NT Exercise: Decimal Expressions

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 × 5^b where a and b are nonnegative integers.

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

NT Exercise: Decimal Expressions

Jawaban Terkonfirmasi

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 $q=n/dinmathbb{Q}$ 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,

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

Because $dnmid n$ and $C$ is an integer, $d$ has to be a factor of $10^k$ , implying $dmid 10^kimplies dmid (2times5)^k$ . Both 2 and 5 are prime numbers. Hence, $d$ has to be in the form of $2^atimes5^b$ 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 number is eventually periodic.

Solution

If a rational number $x$ is eventually periodic, we can write it as
$x=a_0.a_1a_2a_3...a_noverline{b_1b_2b_3...b_k}$
( $a_0$ 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 $xinmathbb{Q}$ ( $x$ is a rational number).

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

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

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