Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:(i) \$\dfrac{13}{3125}\$ (ii) \$\dfrac{17}{8}\$ (iii) \$\dfrac{64}{455}\$ (iv) \$\dfrac{15}{1600}\$ (v) \$\dfrac{29}{343}\$ (vi) \$\dfrac{23}{2^{3}5^{2}}\$ (vii) \$\dfrac{129}{2^{2}5^{7}7^{5}}\$ (viii) \$\dfrac{6}{15}\$ (ix) \$\dfrac{35}{50}\$ (x) \$\dfrac{77}{210}\$

Theorem: Let \$x = \dfrac{p}{q}\$ be a rational number, such that the prime factorisation of q is of the form \$2^{n}5^{m}\$, where n, m are non-negative integers. Then, \$x\$ has a decimal expansion which terminates.(i) \$\dfrac{13}{3125}\$Factorise the denominator, we get\$3125 = 5 \times 5 \times 5 \times 5 \times 5 = 5^5\$So, denominator is in form of \$5^m\$ so, \$\dfrac{13}{3125}\$ is terminating.(ii) \$\dfrac{17}{8}\$Factorise the denominator, we get\$8 = 2 \times 2 \times 2 = 2^3\$So, denominator is in form of \$2^n\$ so, \$\dfrac{17}{8}\$ is terminating.(iii) \$\dfrac{64}{455}\$Factorise the denominator, we get\$455 = 5 \times 7 \times 13 \$So, denominator is not in form of \$2^{n} 5^{m}\$ so, \$\dfrac{64}{455}\$ is not terminating.(iv) \$\dfrac{15}{1600}\$Factorise the denominator, we get\$1600 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5 = 2^{6} 5^{2}\$So, denominator is in form of \$2^{n}5^{m}\$ so, \$\dfrac{15}{1600}\$ is terminating.(v) \$\dfrac{29}{343}\$Factorise the denominator, we get\$343 = 7 \times 7 \times 7 = 7^{3}\$So, denominator is not in form of \$2^{n}5^{m}\$ so, \$\dfrac{29}{343}\$ is not terminating.(vi) \$\dfrac{23}{2^{3}5^{2}}\$Here, the denominator is in form of \$2^{n}5^{m}\$ so, \$\dfrac{23}{2^{3}5^{2}}\$ is terminating.(vii) \$\dfrac{129}{2^{2}5^{7}7^{5}}\$Here, the denominator is not in form of \$2^{n}5^{m}\$ so, \$\dfrac{129}{2^{2}5^{7}7^{5}}\$ is not terminating.(viii) \$\dfrac{6}{15}\$Divide nominator and denominator both by 3 we get \$\dfrac{3}{15}\$So, denominator is in form of \$ 5^{m}\$ so, \$\dfrac{6}{15}\$ is terminating.(ix) \$\dfrac{35}{50}\$ Divide nominator and denominator both by 5 we get \$\dfrac{7}{10}\$Factorise the denominator, we get\$10 = 2 \times 5\$So, denominator is in form of \$2^{n}5^{m}\$ so, \$\dfrac{35}{50}\$ is terminating.(x) \$\dfrac{77}{210}\$Divide nominator and denominator both by 7 we get \$\dfrac{11}{30}\$Factorise the denominator, we get\$30 = 2 \times 3 \times 5\$So, denominator is not in the form of \$2^{n}5^{m}\$ so \$\dfrac{6}{15}\$ is not terminating.

The decimal expansion of the rational number \$\displaystyle\frac{23}{2^{3}5^{2}}\$, will terminate after how many places of decimal?

\$\displaystyle\frac{23}{2^{3}5^{2}}\$ = \$\displaystyle\frac{23}{2(2)^{2}(5)^{2}}\$ = \$\displaystyle\frac{23}{2(2\times5)^{2}}\$ = \$\displaystyle\frac{23}{2(10)^{2}}\$ = \$\displaystyle\frac{11.5}{100} = 0.115\$Therefore, the decimal expansion of the rational number \$\displaystyle\frac{23}{2^{3}5^{2}}\$will terminate after three places of decimal.

The decimal expansion of the rational no. \$\dfrac{43}{2^45^3}\$ will terminate after how many of decimals?

Solution:-If the denominator of a rational number is of the form \${2}^{n}{5}^{m}\$, then it will terminate after \$n\$ places, if \$n > m\$ or \$m\$ places, if \$m > n\$.The given rational number is \$\cfrac{43}{{2}^{4}{5}^{3}}\$As the denominator is of the form \${2}^{n} {5}^{m}\$, where as \$n \left( = 4 \right) > m \left( = 3 \right)\$Hence, the given rational number will terminate after n places, i.e., 4 places.Hence, the correct answer is 4.

\$\displaystyle \frac{13}{3125}\$ is a terminating decimal fraction.

If \$q=2^n \cdot 5^m\$ where \$n, m\$ are non-negative integers, then \$\displaystyle \frac{p}{q}\$ is a terminating decimal fraction.

Write the condition to be satisfied by \$q\$ so that a rational number \$\dfrac {p}{q}\$ has a terminating decimal expansion.

The rational number \$\dfrac {p}{q}\$ where p and q are co-prime q will have a ferminating decimal expansion only if the prime factorisation of q is of the form \$2^n. 5^m\$, where n and m are non-negative integers.

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>>Class 10
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>>Decimal Expansions of Real Numbers
>> $Decimal Expansion of a Number$

Decimal Expansion of a Number

Q: The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational or not. If they are rational, and of them \$\dfrac{p}{q}\$, what can you say about the prime factors of \$q\$?i) \$43.123456789\$ii) \$0.120120012000120000....\$iii) \$43.\overline {123456789} \$

i) \$43.123456789\$ is terminating or rational\$43.123456789=\dfrac{43123456789}{1000000000}=\dfrac{43123456789}{10^9}=\dfrac{43123456789}{(2\times5)^9}=\dfrac{43123456789}{2^9 \times5^9}\$This is in the form of \$\dfrac{p}{q}\$, and the prime factors of \$q\$ are in terms of \$2\$ and \$5\$.ii) \$0.120120012000120000....\$ is non-terminating and non-repeating, it is irrational. Hence cannot be expressed in the form of \$\dfrac{p}{q}\$iii) \$43. \overline{123456789}\$ is non-terminating but repeating. So, it would be rational.Let \$x=43. \overline{123456789}\$ \$\dots(1)\$ \$1000000000x=43123456789,\overline{123456789}\$ \$\dots(2)\$\$(2)-(1)\Rightarrow 999999999x=43123456746\Rightarrow x=\dfrac{43123456746}{999999999}=\dfrac{43123456746}{3^4 \times 37^1 \times 333667^1}\$In a non-termination repeating expansion of \$\dfrac{p}{q}\$, \$q\$ will have factors \$3,37,333667\$.

Q: Look at several examples of rational numbers in the form \$\displaystyle\frac{p}{q}(q\neq 0)\$, where \$p\$ and \$q\$ are integers with no common factors other than \$1\$ and having terminating decimal representaions (expansions). Can you guess what property \$q\$ must satisfy?

The property that \$q\$ must satisfy in order that the rational numbers in the from \$\dfrac{p}{q}\$ , where \$p\$ and \$q\$ are integers with no common factor other than \$1\$, have maintaining decimal representation is prime factorization of \$q\$ has only powers of \$2\$ or power of \$5\$ or both . i.e \$2^{m}\times 5^{n}\$ , where \$m=1,2,3,\cdots\$ or \$n=1,2,3,\cdots\$

Q: Write the denominator of the rational number \$257/5000\$ in the form \$2^m \times 5^n\$ where m, n is non-negative integers. Hence, write its decimal expansion on without actual division.

The fraction \$257/5000\$Since the denominator is \$5000\$,The factors for \$5000\$ are:\$5000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 5\$\$= 23 \times 54\$\$257/5000 = 257/(23 \times 54)\$It is a terminating decimal.So,Let us multiply both numerator and denominator by \$2\$, we get\$257/5000=(257 \times 2)/(5000\times 2)\$\$= 514/10000\$\$= 0.0514\$

Q: Without actually dividing find which of the following are terminating decimals.

We know that \$\cfrac{p}{q}\$ is terminating if p and q are co-prime and q is of the form \${2}^{m} {5}^{n}\$; where m and n are non-negative integers.(A) \$\cfrac{3}{25}\$:Checking co-prime-Since 3 and 25 have no common factors, hence 3 and 25 are co-prime.Now,\$25 = 5 \times 5 = {5}^{2}\$\$\therefore \text{ Denominator} = {5}^{2} = 1 \times {5}^{2} = {2}^{0} \times {5}^{2}\$\$\therefore\$ Denominator is of the form \${2}^{m}{5}^{n}\$, where m = 0 and n = 2.Hence, \$\cfrac{3}{25}\$ is a terminating decimal.(B) \$\cfrac{11}{18}\$:Checking co-prime-Since 11 and 18 have no common factors, hence 11 and 18 are co-prime.Now,\$18 = 2 \times 3 \times 3 = 2 \times {3}^{2}\$\$\therefore\$ Denominator is not of the form \${2}^{m}{5}^{n}\$.Hence, \$\cfrac{11}{18}\$ is not a terminating decimal.(C) \$\cfrac{13}{20}\$:Checking co-prime-Since 13 and 20 have no common factors, hence 13 and 20 are co-prime.Now,\$20 = 2 \times 2 \times 5 = {2}^{2} \times 5\$\$\therefore \text{ Denominator} = {2}^{2} \times 5 = {2}^{2} \times {5}^{1}\$\$\therefore\$ Denominator is of the form \${2}^{m}{5}^{n}\$, where m = 2 and n = 1.Hence, \$\cfrac{13}{20}\$ is a terminating decimal.(D) \$\cfrac{41}{42}\$:Checking co-prime-Since 41 and 42 have no common factors, hence 41 and 42 are co-prime.Now,\$42 = 2 \times 3 \times 7\$\$\therefore\$ Denominator is not of the form \${2}^{m}{5}^{n}\$.Hence, \$\cfrac{41}{42}\$ is not a terminating decimal.

Q: Which of the following will have a terminating decimal expansion?

We know that the divisor of the forms \$2^n5^m\$ always form a terminating decimal number. Let us simplify the expressions and find if the expansion have a terminating decimal expansion. Option A : \$\dfrac{77}{210} \$\$=\dfrac{7\times11}{2\times 3\times 5 \times 7 }\$\$=\dfrac{11}{2\times 3 \times 5}\$Since there is a factor of 3 in the denominator, the decimal expansion will not be terminating.Option B : \$\dfrac{23}{30} \$\$=\dfrac{23}{2\times 3\times 5}\$Since, the denominator contains a power of 3 , it is non-terminating. Option C : \$\dfrac{125}{441} \$\$=\dfrac{5\times5\times5}{3\times3\times7\times7}\$This is also non-terminating. Option D : \$\dfrac{23}{8}\$\$=\dfrac{23}{2\times2\times2}\$This contains power of 2 in the denominator. Hence, the decimal expansion is terminating.

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Important Questions

Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
(i) $\frac{1 3}{3 1 2 5}$ (ii) $\frac{1 7}{8}$ (iii) $\frac{6 4}{4 5 5}$ (iv) $\frac{1 5}{1 6 0 0}$ (v) $\frac{2 9}{3 4 3}$

(vi) $\frac{2 3}{2 ^{3} 5 ^{2}}$ (vii) $\frac{1 2 9}{2 ^{2} 5 ^{7} 7 ^{5}}$ (viii) $\frac{6}{1 5}$ (ix) $\frac{3 5}{5 0}$ (x) $\frac{7 7}{2 1 0}$

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The decimal expansion of the rational number $\frac{2 3}{2 ^{3} 5 ^{2}}$ , will terminate after how many places of decimal?

Medium

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The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational or not. If they are rational, and of them $\frac{p}{q}$ , what can you say about the prime factors of $q$ ?

i) $43.123456789$

ii) $0.120120012000120000 . . . .$

iii) $43 . \overline{123456789}$

Medium

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The decimal expansion of the rational no. $\frac{4 3}{2 ^{4} 5 ^{3}}$ will terminate after how many of decimals?

Medium

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Look at several examples of rational numbers in the form $\frac{p}{q} (q \neq = 0)$ , where $p$ and $q$ are integers with no common factors other than $1$ and having terminating decimal representaions (expansions). Can you guess what property $q$ must satisfy?

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Assertion

$\frac{1 3}{3 1 2 5}$ is a terminating decimal fraction.

Reason

If $q = 2^{n} \cdot 5^{m}$ where $n, m$ are non-negative integers, then $\frac{p}{q}$ is a terminating decimal fraction.

$\frac{1 3}{3 1 2 5}$ is a terminating decimal fraction.

Medium

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Write the denominator of the rational number $257 / 5000$ in the form $2^{m} \times 5^{n}$ where m, n is non-negative integers. Hence, write its decimal expansion on without actual division.

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Without actually dividing find which of the following are terminating decimals.

This question has multiple correct options

Easy

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Which of the following will have a terminating decimal expansion?

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Write the condition to be satisfied by $q$ so that a rational number $\frac{p}{q}$ has a terminating decimal expansion.

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Decimal Expansion of a Number

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QUICK SUMMARY WITH STORIES

Decimal expansion of Rational and Irrational Numbers

The decimal expansion of the rational number 235223​, will terminate after how many places of decimal?

The decimal expansion of the rational no. 245343​ will terminate after how many of decimals?

Look at several examples of rational numbers in the form qp​(q=0), where p and q are integers with no common factors other than 1 and having terminating decimal representaions (expansions). Can you guess what property q must satisfy?

312513​ is a terminating decimal fraction.

If q=2n⋅5m where n,m are non-negative integers, then qp​ is a terminating decimal fraction.

312513​ is a terminating decimal fraction.

Write the denominator of the rational number 257/5000 in the form 2m×5n where m, n is non-negative integers. Hence, write its decimal expansion on without actual division.

Without actually dividing find which of the following are terminating decimals.

Which of the following will have a terminating decimal expansion?

Write the condition to be satisfied by q so that a rational number qp​ has a terminating decimal expansion.

The decimal expansion of the rational number $\frac{2 3}{2 ^{3} 5 ^{2}}$ , will terminate after how many places of decimal?

The decimal expansion of the rational no. $\frac{4 3}{2 ^{4} 5 ^{3}}$ will terminate after how many of decimals?

Look at several examples of rational numbers in the form $\frac{p}{q} (q \neq = 0)$ , where $p$ and $q$ are integers with no common factors other than $1$ and having terminating decimal representaions (expansions). Can you guess what property $q$ must satisfy?

$\frac{1 3}{3 1 2 5}$ is a terminating decimal fraction.

If $q = 2^{n} \cdot 5^{m}$ where $n, m$ are non-negative integers, then $\frac{p}{q}$ is a terminating decimal fraction.

$\frac{1 3}{3 1 2 5}$ is a terminating decimal fraction.

Write the denominator of the rational number $257 / 5000$ in the form $2^{m} \times 5^{n}$ where m, n is non-negative integers. Hence, write its decimal expansion on without actual division.

Write the condition to be satisfied by $q$ so that a rational number $\frac{p}{q}$ has a terminating decimal expansion.