Real Numbers

1.0Types of Numbers

Natural Numbers Counting numbers $1,2,3,4,5$,.....are known as natural numbers. The set of all natural numbers can be represented by : $\mathrm{N}=\{1,2,3,4,5,\dots .$.

Whole Numbers If we include 0 among the natural numbers, then the numbers $0,1,2,3,4,5\dots .$. are called whole numbers. The set of whole numbers can be represented by : $W=\{0,1,2,3,4,5,\dots$. $\}$ Clearly, every natural number is a whole number but 0 is also a whole number which is not a natural number.

Integers All counting numbers and their negatives including zero are known as integers. The set of integers can be represented by: Z or $\mathrm{I}=\{.....,-4,-3,-2,-1,0,1,2,3,4,\dots ..$.

Positive integers The set ${\mathrm{I}}^{+}=\{1,2,3,4,\dots .$.$\}isthesetofallpositiveintegers.Clearly,positiveintegersand$ natural numbers are synonyms.

Negative integers The set ${\mathrm{I}}^{-}=\{\dots ..,-3,-2,-1\}$ is the set of all negative integers.

• 0 is neither negative nor positive.

Rational numbers The numbers of the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q\ne 0$, are known as rational numbers, e.g. $\frac{4}{7},\frac{3}{2},\frac{5}{8},\frac{0}{1},-\frac{2}{3}$, etc. The set of all rational numbers is denoted by Q . i.e. $Q=\left\{x:x=\frac{p}{q};p,q\in I,q\ne 0\right\}$

Since every natural number 'a' can be written as $\frac{a}{1}$, so ' $a$ ' is a rational number. Since 0 can be written as $\frac{0}{1}$ and every non-zero integer 'a' can be written as $\frac{a}{1}$, so 0 and every non zero integer is also a rational number. Every rational number has a peculiar characteristic that when expressed in decimal form is expressible either in terminating decimals or non-terminating repeating decimals.

For example : $\frac{1}{5}=0.2,\frac{1}{3}=0.333\dots ,\dots ,\frac{22}{7}=3.1428714287,\frac{8}{44}=0.18188\dots ..$. , etc. The recurring decimals have been given a short notation as $\mathrm{0.333...}=0.\overline{3}$ 4.1555 $\dots =4.1\overline{5}$ $0.323232\dots =0.\overline{32}$

Irrational numbers

Those numbers which when expressed in decimal form are neither terminating nor repeating decimals are known as irrational numbers, e.g. $\sqrt{2},\sqrt{3},\sqrt{5},\pi$ etc.

• Note, that the exact value of $\pi$ is not $\frac{22}{7}\cdot \frac{22}{7}$ is rational, while $\pi$ is an irrational number. $\frac{22}{7}$ is approximate value of $\pi$. Similarly, 3.14 is not an exact value of $\pi$.

Real numbers

The rational and irrational numbers combined together are called real numbers, e.g. $\frac{13}{21},\frac{2}{5},-\frac{3}{7},\sqrt{3},4+\sqrt{2}$, etc. are real numbers. The set of real numbers is denoted by $R$.

• Note that, the sum, difference or non-zero product of a rational and irrational number is irrational. e.g. $3+\sqrt{2},4-\sqrt{3},\frac{2}{3}-\sqrt{5},4\sqrt{3},-7\sqrt{5}$ are all irrational.
• Prime numbers Except 1, each natural number which is divisible by only 1 and itself is called as prime number e.g., $2,3,5,7,11,13,17,19,23,29,31$,.....etc. There are total 25 prime numbers upto 100 and 45 upto 200. 2 is the only even prime number and the least prime number. $2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97$ are prime numbers upto 100 .

• Co-prime A pair of two natural numbers having no common factor, other than 1, is called a pair of co-prime. For example: $(3,5),(4,5),(5,6),(7,9),(6,7)$ etc, are co-primes.
• Twin primes Prime numbers differing by 2 are called twin primes, e.g. $(3,5),(5,7),(11,13)$ etc, are called twin primes.
• Composite numbers All-natural numbers except 1 , which are not prime are composite numbers. If C is the set of composite numbers then $C=\{4,6,8,9,10,12\dots ..$.
• 1 is neither prime nor composite.

2.0Euclid's division lemma

Euclid's division lemma states that "For any two positive integers a and b, there exist unique integers $q$ and $r$ such that $a=bq+r,0\le r<b$." e.g. (i) Consider number 23 and 5, then : $23=5\times 4+3$ Comparing with $\mathrm{a}=\mathrm{bq}+\mathrm{r}$ we get, $\mathrm{a}=23,\mathrm{b}=5,\mathrm{q}=4,\mathrm{r}=3$ and $0\le \mathrm{r}<\mathrm{b}$ (as $0<3<5$ ) (ii) Consider positive integers 18 and 4 $18=4\times 4+2$ For $18(=\mathrm{a})$ and $4(=\mathrm{b})$ we have $\mathrm{q}=4,\mathrm{r}=2$ and $0\le \mathrm{r}<\mathrm{b}$

In the relation $\mathrm{a}=\mathrm{bq}+\mathrm{r}$, where $0\le \mathrm{r}<\mathrm{b}$ is nothing but a statement of the long division of number a by b in which q is the quotient obtained and r is the remainder. e.g. Long division of 25 by 7 . So, $25=7\times 3+4$ Here, quotient $=3$, remainder $=4$

Building Concepts 1 Prove that every odd integer is of the form $2q+1$, where $q$ is an integer. Explanation Let a be any positive integer. By using Euclid division lemma with a and $b=2$ we have $\mathrm{a}=2\mathrm{q}+\mathrm{r}$, where q is integer $\ge 0$ and $0\le \mathrm{r}<2$ Since $0\le r<2,\therefore r=0,1$ For $r=0,a=2q+r=2q$ (even integer) and for $r=1,a=2q+r=2q+1$ Now, we know that an integer is either even or odd and $\mathrm{a}=2\mathrm{q}+1$ is not an even integer $\therefore a=2q+1$ is an odd integer.

Building Concepts 2 "The product of two consecutive positive integers is divisible by 2". Is this statement true or false? Give reasons. Explanation: True, because $n(n+1)$ will always be even, as one of the factors is odd and other is even.

Building Concepts 3 A positive integer is of the form $3q+1,q$ being a natural number. Can you write its square in any form other than $3m+1$, i.e., $3m$ or $3m+2$ for some integer $m$ ? Justify your answer. Explanation No, because $(3q+1{)}^2=9q^2+6q+1=3\left(3q^2+2q\right)+1=3m+1$ {where $m=3q^2+2q\}$

Building Concepts 4 Show that the square of any positive integer cannot be of the form $5q+2$ or $5q+3$ for any integer $q$. Explanation: We know that any positive integer can be of the form $5\mathrm{m},5\mathrm{m}+1$, $5m+2,5m+3$ or $5m+4$ for some integer $m$. Thus, we have: Case (i) : $(5\mathrm{m}{)}^2=25{\mathrm{m}}^2=5\left(5{\mathrm{m}}^2\right)=5\mathrm{q},\mathrm{q}$ is an integer. Case (ii) : $(5\mathrm{m}+1{)}^2=25{\mathrm{m}}^2+10\mathrm{m}+1=5\left(5{\mathrm{m}}^2+2\mathrm{m}\right)+1=5\mathrm{q}+1$, q is an integer. Case (iii) : $(5m+2{)}^2=25{\mathrm{m}}^2+20\mathrm{m}+4=5\left(5{\mathrm{m}}^2+4\mathrm{m}\right)+4=5\mathrm{q}+4$, q is an integer. Case (iv) : $(5m+3{)}^2=25m^2+30m+9=5\left(5m^2+6m+1\right)+4=5q+4,q$ is an integer. Case (v) : $(5m+4{)}^2=25m^2+40m+16=5\left(5m^2+8m+3\right)+1=5q+1$, $q$ is an integer. Thus, the square of a positive integer cannot be of the form $5q+2$ or $5q+3$ for any integer $q$.

Building Concepts 5 Prove that if $x$ and $y$ are both odd positive integers, then $x^2+y^2$ is even but not divisible by 4. Explanation Let $\mathrm{x}=2\mathrm{m}+1$ and $\mathrm{y}=2\mathrm{n}+1$ for some integers m and n . $$\begin{gathered} \therefore x^2+y^2=(2m+1{)}^2+(2n+1{)}^2 \\ =4m^2+4m+1+4n^2+4n+1=4\left(m^2+n^2\right)+4(m+n)+2=4q+2 \\ \text{ where }q=\left(m^2+n^2\right)+(m+n) \end{gathered}$$ $\Rightarrow x^2+y^2$ is even and leaves remainder 2 on dividing by 4 $\Rightarrow x^2+y^2$ is not divisible by 4 .

3.0HCF by Euclid's division algorithm

If 'a' & 'b' are positive integers such that $\mathrm{a}=\mathrm{bq}+\mathrm{r}$, then, every common divisor of 'a' & 'b' is a common divisor of ' b ' & ' r ' and vice-versa.

In mathematics, the Euclid's Algorithm is an efficient method for computing the greatest common divisor (GCD) or highest common factor (HCF). So, let us state Euclid's division algorithm clearly. To obtain the HCF of two positive integers, say cand d, with $\mathrm{c}>\mathrm{d}$ follow the steps below:

Step-1: Apply Euclid's division lemma, to c and d . So, we find whole numbers, q and r such that $\mathrm{c}=\mathrm{dq}+\mathrm{r},0\le \mathrm{r}<\mathrm{d}$.

Step-2: If $r=0,d$ is the HCF of $c$ and $d$. If $r\ne 0$, apply the division lemma to $d$ and $r$. Step-3: Continue the process till the remainder is zero. The divisor at this stage will be the required HCF. This algorithm works because HCF ( $c,d$ ) = HCF ( $d,r$ ) where the symbol HCF (c, d) denotes the HCF of c and d .

Building Concepts 6 Explain the principle behind Euclid's division algorithm. Explanation: The Euclid's division algorithm is based on the principle that the greatest common divisor of two numbers does not change if the smaller number is subtracted from the larger number. Consider two numbers 252 and 105 whose HCF is 21 . $252=21\times 12$ and $105=21\times 5$ $252-105=147$, the $\mathrm{HCF}(147,105)$ is also 21 . Since the larger of the two numbers is reduced, repeating this process gives successively smaller numbers until one of them is zero. When that occurs, the GCD is the remaining nonzero number Step-1: $252=105\times 2+42\left\{\begin{array}{l}252-105=147\\ 147-105=42\end{array}\right\}$ Step-2 : $105=42\times 2+21\left\{\begin{array}{l}105-42=63\\ 63-42=21\end{array}\right\}$ Step-3: $42=21\times 2+0\left\{\begin{array}{l}42-21=21\\ 21-21=0\end{array}\right\}$ Required HCF $(252,105)=21$

Numerical Ability 1 Use Euclid's division algorithm to find the HCF of 441, 567, 693. Solution: In order to find the HCF of 441, 567 and 693, we first find the HCF of 441 and 567 by Euclid's division algorithm. Using division algorithm, we get $567=441\times 1+126$ $441=126\times 3+63$ $126=63\times 2+0$ So, $\mathrm{HCF}(567,441)=63$ Now, we find the HCF of 63 and 693 $693=63\times 11+0$ $\therefore \mathrm{HCF}(63,693)=63$ Hence HCF $(441,567,693)=63$

4.0Fundamental theorem of arithmetic

The Fundamental Theorem of Arithmetic tells us something important about the relationship between composite numbers and prime numbers. It is usually stated as follows : "Every composite number can be expressed as a product of primes, and their decomposition is unique, apart from the order in which the prime factors occur." e.g. $12600=2^3\times 3^2\times 5^2\times 7$

Thus we have expressed the composite number 12600 as product of powers of primes in ascending order and this decomposition is unique.

HCF and LCM of numbers

HCF (Highest common factor) The HCF of two or more numbers is the greatest number that divides each one of them exactly.

LCM (Least common Multiple) The least number which is exactly divisible by each one of the given numbers is called their LCM.

In order to find HCF and LCM of given numbers by Prime Factorisation Method or Fundamental Theorem of Arithmetic, express each number as the product of primes, then HCF = Product of least powers of common factors.

LCM = Product of highest powers of all the factors.

Important relation Product of two numbers $=($ Their HCF $)\times ($ Their LCM $)$

Building Concepts 7 The numbers 525 and 3000 are both divisible only by $3,5,15,25$ and 75 . What is HCF (525, 3000)? Justify your answer. Explanation: Since HCF of given numbers is the greatest number that divides each one of them exactly. Therefore, $\mathrm{HCF}(525,3000)=75$.

Numerical Ability 2 Find the LCM and HCF of 1296 and 2520 by applying the fundamental theorem of arithmetic method i.e. using the prime factorisation method. Solution: $1296=2\times 2\times 2\times 2\times 3\times 3\times 3\times 3=2^4\times 3^4$ $2520=2\times 2\times 2\times 3\times 3\times 5\times 7=2^3\times 3^2\times 5\times 7$ LCM $=2^4\times 3^4\times 5\times 7=45360$ HCF $=2^3\times 3^2=72$

Numerical Ability 3 Given that HCF $(306,657)=9$. Find LCM $(306,657)$. Solution $\mathrm{HCF}(306,657)=9$ means HCF of 306 and $657=9$ Required LCM $(306,657)$ means required. LCM of 306 and 657. For any two positive integers; their $\mathrm{LCM}(a,b)=\frac{\text{ Product of the numbers }}{\mathrm{HCF}(a,b)}$ i.e., $\mathrm{LCM}(306,657)=\frac{306\times 657}{9}=22,338$

What is the relation between the LCM, HCF (of 3 numbers) and the HCF's of them taken by pairs? Exploring the Concept Consider any three natural numbers (say ${\mathrm{N}}_1,{\mathrm{N}}_2$ and ${\mathrm{N}}_3$ ). Find their LCM and HCF. Find HCF of $N_1$ and $N_2$, the HCF of $N_2$ and $N_3$ and the HCF of $N_1$ and $N_3$. Find the product of the three numbers. Drawing conclusions Study some more examples draw up a table and discover.

State your discovery.

Important relation $\mathrm{LCM}(p,q,r)=\frac{p\cdot q\cdot r\cdot \mathrm{HCF}(p,q,r)}{\mathrm{HCF}(p,q)\cdot \mathrm{HCF}(q,r)\cdot \mathrm{HCF}(p,r)}$ $\mathrm{HCF}(p,q,r)=\frac{p\cdot q\cdot r\cdot \mathrm{LCM}(p,q,r)}{\mathrm{LCM}(p,q)\cdot \mathrm{LCM}(q,r)\cdot \mathrm{LCM}(p,r)}$

Building Concepts 8 Can two numbers have 18 as their HCF and 380 as their LCM? Give reason. Explanation No, because HCF (18) does not divide LCM (380).

• LCM of two numbers must be divided by their HCF.

Building Concepts 9 Find the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3, respectively. Explanation: Consider the numbers 1251-1, 9377-2 and 15628-3 or 1250, 9375 and 15625. Required number is the HCF of 1250, 9375 and 15625. Now, $1250=2\times 5^4$ $9375=3\times 5^5$ $15625=5^6$ $\mathrm{HCF}=5^4=625$

Building Concepts 10 On a morning walk, three persons step off together and their steps measure 40 cm , 42 cm and 45 cm , respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps? Explanation: $40=2^3\times 5^1$ $42=2^1\times 3^1\times 7^1$ $45=3^2\times 5^1$ Minimum distance covered $=$ LCM of 40, 42, 45 $2^3\times 3^2\times 5^1\times 7^1=8\times 9\times 35=2520\mathrm{cm}$

Building Concepts 11 Explain why $7\times 11\times 17+17$ is a composite number. Explanation: Let $7\times 11\times 17+17=(7\times 11+1)\times 17$ $(77+1)\times 17=78\times 17$ $\Rightarrow 7\times 11\times 17+17=2\times 3\times 13\times 17$ $2\times 3\times 13\times 17$ is a composite number as having more than two distinct factors.

Building Concepts 12 Show that $12^{\mathrm{n}}$ cannot end with digit $0$ or 5 for any natural number n . Explanation: $12^{\mathrm{n}}=(2\times 2\times 3{)}^{\mathrm{n}}=2^{2\mathrm{n}}\times 3^{\mathrm{n}}$ So, the only primes in the factorisation of $12^{\mathrm{n}}$ are 2 and 3 . For $12^{\mathrm{n}}$ to end with digit 0 or 5 , it must have 5 as a prime factor. But by fundamental theorem of arithmetic, it cannot have 5 as a prime factor. Hence $12^{\mathrm{n}}$ cannot end with digit 0 or 5 for any natural number n .

5.0Revisiting irrational numbers

An irrational number is a real number that cannot be expressed as the ratio of two integers. The numbers $\sqrt{2},\sqrt{3},\sqrt{15},\pi$, e, etc. are examples of irrational numbers. The decimal representation of irrational numbers is non-terminating and non-repeating. e.g. 0.101001000100001....

Theorem : Let p be a prime number. If p divides ${\mathrm{a}}^2$, then p divides a also, where a is a positive integer. Proof : Let the prime factorisation of a be as follows : $\mathrm{a}={\mathrm{p}}_1{\mathrm{p}}_2\dots {\mathrm{p}}_{\mathrm{n}}$, where ${\mathrm{p}}_1{\mathrm{p}}_2,\dots ,{\mathrm{p}}_{\mathrm{n}}$ are primes, not necessarily distinct. Therefore, $a^2=\left(p_1,p_2\dots p_n\right)\left(p_1,p_2\dots p_n\right)=p_1^2{p}_2^2\dots p_n^2$. Now, we are given that $p$ divides $a^2$. Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of ${\mathrm{a}}^2$. However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of $a^2$ are $p_1^2{p}_2^2\dots p_n^2$. So $p$ is one of $p_1{p}_2\dots p_n$. Now, since $a=p_1{p}_2\dots p_n,p$ divides $a$. Theorem : Prove that square root of 2 is irrational Proof: We shall prove this by the method of contradiction. If possible, let us assume that $\sqrt{2}$ is a rational number. Then $\sqrt{2}=\frac{a}{b},b\ne 0$ where $\mathrm{a},\mathrm{b}$ are integers having no common factor other than 1 . $$\begin{gathered} \Rightarrow (\sqrt{2}{)}^2={\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^2 \\ \Rightarrow 2=\frac{{\mathrm{a}}^2}{{\mathrm{b}}^2} \\ \Rightarrow {\mathrm{b}}^2=\frac{{\mathrm{a}}^2}{2} \\ \Rightarrow 2\text{ divides }{\mathrm{a}}^2 \\ \Rightarrow 2\text{ divides }\mathrm{a}\text{, therefore let }\mathrm{a}=2\mathrm{c}\text{ for some integer }\mathrm{c}\text{. } \\ \Rightarrow {\mathrm{a}}^2=4{\mathrm{c}}^2 \\ \Rightarrow {\mathrm{b}}^2=\frac{4{\mathrm{c}}^2}{2} \\ \Rightarrow \frac{{\mathrm{b}}^2}{2}={\mathrm{c}}^2 \\ \Rightarrow 2\text{ divides }{\mathrm{b}}^2 \\ \Rightarrow 2\text{ divides }\mathrm{b} \\ \text{ Thus, }2\text{ is a common factor of a and }\mathrm{b}\text{. } \\ \text{ But it contradicts our assumption that a and }\mathrm{b}\text{ have no common factor other (1)) } \\ \text{ So, our assumption that }\sqrt{2}\text{ is a rational, is wrong. } \\ \text{ Hence, }\sqrt{2}\text{ is irrational. } \end{gathered}$$

Building Concepts 13 Show that $3\sqrt{2}$ is irrational. Explanation: Let us assume, to the contrary, that $3\sqrt{2}$ is rational. $\Rightarrow 3\sqrt{2}=\frac{\mathrm{a}}{\mathrm{b}}(\mathrm{b}\ne 0,\mathrm{a}$ and b are coprime $)$ $\Rightarrow \sqrt{2}=\frac{\mathrm{a}}{3\mathrm{b}}$ Since 3, $a$ and $b$ are integers, $\Rightarrow \frac{\mathrm{a}}{3\mathrm{b}}$ is rational, and so $\sqrt{2}$ is rational. But this contradicts that $\sqrt{2}$ is irrational. So, we conclude that $3\sqrt{2}$ is irrational.

Building Concepts 14 Prove that $2+\sqrt{3}$ is irrational. Explanation: Let $2+\sqrt{3}$ be a rational number equals to $r$ $\therefore 2+\sqrt{3}=r$ $\sqrt{3}=r-2$ Here L.H.S is an irrational number while R.H.S. $\mathrm{r}-2$ is rational. $\therefore$ L.H.S $\ne$ R.H.S Hence it contradicts our assumption that $2+\sqrt{3}$ is rational. $\therefore 2+\sqrt{3}$ is irrational.

• $2+\sqrt{3}$ is not equal to $2\sqrt{3}$

Building Concepts 15 Prove that $\sqrt{2}+\sqrt{3}$ is irrational. Explanation Let $\sqrt{2}+\sqrt{3}$ be rational number say ' x ' $\Rightarrow \mathrm{x}=\sqrt{2}+\sqrt{3}$ Squaring both side $x^2=2+3+2\sqrt{6}=5+2\sqrt{6}$ $\Rightarrow x^2=5+2\sqrt{6}\Rightarrow \sqrt{6}=\frac{x^2-5}{2}$ As $x,5$ and 2 are rational. $\Rightarrow \frac{x^2-5}{2}$ is a rational number. $\Rightarrow$ So, $\sqrt{6}$ also is a rational number. Which is contradiction of the fact that $\sqrt{6}$ is an irrational number. Hence our assumption is wrong $\Rightarrow \sqrt{2}+\sqrt{3}$ is an irrational number.

• $\sqrt{2}+\sqrt{3}$ is not equal to $\sqrt{5}$

Building Concepts 16 Prove that $\sqrt{p}+\sqrt{q}$ is irrational, where $p,q$ are primes. Explanation: Let us suppose that $\sqrt{\mathrm{p}}+\sqrt{\mathrm{q}}$ be rational. Let $\sqrt{\mathrm{p}}+\sqrt{\mathrm{q}}=\mathrm{r}$, where r is rational. Therefore $\sqrt{\mathrm{p}}=\mathrm{r}-\sqrt{\mathrm{q}}$ Squaring both sides, we get $\mathrm{p}={\mathrm{r}}^2+\mathrm{q}-2\mathrm{r}\sqrt{\mathrm{q}}$. Therefore $\sqrt{\mathrm{q}}=\frac{{\mathrm{r}}^2+\mathrm{q}-\mathrm{p}}{2\mathrm{r}}$ Now, R.H.S. is a rational number, since $r$ is rational. $\Rightarrow r^2$ is rational, $q,p$ are both rational (since $p,q$ are primes). But $\sqrt{\mathrm{q}}$ is irrational, since q is prime. $\therefore$ We arrive at a contradiction $\therefore$ Our supposition is wrong. Hence $\sqrt{\mathrm{p}}+\sqrt{\mathrm{q}}$ is irrational.

6.0Revisiting rational numbers and their decimal expansion

In Mathematics, a rational number is any number that can be expressed as the quotient $\frac{\mathrm{a}}{\mathrm{b}}$ of two integers, with the denominator $b$ not equal to zero.

• The decimal representation of rational number is either terminating or nonterminating but repeating. e.g. $\frac{3}{7},2,0,-5,2.6,2.7777$... etc.

Nature of the decimal expansion of rational numbers

Theorem 1: Let x be a rational number whose decimal expansion terminates. Then we can express $x$ in the form $\frac{p}{q}$, where $p$ and $q$ are co-primes, and the prime factorisation of $q$ is of the form $2^m\times 5^n$, where $m,n$ are non-negative integers. Theorem 2: Let $x=\frac{p}{q}$ be a rational number, such that the prime factorisation of $q$ is of the form $2^m\times 5^n$, where $m,n$ are non-negative integers. Then, $x$ has a decimal expansion which terminates. Theorem 3: Let $x=\frac{p}{q}$ be a rational number, such that the prime factorisation of $q$ is not of the form $2^m\times 5^n$, where $m,n$ are non-negative integers. Then, $x$ has a decimal expansion which is non-terminating repeating. e.g. (i) $\frac{189}{125}=\frac{189}{5^3}=\frac{189}{2^0\times 5^3}$ we observe that the prime factorisation of the denominator of this rational number is of the form $2^{\mathrm{m}}\times 5^{\mathrm{n}}$, where $\mathrm{m},\mathrm{n}$ are non-negative integers. Hence, $\frac{189}{125}$ has terminating decimal expansion. (ii) $\frac{17}{6}=\frac{17}{2\times 3}$ we observe that the prime factorisation of the denominator of this rational number is not of the form $2^{\mathrm{m}}\times 5^{\mathrm{n}}$, where $\mathrm{m},\mathrm{n}$ are non-negative integers. Hence $\frac{17}{6}$ has nonterminating and repeating decimal expansion. (iii) $\frac{17}{8}=\frac{17}{2^3\times 5^0}$ So, the denominator 8 of $\frac{17}{8}$ is of the form $2^{\mathrm{m}}\times 5^{\mathrm{n}}$, where $\mathrm{m},\mathrm{n}$ are non-negative integers. Hence $\frac{17}{8}$ has terminating decimal expansion. (iv) $\frac{64}{455}=\frac{64}{5\times 7\times 13}$ Clearly, 455 is not of the form $2^{\mathrm{m}}\times 5^{\mathrm{n}}$. So, the decimal expansion of $\frac{64}{455}$ is non-terminating repeating.

Building Concepts 17 Without actually performing the long division, find if $\frac{987}{10500}$ will have terminating or non-terminating (repeating) decimal expansion. Give reason for your answer. Explanation: Terminating decimal expansion, because $\frac{987}{10500}=\frac{47}{500}$ and $500=5^3\times 2^2$. $\frac{987}{10500}=\frac{329}{2^2.5^3.7}=\frac{47}{2^2.5^3}=0.094$

Building Concepts 18 A rational number in its decimal expansion is 327.7081. What can you say about the prime factors of $q$, when this number is expressed in the form $\frac{p}{q}$ ? Give reasons. Explanation: Since 327.7081 is a terminating decimal number, so q must be in the form of $2^{\mathrm{m}}\times 5^{\mathrm{n}}$ where $\mathrm{m},\mathrm{n}$ are natural numbers.

Numerical Ability 4 The decimal expansion of the rational number $\frac{43}{\left(2^4\right)\left(5^3\right)}$ will terminate after how many places of decimals? Solution $\frac{43}{\left(2^4\right)\left(5^3\right)}=\frac{(43)(5)}{\left(2^4\right)\left(5^4\right)}=\frac{215}{10^4}=0.0215$ $\therefore$ Given rational number will terminate after four places of decimals.

Building Concepts 19 Why is it so that the denominator of the rational number must be in the form $2^m\times 5^n$ (where $\mathbf{m}$ and $\mathbf{n}$ are non-negative integers) so as to have the decimal expansion of that rational number as terminating? Explanation: Let $\frac{p}{q}$ be a rational number $(\mathrm{HCF}(p,q)=1)$ Let $q$ be not in the form of $2^{\mathrm{m}}.5^{\mathrm{n}}$ then it can be decomposed into prime factors other than 2 and 5 . Let's say $\mathrm{q}=3^{\mathrm{m}}.7^{\mathrm{n}}$ Then the rational number will not terminate because for any terminating decimal its denominator should be of the form $10^n$, where $\mathrm{n}\in {\mathrm{I}}_{+}$ e.g. $\frac{17\times 19}{3\times 7}=15.\overline{380952}$

7.0Memory Map