NT Exercise: Decimal Expressions1. Prove that a number's decimal expression

Berikut ini adalah pertanyaan dari nursahel26 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

NT Exercise: Decimal Expressions1. Prove that a number's decimal expression has finitely many decimal places if and only if the denominator (after simplification) is in the form of 2^a × 5^b where a and b are nonnegative integers.

2. Prove that the decimal representation of every rational number is eventually periodic.

Jawaban dan Penjelasan

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Problem 1
Prove that a number's decimal expression has finitely many decimal places if and only if the denominator (after simplification) is in the form of $2^a \times 5^b$ where $a$ and $b$ are nonnegative integers.

Solution

Let $q=n/d\in\mathbb{Q}$ where $n$ and $d$ are relatively prime numbers, so that $n/d$ can not be further simplified. If a decimal expression of $q$ has FINITELY many decimal places, without loss of generality, we can arbitrarily choose an odd number $k \in \mathbb{N}$ , so that
$q=\dfrac{n}{d}=c_0.c_1c_2c_3{\,\dots\,}c_k$
( $c_0$ is the integer part, while $.c_1c_2c_3{\,\dots\,}c_k$ is the fraction part of $q$ )

Thus,

$\begin{aligned}\frac{n}{d}&=10^0c_0+10^{-1}c_1+10^{-2}c_2+10^{-3}c_3+{\dots}+10^{-k}c_k\\&=\frac{c_0}{10^0}+\frac{c_1}{10^1}+\frac{c_2}{10^2}+\frac{c_3}{10^3}+{\dots}+\frac{c_k}{10^k}\\\frac{n}{d}&=\frac{c_0}{10^0}+\frac{c_k}{10^k}+\frac{c_1}{10^1}+\frac{c_{k-1}}{10^{k-1}}+\frac{c_2}{10^2}+\frac{c_{k-2}}{10^{k-2}}\\&\:\;\vdots\ +\frac{c_3}{10^3}+\frac{c_{k-3}}{10^{k-3}}+{\dots}+\frac{c_{(k-1)/2}}{10^{(k-1)/2}}+\frac{c_{(k+1)/2}}{10^{(k+1)/2}}\\\end{aligned}$
$\begin{aligned}\frac{n}{d}&=\frac{10^kc_0+10^{k-1}c_1+10^{k-2}c_2+{\dots}+10^2c_{k-2}+10c_{k-1}+c_k}{10^k}\\\frac{n}{d}&=\frac{C}{10^k}\implies \frac{10^{k}n}{d}=C\,,\ C\in\mathbb{Z}\\\end{aligned}$

Because $d\nmid n$ and $C$ is an integer, $d$ has to be a factor of $10^k$ , implying $d\mid 10^k\implies d\mid (2\times5)^k$ . Both 2 and 5 are prime numbers. Hence, $d$ has to be in the form of $2^a\times5^b$ where $a$ and $b$ are nonnegative integers ( $a$ or $b$ may possibly be 0), and $a,b\ \le \ k$ . $\blacksquare$

__________________

Problem 2
Prove that the decimal representation of every rational numberiseventually periodic.

Solution

If a rational number $x$ is eventually periodic, we can write it as
$x=a_0.a_1a_2a_3...a_n\overline{b_1b_2b_3...b_k}$
( $a_0$ is the integer part, $a_1a_2a_3...a_n$ is the non-repeating part, and $\overline{b_1b_2b_3...b_k}$ is the repeating part of $x$ )

Backward proof: If the decimal representation of $x$ is eventually periodic, then $x\in\mathbb{Q}$ ( $x$ is a rational number).

$\begin{aligned}x&=a_0.a_1a_2a_3...a_n\overline{b_1b_2b_3...b_k}\\&=a_0.a_1a_2a_3...a_n+\frac{1}{10^n} (0.\overline{b_1b_2b_3...b_k})\\\end{aligned}$

$a_0.a_1a_2a_3...a_n \in\mathbb{Q}$ , so it is enough to show that $y=0.\overline{b_1b_2b_3...b_k} \in \mathbb{Q}$ .

$\begin{aligned}y&=0.\overline{b_1b_2b_3...b_k}\\&=b_1b_2b_3...b_k\left(\frac{1}{10^k}+\frac{1}{10^{2k}}+\frac{1}{10^{3k}}+{\dots}\right)\\&=\frac{b_1b_2b_3...b_k}{10^k}\underbrace{\left(1+\frac{1}{10^k}+\frac{1}{10^{2k}}+{\dots}\right)}_{\sf geometric\ series}\\&=\frac{b_1b_2b_3...b_k}{10^k}\left(\frac{1}{1-\dfrac{1}{10^k}}\right)\\y&=\frac{b_1b_2b_3...b_k}{10^k-1}\end{aligned}$

Both numerator and denominator are positive integers, so $y\in\mathbb{Q}$ , and it proves that $x\in\mathbb{Q}$ . $\blacksquare$

Forward proof: If $x\in\mathbb{Q}$ ( $x$ is a rational number), then the decimal representation of $x$ is eventually periodic.

Let $x=\dfrac{a}{b}\in\mathbb{Q},\ a,b\in\mathbb{Z}$ .

We can take $\displaystyle a_1=\left \lfloor \frac{a}{b} \right \rfloor$ and $c_0=\dfrac{a}{b}-a_0$ , so that

$\begin{aligned}&\frac{a}{b}=a_0+c_0\ \Rightarrow a=a_0b+r_0\\&\quad{\sf where}\ r_0=bc_0\\\end{aligned}$

Take $a_1=\left \lfloor 10c_0 \right \rfloor$ and $c_1=10c_0-a_1$ , so that

$\begin{aligned}10r_0&=b(10c_0)\\&=b(a_1+c_1)\\&=a_1b+bc_1\\10r_0&=a_1b+r_1\\&\ {\sf where}\ r_1=bc_1\\\end{aligned}$

Trying to do an inductive approach, let's define:

$a_k=\left \lfloor 10c_{k-1} \right \rfloor$ and $c_k=10c_{k-1}-a_k$ , $\forall\,k\in\mathbb{N}$ .

By the same logic as $10r_0$ , we get:

$10r_{k-1}=a_kb+r_k,$ where $r_k=bc_k$

Note that $0 \le 10c_k < 10$ , implying $0 \le a_k \le 9$ , $\forall k\in\mathbb{N}$ .

Because $0 \le r_k < b$ $\forall\,k$ , there exists $m,n\in\mathbb{N}$ , so that $n > m$ , and $r_n = r_m$ .

Hence, we get

$\begin{cases}10r_m=a_{m+1}+r_{m+1}\\10r_n=a_{n+1}+r_{n+1}\end{cases}$

By division algorithm, we get $a_{m+1}=a_{n+1}$ and $r_{m+1}=r_{n+1}$ .

So, $a_{m+i}=a_{n+i}$ $\forall\,i\in\mathbb{N}$ .

Thus $x=a_0.a_1a_2a_3...a_m\overline{a_{m+1}a_{m+2}...a_n}$ is eventually periodic, and it completes the proof. $\blacksquare$

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Last Update: Fri, 14 Oct 22

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NT Exercise: Decimal Expressions1. Prove that a number's decimal expression

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