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# 数学代写|数论代写Number Theory代考|MA3150 Worked out Exercises

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## 数学代写|数论代写Number Theory代考|Worked out Exercises

Problem 11.3.1. Find the rational number, expressed in lowest terms, represented by each of the following simple continued fractions: $(i)[3 ; 7,15,1] \quad(i i)[2 ; 1,2,1,1,4]$
Solution 11.3.1. (i) By virtue of Definition(11.2.1), we obtain
$$[3 ; 7,15,1]=3+\frac{1}{7+\frac{1}{15+1}}=3+\frac{1}{7+\frac{1}{16}}=\frac{355}{113} .$$
Remark 11.3.1. $[3 ; 7,15,1]$ is a good approximation for $\pi$.
(ii)
$$[2 ; 1,2,1,1,4]=2+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{1+\frac{1}{4}}}}}=\frac{87}{32} .$$
Problem 11.3.2. Find the simple continued fraction, not terminating with the partial quotient of 1, of each of the following rational numbers: (i) $\frac{17}{9}$ $-\frac{746}{830}$.
Solution 11.3.2. Here we need to apply the proof of Theorem(11.2.2) to find the simple continued fraction expansion. The tool required here is Euclidean algorithm.
(i) Applying Euclidean algorithm on 17 and 9 we have,
\begin{aligned} 17 &=1 \cdot 9+8 \ 9 &=1 \cdot 8+1 \ 8 &=8 \cdot 1 \end{aligned}
Thus the sequence of quotients give the continued fraction expansion $[1 ; 1,8]$.

## 数学代写|数论代写Number Theory代考|Infinite Continued Fractions

In the present section, we will study infinite continued fractions and establish how to express a real quantity with the help of an infinite continued fraction. We will depict the utilization of the continued fraction representation of a real number to generate rational numbers that are approximations of this real number.

In the subsequent system, we will study infinite continued fractions of quadratic irrationalities.

Let us begin with an infinite sequence of positive integers $a_{0}, a_{1}, a_{2}, \ldots, \ldots, .$ To define and to study infinite continued fraction, we need few results from mathematical analysis which we have covered in prerequisites. The following theorem is of fundamental importance, based on the relation between Fibonacci sequence and infinite continued fraction. To know Fibonacci sequence in details, refer to Chapter14(sec 14.1).

Theorem 11.4.1. For the simple infinite continued fraction $\left[b_{0} ; b_{1}, b_{2}, b_{3}, \ldots, b_{n}\right]$, the inequality $q_{k} \geq U_{k}(k=1,2, \ldots)$ holds where $C_{k}=\frac{p_{k}}{q_{k}}$ is the $k$-th convergent of the fraction and $U_{k}$ denotes the $k$-th Fibonacci number(See Chapter 14).
Proof. We establish by mathematical induction. For $k=1,2$ we have $q_{1}=b_{1} \geq$ $1=U_{1}$ and $q_{2}=b_{2} q_{1}+q_{0}=b_{2} b_{1}+b_{0} \geq U_{2}$ respectively. So the result is true for $k=1,2$. Let us assume the result be true for $k=m$. Then $q_{m} \geq U_{m}$. We are to prove the result for $k=m+1$. Now
\begin{aligned} q_{m+1} &=b_{m+1} q_{m}+q_{m-1}, \ & \geq b_{m+1} U_{m}+U_{m-1}, \ & \geq U_{m}+U_{m-1}, \ &=U_{m+1}, \text { from the definition of Fibonacci sequence. } \end{aligned}
Hence we are done.
The following theorem define infinite continued fractions as limits of finite continued fractions. The limit $\eta$ described in the statement of the theorem is called the value of the infinite simple continued fraction $\left[b_{0} ; b_{1}, b_{2}, b_{3}, \ldots, b_{k}, \ldots\right]$.

## 数学代写|数论代写Number Theory代考|Worked out Exercises

$$[3 ; 7,15,1]=3+\frac{1}{7+\frac{1}{15+1}}=3+\frac{1}{7+\frac{1}{16}}=\frac{355}{113} .$$

(二)
$$[2 ; 1,2,1,1,4]=2+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{1+\frac{1}{4}}}}}=\frac{87}{32} .$$

(i) 对 17 和 9 应用欧几里得算法，
$$17=1 \cdot 9+89=1 \cdot 8+18=8 \cdot 1$$

## 数学代写数论代写Number Theory代考|lnfinite Continued Fractions

$q_{m+1}=b_{m+1} q_{m}+q_{m-1}, \quad \geq b_{m+1} U_{m}+U_{m-1}, \geq U_{m}+U_{m-1}, \quad=U_{m+1}$, from the definition of Fibonacci sequence.

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