1}(x_1,x_2,\ldots,x_n)+q_{n-1}(x_1,x_2,\ldots,x_n)$ and

$\displaystyle q_n=p_{n-1}(x_1,x_2,\ldots,x_n)$ .

Example. We have

$\displaystyle p_1=x_0x_1+1$ and $\displaystyle q_1=x_1$ , therefore $\displaystyle \frac{p_1}{q_1}=x_0+\frac{1}{x_1}$ .

$\displaystyle p_2=x_0x_1x_2+x_0+x_2$ and $\displaystyle q_2=x_1x_2+1$ , hence

$\displaystyle \frac{p_2}{q_2}=x_0+\frac{x_2}{x_1x_2+1}=x_0+\cfrac{1}{x_1+\cfrac{1}{x_2}}. \Box$

For all non negative integer $\displaystyle n$ we have

$\displaystyle [x_0;x_1,x_2,\ldots,x_n]:=\frac{p_n}{q_n}=x_0+\cfrac{1}{x_1+\cfrac{1}{x_2+\cfrac{1}{x_3+\cdots+\cfrac{1}{x_{n-1}+\cfrac{1}{x_n}}}}}.$

A rational function of this form is called a continued fraction.

Proposition 1. For all $\displaystyle n>1$ , we have $\displaystyle p_n=x_np_{n-1}+p_{n-2}$ and $\displaystyle q_n=x_nq_{n-1}+q_{n-2}.$

Proof. We use induction on $\displaystyle n$ . From the above example we have the assertion is true for $\displaystyle n=2$ . Now suppose that the assertion has been established for $\displaystyle n-1$ ( $\displaystyle n>2$ ). Then we have

$\displaystyle p_{n-1}(x_1,x_2,\ldots,x_n)=x_np_{n-2}(x_1,x_2,\ldots,x_{n-1})+p_{n-3}(x_1,x_2,\ldots,x_{n-2})$

and

$\displaystyle q_{n-1}(x_1,x_2,\ldots,x_n)=x_nq_{n-2}(x_1,x_2,\ldots,x_{n-1})+q_{n-3}(x_1,x_2,\ldots,x_{n-2}).$

Therefore by define of $\displaystyle \{p_n\}$ and $\displaystyle \{q_n\}$ ,

$\displaystyle p_n=x_0p_{n-1}(x_1,x_2,\ldots,x_n)+q_{n-1}(x_1,x_2,\ldots,x_n)=x_np_{n-1}+p_{n-2},$

and $\displaystyle q_n=x_nq_{n-1}+q_{n-2}$ . Thus the assertion is true for $\displaystyle n$ . $\Box$

For convenience, we define $\displaystyle p_{-2}=0$ , $\displaystyle p_{-1}=1$ , $\displaystyle q_{-2}=1$ , and $\displaystyle q_{-1}=0$ .

Proposition 2. For all $\displaystyle n>-2$ , we have $\displaystyle p_nq_{n-1}-q_np_{n-1}=(-1)^{n-1}$ .

Proof. We prove by induction on $\displaystyle n$ . The case $\displaystyle n=-1$ is trivial. Now suppose that the assertion is true for $\displaystyle n-1$ ( $\displaystyle n\geq 0$ ). Then by proposition 1, we have

$\displaystyle p_nq_{n-1}-q_np_{n-1}=(x_np_{n-1}+p_{n-2})q_{n-1}-(x_nq_{n-1}+q_{n-2})p_{n-1}$

$=-(p_{n-1}q_{n-2}-q_{n-1}p_{n-2})=(-1)^{n-1}.$

Therefore the assertion is true for $\displaystyle n-1$ . $\Box$

Proposition 3. For all $\displaystyle n>-1$ , we have $\displaystyle p_nq_{n-2}-q_np_{n-2}=(-1)^nx_n$ .

Proof. Assume $\displaystyle n>-1$ is an integer number. By propositions 1 and 2, we have

$\displaystyle p_nq_{n-2}-q_np_{n-2}=(x_np_{n-1}+p_{n-2})q_{n-2}-(x_nq_{n-1}+q_{n-2})p_{n-2}$ $=x_n(p_{n-1}q_{n-2}-q_{n-1}p_{n-2})=(-1)^nx_n. \Box$

Proposition 4. If $\displaystyle n\geq 0$ and $\displaystyle \theta:=[x_0;x_1,x_2,\ldots,x_{n+1}]$ then $\displaystyle p_n-\theta q_n=\frac{(-1)^{n-1}}{q_{n+1}}.$

Proof. By proposition 2, we have

$\displaystyle p_n-\theta q_n=p_n-\frac{p_{n+1}}{q_{n+1}}\cdot q_n=\frac{-(p_{n+1}q_n-q_{n+1}p_n)}{q_{n+1}}=\frac{(-1)^{n-1}}{q_{n+1}}. \Box$

Stern–Brocot tree

Có một dãy tương tự như dãy Farey (xem [1] và [2]), tên chúng là dãy Stern-Brocot. Dãy này được tìm ra một cách độc lập bởi Moritz Stern (1858) và Achille Brocot (1861). Stern là một nhà Toán học Đức còn Brocot là một nhà thiết kế đồng hồ người Pháp.

Trong định nghĩa sau, $\displaystyle \dfrac{1}{0}$ là số hữu tỷ hình thức, ta xem như nó lớn hơn mọi số hữu tỷ. Dãy Stern-Brocot thứ $\displaystyle n\, (n\in\mathbb{N})$ , ký hiệu $\displaystyle SB_n$ , được xác định như sau: $\displaystyle SB_0$ là dãy $\displaystyle \frac{0}{1},\frac{1}{0}$ và với mỗi số nguyên dương $\displaystyle n$ , $\displaystyle SB_n$ được tạo ra bằng cách chép lại toàn bộ (giữ nguyên thứ tự) các số hạng của $\displaystyle SB_{n-1}$ và chèn vào giữa hai số hạng liên tiếp phân số trung gian ở dạng tối giản của chúng.

Một số dãy Stern-Brocot:

$\displaystyle SB_0:\frac{0}{1},\frac{1}{0}$ .

$\displaystyle SB_1:\frac{0}{1},\frac{1}{1},\frac{1}{0}$ .

$\displaystyle SB_2:\frac{0}{1},\frac{1}{2},\frac{1}{1},\frac{2}{1},\frac{1}{0}$ .

Dễ thấy rằng với mỗi số tự nhiên $\displaystyle n$ , $\displaystyle SB_n$ là một dãy tăng gồm $\displaystyle 2^n+1$ số hữu tỷ không âm, và hai số cách đều số ở giữa là nghịch đảo của nhau.

Ta có thể nhúng các dãy Stern-Brocot vào một cây như hình , gọi là cây Stern-Brocot. Ta thấy mỗi số hữu tỷ không âm có mặt đúng một lần trong cây. Thật vậy, vì mỗi dãy Stern-Brocot là một dãy tăng nên mọi số hữu tỷ không âm xuất hiện nhiều nhất một lần trong cây, bây giờ ta chứng minh mọi số hữu tỷ không âm đều xuất hiện trong cây. Xét một số hữu tỷ không âm ở dạng tối giản $\displaystyle \dfrac{m}{n}$ . Tồn tại số tự nhiên $\displaystyle p$ và hai số hạng $\displaystyle \dfrac{a}{b},\dfrac{a'}{b'}$ của $\displaystyle SB_p$ sao cho $\displaystyle \dfrac{a}{b}<\dfrac{m}{n}<\dfrac{a'}{b'}$ . Nếu $\displaystyle \dfrac{m}{n}$ là phân số trung gian của $\displaystyle \dfrac{a}{b},\dfrac{a'}{b'}$ thì tất nhiên nó xuất hiện trong cây, nếu không sẽ xảy ra một trong hai trường hợp: $\displaystyle \dfrac{a}{b}<\dfrac{m}{n}<\dfrac{a+a'}{b+b'}$ , ta thay phân số $\displaystyle \dfrac{a'}{b'}$ bởi $\displaystyle \dfrac{a+a'}{b+b'}$ ; $\displaystyle \dfrac{a+a'}{b+b'}<\dfrac{m}{n}<\dfrac{a'}{b'}$ , ta thay phân số $\displaystyle \dfrac{a}{b}$ bởi $\displaystyle \dfrac{a+a'}{b+b'}$ . Quá trình này không thể tiếp tục mãi vì với dãy dạng $\displaystyle \dfrac{a}{b}<\dfrac{m}{n}<\dfrac{a'}{b'}$ ta luôn có $\displaystyle m+n\geq a+a'+b+b'$ , suy ra $\displaystyle \dfrac{m}{n}$ xuất hiện trong cây.

Khi thay $\displaystyle \dfrac{0}{1}<\dfrac{1}{0}$ bởi hai số hữu tỷ không âm $\displaystyle \dfrac{a}{b}<\dfrac{c}{d}$ với $\displaystyle bc-ad=1$ ta sẽ được các dãy mới, gọi là các dãy Stern-Brocot suy rộng. Có thể chứng minh được rằng mọi số hữu tỷ nằm giữa $\displaystyle \dfrac{a}{b}$ và $\displaystyle \dfrac{c}{d}$ đều xuất hiện trong một dãy Stern-Brocot suy rộng nào đó.