Rational Numbers: Terminating and Recurring Decimals

Pure recurring decimal is a decimal fraction in which all the figures after the decimal point are repeated.
$\displaystyle \frac { 1 }{ 6 }= 0.6666666666..$ is $ 0.\overset { \ _ \ _  }{ 6 } $.

A terminating decimal is a decimal that ends. It's a decimal with a finite number of digits.

On dividing 4 by 9 we get

A terminating decimal is a decimal that ends. It's a decimal with a finite number of digits.

A terminating decimal is a decimal that ends. It's a decimal with a finite number of digits.

A terminating decimal is a decimal that ends. It's a decimal with a finite number of digits. 
$\displaystyle \frac {1}{20}= 0.05$ 

Factorize the denominator we get $1600=2 \times 2 \times 2 \times2 \times 2 \times 2 \times 5 \times 5  = 2^{6} \times 5^{2}$
so denominator is in form of $2^n \times 5^m$
Hence $\frac{15}{1600}$  is  terminating.

correct option is B..

$\cfrac{6}{11}$ $=$ $ 0.545454....$ 

Simplify it by dividing nominator and denominator both by 7 we get $\frac{1}{30}$
Factorize the denominator we get  $30=2 \times 3 \times 5$
Denominator has 3 also in denominator
So denominator is not in form of $2^{n} \times 5^{n}$
 Hence it is non terminating.

Given that, $0.\overline{585}$.

Let,

$ x=0.\overline{585} $

$ x=0.585585585........ $

Multiply by 1000 on both sides,

$ 1000x=1000\times 0.585585585........ $

$ =585.585585585.... $

$ =585+0.585585585........ $

$ 1000x=585+x $

$ 999x=585 $

$ x=\dfrac{585}{999} $

Hence, this is the answer.

A terminating decimal is a decimal that ends. It's a decimal with a finite number of digits.
Therefore, $C$ is the correct answer.

$\dfrac {1}{3} = 0.33333333333333333333333333....$ 

A non- terminating representation can be repeating $\dfrac{7}{22} = 0.31818181....$
or non-repeating decimal $\pi = 3.1415....$
Therefore, $B$ is the correct answer.

The decimal expansion of a number may terminate in which case the number is called a regular number or finite decimal. In the given options:

a. $\dfrac{77}{210} = 0.366666666....... \; or  \; 0.3\bar6$

b. $\dfrac{23}{30} = 0.766666666....... \; or  \; 0.7\bar6$

c. $\dfrac{125}{441} = 0.283446712......$ i.e. it goes on infintely

d. $\dfrac{23}{8} = 2.875$

A number having non-terminating and recurring decimal expansion is  a Rational Number

$1.23\bar{48}$ can be written as $1.23484848484848....$ and so on 

$\dfrac{3}{5} = 0.6$ which is terminating decimal. 

$3.24636363.......$ is a non-terminating, but repeating number as after one-hundredth place, digits are recurring.

Let $x$ = .33333......

$35/50=7/10=0.7$
This is a terminating decimal number.

Checking the termination of $\cfrac{93}{1500}$ is same as checking the termination of $\cfrac{31}{500}$ which is equal to $\cfrac{62}{1000}$

$\cfrac { 1 }{ 3 } -0.33333333=\cfrac { 1 }{ 3 } -\cfrac { 33333333 }{ { 10 }^{ 8 } } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\cfrac { { 10 }^{ 8 }-99999999 }{ 3\cdot { 10 }^{ 8 } } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\cfrac { 1 }{ 3\cdot { 10 }^{ 8 } } $

The form of q is $2^n*5^m$
q can be $1,2,5,10,20,40....$
Any integer divided by these numbers will always give a terminating decimal number.

The given value is $\dfrac{13}{3125}$ the denominator is 3125 which can be written as:

Given 

A fraction represents a terminating decimal if the denominator's prime factorization contains only 2s and 5s. 3125 = 5^5, so 317/3125 is a terminating decimal.

Expanding the decimals: x = 0.123444..., y = 0.123434..., and z = 0.1234234... Comparing the digits from left to right shows that x is the largest, followed by y, and then z.

 The rational no having denominator $3, 7, 9, 11, 13, 17, 23, 27$.............. and multiple of these number will have non terminating decimal .
(1) $\dfrac{7}{16}$ the denominator of this rational number is not having these above number and multiple of these number, so this will have  terminating decimal.
(2) $\dfrac{23}{125}$ -- the denominator of this rational number is not having these above number and multiple of these number, so this will have  terminating decimal.
(3) $\dfrac{9}{14}$ --he denominator of this rational number is having these above number multiple of $7$, so this will have non terminating decimal.
(4)$\dfrac{32}{45}$--he denominator of this rational number is having these above number multiple of 9, so this will have non terminating decimal.
(5) $\dfrac{43}{50}$-- the denominator of this rational number is not having these above number and multiple of these number, so this will have  terminating decimal.
(6)$\dfrac{17}{40}$ -- the denominator of this rational number is not having these above number and multiple of these number, so this will have  terminating decimal.
(7)$\dfrac{61}{75}$-- he denominator of this rational number is having these above number multiple of 3, so this will have non terminating decimal.
(8)$\dfrac{123}{250}$--the denominator of this rational number is not having these above number and multiple of these number, so this will have  terminating decimal.
(i), (ii), (v), (vi) and (viii) will have  terminating decimal.

We know that The rational no having denominator 3, 7, 9, 11, 13, 17, 23, 27.............. and multiple of these number will have non terminating decimal .
As  decimal expansion is 327.7081 which is terminating.
prime factors of q, when this number is expressed in the form p/q will not be above number, it will be 2 or 5 or both.

$\displaystyle \frac{1}{22} = 0.04545454545$ is not a terminating decimal.

$\dfrac{1}{90} = 0.0111111111 = 0.0\bar{1}$

First change the mixed fraction into proper fraction.
$5\dfrac{3}{8}=\dfrac{43}{8}$

We need to arrange the numbers from smallest to largest.
So, $0.005, 0.055, 0.55, 5.5$ is in ascending order

$\dfrac {1}{4} = 0.25$ is a terminating decimal.

If there are prime factors in the denominator other than $2$ or $5$, then the decimals repeat.
$\dfrac {1}{24} = \dfrac {1}{3\times 2\times 2\times 2}$ (there is a factor of $3$, the decimal will repeat.)
Therefore, $A$ is the correct answer.

When the prime factorization of the denominator of a fraction has only factors of $2$ and factors of $5$, we can always express the decimal as terminating decimal. 
For examples $\dfrac {1}{25} = \dfrac {1}{5\times 5}$ repeats (just powers of $5$, the decimal terminates.)
Therefore, $D$ is the correct answer.

The division is completed when we get the remainder zero. In this division process we do not get a zero and it is never ending. This process of division is called a non- terminating decimal.
Therefore, $B$ is the correct answer.

Recall that in order for the decimal version of a fraction to terminate, the fraction's denominator in fully reduced form must have a prime factorization that consists of only 2's and/or 5's. 

An fraction will be terminating decimal only if denominator will be in form of 

Any rational number its denominator is in the form of ${ 2 }^{ m }\times { 5 }^{ n }$,where m,n are positive integer s are terminating decimals.

We know that a terminating decimal is a decimal that ends. It's a decimal with a finite number of digits. For example $\dfrac {1}{4}=0.25$, it has only two decimal digits.

$\dfrac{2^{2}\times3^{2}\times7^{2}}{2^{5}\times5^{3}\times3^{2}\times7} = 2^{2-5}\times3^{2-2}\times5^{-3}\times7^{2-1}$

Terminating decimal expansion, because

$ \dfrac{987}{10500}=\dfrac{329}{3500}=\dfrac{329}{2^2.5^3.7}=\dfrac{47}{2^2.5^3}=.094. $

$1^{st}$ number
$x _{1} = 7$
$\Rightarrow \dfrac {1}{x _{1}} = 0.\overline {142857}$
such that,
$2^{nd}$ number
$x _{2} = 13$
$\Rightarrow \dfrac {1}{x _{2}} = 0.\overline {076923}$
$x _{3} = 21$
$\Rightarrow \dfrac {1}{x _{3}} = \dfrac {1}{21} = 0.\overline {047619}$
$x = x _{1} + x _{2} + x _{3}$
$\Rightarrow 7 + 13 + 21 = 41$.

If q is not of the form $2^n*5^m$ then definitely q can take any of the values 3,6,9,12,15,...etc.
As we know 4/3,20/6 all are non-terminating, thus required decimal expansion is non terminating.

In 7.478478.......  we can see that after decimal the digits 478 are repeating itself again and again.