Chapter 1: Real Numbers

Algorithm: An algorithm is a series of well-defined steps which gives a procedure for solving a type of problem.
Lemma: A lemma is a proven statement used for proving another statement.
Theorem 1.1: (Euclid’s Division Lemma) Given positive integers a and b, there exist unique integers q and r satisfying a=bq + r, 0 ≤ r < b.

OR

Any positive integer a can be divided by another positive integer b in such a way that it leaves a remainder r that is smaller than b.

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1.         Show that every positive even integer is of the form 2q, and that every positive odd integer is of the form 2q + 1, where q is some integer.
Solution:
Let, a be any positive integer and b = 2.
Then, by Euclid’s algorithm,
a = 2q + r, for some integer q ≥ 0, and r = 0 or r = 1, because 0 ≤ r < 2.
So, a = 2q or 2q + 1.
If a is of the form 2q, then a is an even integer.
Also, a positive integer can be either even or odd.
⸫Any positive odd integer is of the form 2q + 1.

2.         Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.
Solution:
Let, a be a positive odd integer and b = 4.
Then, by Euclid’s algorithm,
a = 2q + r, for some integer q ≥ 0, and r = 0 or r = 1 or r = 2 or r = 3, because 0 ≤ r < 4.
That is, a can be 4q, or 4q + 1, or 4q + 2, or 4q + 3, where q is the quotient.
However, since a is odd, a cannot be 4q or 4q + 2 (since they are both divisible by 2).
Therefore, any odd integer is of the form 4q + 1 or 4q + 3.

3.         Show that any positive odd integer is of the form 6q+1, or 6q+3, or 6q+5, where q is some integer.
Solution:
Let a be any positive integer and b=6. Applying division Lemma with a and b=3, we have,
            a=6q+r, where 0≤r<6 and q is some integer
⸫r=0, 1, 2, 3, 4, 5
⇒a=6q+0 or a=6q+1 or a=6q+2 or a=6q+3 or a=6q+4 or a=6q+5
⇒a=6q or a=6q+1 or a=6q+2 or a=6q+3 or a=6q+4 or a=6q+5

Now,
6q=2×3q
6q+2=2(3q+1)
6q+4=2(3q+2)
⸫6q, 6q+2 and 6q+4 are even integers
Since, a is an odd positive integer, therefore a cannot be 6q, or 6q+2, or 6q+4
⸫a=6q+1 or a=6q+3 or a=6q+5.

4.         Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 for some integer m.
Solution:
Let a be any positive integer. Then, it is of the form 3q or 3q+1, or 3q+2. So, we have the following cases:

Case I: When a=3q
In this case, we have
            a² = (3q)² = 9q² = 3.3q² = 3m, where m=3q²
Case II: When a=3q+1
In this case, we have
            a² = (3q + 1)² = 9q² + 6q + 1 = 3(3q² + 2q) + 1= 3m + 1, where m=3q²+2q
Case III: When a=3q+2
In this case, we have
            a² = (3q + 2)² = 9q² + 12q + 4= 9q² + 12q + 3+1 = 3(3q² + 2q + 1) + 1= 3m + 1, where m=3q²+2q+1
⸫Any positive integer is either of the form 3m or 3m+1 for some integer m.

5.         Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, or 9m+1, or 9m+8.
Solution:
Let a be any positive integer. Then, it is of the form 3q or 3q+1, or 3q+2. So, we have the following cases:
Case I: When a=3q
In this case, we have
            a³ = (3q)³ = 27q³ = 9.3q³ = 9m, where m=3q³
Case II: When a=3q+1
In this case, we have
            a³ = (3q + 1)³ = 27q³ + 27q²+ 9q + 1 = 9(3q³ + 3q² + q) + 1= 3m + 1,
where m = 3q³ + 3q² + q
Case III: When a =3q+2
In this case, we have
            a³ = (3q + 1)³ = 27q³ + 54q²+ 36q + 8 = 9(3q³ + 6q² +4q) + 8= 3m + 8,
where m= 3q³ + 6q² +4q
⸫The cube of any positive integer is of the form 9m, or 9m+1, or 9m+8.

6.         Prove that the product of two consecutive positive integers is divisible by 2.
Solution:
Let a and b be any two consecutive positive integers.
Let a = m and b = m+1, for some integer m.
Now,
a×b=m(m+1)
We know that, any positive integer in of the form 2q or 2q+1 for some integer q.
Now, if m=2q, then
a×b=m(m+1) =2q(2q+1)
⸫a×b is divisible by 2.
Again, if m=2q+1, then
a×b=m(m+1)
      =(2q+1)(2q+1+1)
      =(2q+1)(2q+2)
      =2(2q+1)(q+1)
⸫a×b is divisible by 2.
⸫The product of two consecutive positive integers is divisible by 2.

7.         If a and b are two odd positive integers such that a > b , the prove that one of the two numbers $\frac{a+b}{2}$ and $\frac{a-b}{2}$ is odd and other is even.
Solution:
Given, a and b are two odd positive numbers and a>b.
Let, a=2m+1 and b=2n+1, where m and n are some positive integers.
Now,
$\frac{a+b}{2}=\frac{\left(2m+1\right)+\left(2n+1\right)}{2}=\frac{2m+1+2n+1}{2}=\frac{2m+2n+2}{2}=\frac{2\left(m+n+1\right)}{2}=m+n+1$
And
$\frac{a-b}{2}=\frac{\left(2m+1\right)-\left(2n+1\right)}{2}=\frac{2m+1-2n-1}{2}=\frac{2m-2n}{2}=\frac{2\left(m-n\right)}{2}=m-n$
Now, if m and n both are odd integers, then
$\begin{array}{l}m+n\text{ is even}\text{. }\left[\because \text{sum of two odd integers is an even integer}\right]\\ \Rightarrow m+n+1\text{ is odd }\left[\because \text{sum of a even number and 1 is odd}\right]\\ \therefore \frac{a+b}{2}\text{ is a odd number}\text{. And}\\ \therefore m-n\text{ is even }\left[\because \text{Difference of two odd number is an even number}\text{.}\right]\\ \therefore \frac{a-b}{2}\text{ is an even number}\text{.}\end{array}$
If m and n both are even integers, then
m + n is even.              [⸪Sum of two even integers is an even integer]
⇒ m + n + 1 is odd.    [⸪Sum of two even integers and 1 is an odd integer]
⇒$\frac{a+b}{2}$  is odd.
$\begin{array}{l}\therefore m-n\text{ is even }\left[\because \text{Difference of two even number is an even number}\text{.}\right]\\ \therefore \frac{a-b}{2}\text{ is an even number}\text{.}\end{array}$
If m is even and n is odd, then
$\begin{array}{l}\therefore m+n\text{ is odd}\text{. }\left[\because \text{sum of an even integer and a odd integer is an odd integer}\right]\\ \therefore m+n+1\text{ is even}\text{. }\left[\because \text{sum of a odd number and 1 is even}\right]\\ \therefore \frac{a+b}{2}\text{ is a even number}\text{. And}\\ \therefore m-n\text{ is odd }\left[\because \text{Difference of an even integer and a odd integer is an odd integer}\text{.}\right]\\ \therefore \frac{a-b}{2}\text{ is an odd number}\text{.}\end{array}$
If m is odd and n is even, then
$\begin{array}{l}\therefore m+n\text{ is even}\text{. }\left[\because \text{sum of an even integer and a odd integer is an odd integer}\right]\\ \therefore m+n+1\text{ is odd }\left[\because \text{sum of a odd number and 1 is even}\right]\\ \therefore \frac{a+b}{2}\text{ is a even number}\text{. And}\\ \therefore m-n\text{ is even }\left[\because \text{Difference of an even integer and a odd integer is an odd integer}\text{.}\right]\\ \therefore \frac{a-b}{2}\text{ is an odd number}\text{.}\end{array}$
Therefore, one of the two numbers $\frac{a+b}{2}$ and $\frac{a-b}{2}$  is odd and other is even.

8.         Show that the square of an odd positive integer is of the form 8q + 1, for some integer q.
Solution:
Let a be an odd integer and b=4. By division lemma there exists integers m and r such that,
a=4m+r, where 0≤r<4.
Therefore, r=0, 1, 2, 3. Then,
$a=4m\text{ or }4m+1\text{ or }4m+2\text{ or 4m+3}$
$\therefore a=4m+1\text{ or 4}m\text{+3}$                              [⸪4m and 4m+2 and even numbers.]
Now,
$a^2={\left(4m+1\right)}^{\text{2}}=16m^2+8m+1=8\left(2m^2+m\right)+1=8q+1,\text{ where q=}2m^2+m$
Or
$a^2={\left(\text{4}m\text{+3}\right)}^2=16m^2+24m+9=8\left(2m^2+3m+1\right)+1=8q+1,\text{ where }q\text{=}2m^2+3m+1$
⸫The square of an odd positive integer is of the form 8q+1, for some integer q.

9.         Show that any positive integer is of the form 3q or, 3q + 1 or 3q + 2, where q is some integer.
Solution:
Let a be an odd integer and b=3.
By division lemma there exists integers q and r such that,
a=3q+r, where 0≤r<3.
Therefore, r=0, 1, 2. Then,
$\Rightarrow a=3q\text{ or }3q+1\text{ or }3q+2$ , for some integer q.

10.       Show that n² – 1 is divisible by 8, if n is an odd positive integer.
Solution:
Given n is an odd integer and let, b=4.
By division lemma there exists integers m and r such that,
n=4m+r, where 0≤r<4.
Therefore, r=0, 1, 2, 3. Then,
$n=4m\text{ or }4m+1\text{ or }4m+2\text{ or 4m+3}$
$\therefore n=4m+1\text{ or 4}m\text{+3}$                               [⸪4m and 4m+2 and even numbers.]
If, n=4m + 1, then
n²– 1= (4m + 1)² – 1
         = 16m² + 8m + 1 – 1
         = 16m² + 8m
         = 8(2m² + m)
n²– 1 is divisible by 8.
If, n=4m + 3, then
n²– 1= (4m + 3)² – 1
         = 16m² + 24m + 9 – 1
         = 16m² + 24m +8
         = 8(2m² + 3m + 1)
⸫n² – 1 is divisible by 8.

11.       Prove that if x and y are both odd positive integers, then x² + y² is even but not divisible by 4.
Solution:
We know that any odd positive integer is of the form 2q + 1, for some integer q.
Let, x = 2m + 1 and y = 2n + 1, for some integers m and n.
Now,
x² + y² = (2m + 1)² + (2n + 1)² = 4m² + 4m + 1 + 4n² + 4n + 1  = 4(m² + m  + n² + n) + 2
⸫x² + y² is even and leaves remainder 2 when divided by 4.
⸫x² + y² is even but not divisible by 4.

12.       Show that the square of an odd positive integer is of the form 8m + 1, for some whole number m.
Solution:
Any positive odd integer is of the form 2q + 1, where q is an integer.
⸫(2q + 1)² = 4q² + 4q + 1 = 4q (q + 1) + 1 ………………………..(1)
q (q + 1) is either 0 or even. So, it is 2m, where m is a whole number.
⸫(2q + 1)² = 4.2 m + 1 = 8m + 1. [From (1)]

13.       Prove that n²– n is divisible by 2 for every positive integer n.
Solution:
We know that any positive integer n is of the form 2q or 2q + 1, for some integer q.
If n=2q, then
n²– n = (2q)² - 2q
          =4q² - 2q
          =2(2q²– q)
⸫n²– n is divisible by 2.
If n =2q + 1, then
n²– n = (2q + 1)² – (2q + 1)
          =4q² + 4q +1 - 2q - 1
          =4q² + 2q
          =2(2q² + q)
⸫n²– n is divisible by 2.
⸫n²– n is divisible by 2 for every positive integer n.

14.       Prove that one of every three consecutive positive integers is divisible by 3.
Solution:
Let, n, n +1, n + 2 be three consecutive positive integers.
We know that n is of the form of 3q, or 3q + 1, or 3q + 2
If n = 3q, then
n is divisible by 3 but n + 1 and n + 2 are not divisible by 3.
If n = 3q + 1, then
n is not divisible by 3
n + 1 = 3q + 1 +1
         = 3q + 2
⸫n + 1 is not divisible by 3
n + 2 = 3q + 2 +1
         = 3q + 3
         = 3(q + 1)
⸫n + 2 is divisible by 3.
If n = 3q + 2, then
n is not divisible by 3
n + 1 = 3q + 2 +1
         = 3q + 3
         = 3(q + 1)
⸫n + 1 is divisible by 3
n + 2 = 3q + 2 + 2
         = 3q + 4
⸫n + 2 is not divisible by 3.
⸫One of every n, n +1, n + 2 is divisible by 3.

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To obtain the HCF of two positive integers, say c and d, with c > 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 c=dq + r, 0≤ r < d.
Step 2: If r = 0, d is the HCF of c and d. If r ≠ 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.
Where, HCF (c, d) denotes the HCF of c and d.
1.         Use Euclid’s division algorithm to find the HCF of:
(i)         135 and 225
Solution:
Given integers are 135 and 225.
Applying Euclid’s Division lemma to 135 and 225, we get
225 = 135 × 1 + 90
135 = 90 × 1 + 45
            90 = 45 × 2 + 0
The remainder at the last stage is zero.
Therefore, 45 is the H. C. F. of 135 and 225
(ii)        196 and 38220
Solution:
Given integers are 196 and 38220.
Applying Euclid’s Division lemma to 196 and 38220, we get
38220 = 196 × 195 + 0
The remainder at the last stage is zero.
⸫196 is the H. C. F. of 196 and 38220.
(iii)       867 and 255
Solution:
Given integers are 867 and 255.
Applying Euclid’s Division lemma to 867 and 255, we get
867 = 255 × 3 + 102
255 = 102 × 2 + 51
102 = 51 × 2 + 0
The remainder at the last stage is zero.
⸫51 is the H. C. F. of 867 and 255.
(iv)       4052 and 12576
Solution:
Given integers are 4052 and 12576.
Applying Euclid’s Division lemma to 4052 and 12576, we get
12576 = 4052 × 3 + 420
4052 = 420 × 9 + 272
            420 = 272 × 1 + 148
            272=148 × 1+124
            148=124 × 1+24
            124=24 × 5+4
            24=4 × 6+0
The remainder at the last stage is zero.
⸫4 is the H. C. F. of 4052 and 12576

3.         An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of column in which they can march?
Solution:
The maximum number of column is the HCF of 616 and 32. Using Euclid’s Algorithm, we have
616 = 32 × 19 + 8
32 = 8 × 4 + 0
⸫The maximum number of column is 8.

4.         A sweetseller has 420 kaju barfis and 130 badam barfis. She wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray. What is the number of that can be placed in each stack for this purpose?
Solution:
The maximum number of barfis in each stack: HCF (420, 130).
Now, Applying Euclid’s algorithm, we have:
420 = 130 × 3 + 30
130 = 30 × 4 + 10
30 = 10 × 3 + 0
So, the HCF of 420 and 130 is 10.
⸫The sweetseller can make stacks of 10 for both kinds of barfi

Theorem 1.2: (Fundamental Theorem of Arithmetic) Every composite number can be expressed as a product of primes in a unique way

** The prime factorisation of a natural number is unique, except for the order of its factors.

HCF(a, b) = Product of the smallest power of each common prime factor in a and b.
LCM(a, b) = Product of the greatest power of each prime factor in a and b.

LCM × HCF = product of the two numbers.

Example 5: Consider the numbers 4ⁿ, where n is a natural number. Check whether there is any value of n for which 4ⁿ ends with the digit zero.
Solution:
If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5
4ⁿ = (2²)ⁿ = 2²ⁿ
The only prime in the factorization of 4ⁿ is 2 not 5.
Hence, it cannot end with the digit 0.

Q.        Can the number 6ⁿ, n being a natural number, end with the digit 5? Give reasons.
Solution:
No, because 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ, so the only primes in the factorization of 6ⁿ are 2 and 3, and not 5.
Hence, it cannot end with the digit 5.

Q.        Check whether 6ⁿ can end with the digit 0 for any natural number n.
Solution:
If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5
Prime factorisation of 6ⁿ = (2 ×3)ⁿ
It can be observed that 5 is not in the prime factorisation of 6ⁿ.
Hence, for any value of n, 6ⁿ will not be divisible by 5.
Therefore, 6ⁿ cannot end with the digit 0 for any natural number n.

Example 6: Find the LCM and HCF of 6 and 20 by the prime factorisation method.
Solution:
We have,
6 = 2 × 3
20 = 2 × 2 × 5 = 22 × 5.
HCF(6, 20) = 2
LCM(6, 20) = 2 × 2 × 3 × 5 = 60

Example 7: Find the HCF of 96 and 404 by the prime factorisation method. Hence, find their LCM.
Solution:
We have,
96 = 25 × 3
404 = 22 × 101
Therefore, the HCF of these two integers is 22 = 4.
Also, LCM (96, 404) = $LCM(96,404)=\frac{96\times 404}{HCF(96,404)}=\frac{96\times 404}{4}=9696$

Example 8: Find the HCF and LCM of 6, 72 and 120, using the prime factorization method.
Solution:
We have,
6 = 2 × 3
72 = 2³ × 3²
120 = 2³× 3 × 5
Here, 2¹ and 3¹ are the smallest powers of the common factors 2 and 3, respectively.
So, HCF (6, 72, 120) = 2¹ × 3¹ = 2 × 3 = 6
2³, 3² and 5¹ are the greatest powers of the prime factors 2, 3 and 5 respectively
So, LCM (6, 72, 120) = 2³× 3²× 5¹ = 360

EXERCISE 1.2

1.         Express each number as a product of its prime factors:
(i)         140
Solution:
140 = 2²× 5 × 7

2.         Find the LCM and HCF of the following pairs of integers and verify that,
            LCM × HCF = product of the two numbers.
(i)         26 and 91
Solution:
26 = 2 × 13
91 = 7 × 13
HCF( 26, 91) = 13
LCM(26, 91) = 2 × 13 × 7 = 182
LCM × HCF = 13 × 182 = 2366
26 × 91 = 2366
⸫LCM × HCF = product of the two numbers.

3.         Find the LCM and HCF of the following integers by applying the prime factorization method.
(i)         12, 15 and 21
Solution:
12 = 2 × 2 × 3
15 = 3 × 5
21 = 3 × 7
HCF (12, 15, 21) = 3
LCM (12, 15, 21) =2 × 2 × 3 × 5 × 7 = 420
4.         Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution:
Given, HCF (306, 657) = 9
$\therefore \text{LCM }\left(\text{3}0\text{6},\text{ 657}\right)=\frac{306\times 657}{HCF(306,657)}=\frac{201042}{9}=22338$

6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution:
7 × 11 × 13 + 13 = (7 × 11 + 1)13
⸪13 is a factor of 7 × 11 × 13 + 13.
⸫7 × 11 × 13 + 13 is a composite number.
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = (7 × 6 × 4 × 3 × 2 × 1 +1) 5
⸪5 is a factor of 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5.
⸫7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number.

7.         There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Solution:
The required number of minutes is LCM (18, 12).
18 = 2 × 3 × 3
12 = 2 × 2 × 3
⸫LCM (18, 12) = 2² × 3² = 36.
⸫Sonia and Ravi will meet the starting point after 36 minutes.

***

Irrational numbers: A number is called irrational if it cannot be written in the form $\frac{p}{q}$ , where p and q are integers and q ≠ 0.

Theorem 1.3: Let p be a prime number. If p divides a², then p divides a, where a is a positive integer.

Theorem 1.4: √2 is irrational.
Proof: Let us assume, to the contrary, that √2 is rational.
So, we can find integers a and b (s≠ 0) such that,
√2 =$\frac{a}{b}$ , where a and b are coprime.
⇒b√2=a
⇒2b²=a² ……………………….. (1)
⸫2 is a factor of a².
⇒2 is a factor of a.
Let, a = 2c for some integer c.
From (1) we get,
2b² = 4c²
⇒b² = 2c².
⸫2 is a factor of b²
⇒2 is a factor of b
⸫a and b have at least 2 as a common factor.
But this contradicts the fact that a and b have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that √2 is rational.
So, we conclude that √2 is irrational.
Example 9: Prove that √3 is irrational.
Solution:
Let us assume, to the contrary, that √3 is rational.
So, we can find integers a and b (s≠ 0) such that,
√3 =$\frac{a}{b}$ , where a and b are coprime.
⇒b√3=a
⇒3b²=a² ……………………….. (1)
⸫3 is a factor of a².
⇒3 is a factor of a.
Let, a = 3c for some integer c.
From (1) we get,
3b² = 9c²
⇒b² = 3c².
⸫3 is a factor of b²
⇒3 is a factor of b
⸫a and b have at least 3 as a common factor.
But this contradicts the fact that a and b have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that √3 is rational.
So, we conclude that √3 is irrational.

Example 10: Show that 5 – √3 is irrational.
Solution:
Let us assume, to the contrary, that 5 – √3 is rational.
That is, we can find coprime a and b (b ≠ 0) such that,
5 – √3=$\frac{a}{b}$ , where a and b are coprime
⇒√3=5-$\frac{a}{b}$
⇒√3=$\frac{\mathrm{5b\; -\; a}}{b}$
Since 5, a and b are integers, we get,
$\frac{\mathrm{5b\; -\; a}}{b}$is rational
⇒√3 is rational.
But this contradicts the fact that √3 is irrational.
This contradiction has arisen because of our incorrect assumption that 5 – √3 is rational.
So, we conclude that 5 − √3 is irrational.

Example 11: Show that 3√2 is irrational.
Solution:
Let us assume, to the contrary, that 3√2 is rational.
⸫We can find coprime a and b (b ≠ 0), such that,
3√2==$\frac{a}{b}$
⇒√2=$\frac{a}{3b}$
⸪3, a and b are integers,
⸫$\frac{a}{3b}$ is rational
⇒√2 is rational.
But this contradicts the fact that √2 is irrational.
So, we conclude that 3√2 is irrational.

12.       Prove that √2 + √3 is irrational.
Solution:
Let us suppose that √2 + √3 is rational. Let √2 + √3 = a, where a is rational.
Therefore, √2 = a − √3
Squaring on both sides, we get
2 = a² + 3 – 2a√3
⇒2 – a²– 3 = – 2a√3
⇒– a²– 1 = – 2a√3
$\Rightarrow \frac{a^2+1}{2a}=\sqrt{3}$
⸪ a, 1 and 2 are integers.
⸫$\frac{a^2+1}{2a}$ is a rational number.
Therefore, $\sqrt{3}$  is a rational number, which is a contradiction as $\sqrt{3}$ is a irrational number
Hence, √2 + √3 is irrational.
                              
Theorem 1.5: Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form $\frac{p}{q}$ , where p and q are coprime, and the prime factorisation of q is of the form 2ⁿ5^m, where n, m are non-negative integers.

Theorem 1.6: Let x = $\frac{p}{q}$  be a rational number, such that the prime factorization of q is of the form 2ⁿ5^m, where n, m are non-negative integers. Then x has a decimal expansion which terminates.

Theorem 1.7: Let x =$\frac{p}{q}$  be a rational number, such that the prime factorization of q is not of the form 2ⁿ5^m, where n, m are non-negative integers. Then, x has a decimal expansion which is non-terminating repeating (recurring).

EXERCISE 1.4

1.         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)           $\frac{13}{3125}$
Answer:
$\frac{13}{3125}$ =$\frac{13}{5^5}$
Since, the denominator has only 5 as its factor; it has a terminating decimal expansion.
(ii)          $\frac{17}{8}$
Answer:
$\frac{17}{8}$ =$\frac{17}{2^3}$
Since, the denominator has only 2 as its factor; it has a terminating decimal expansion.
(iii)         $\frac{64}{455}$
Answer:
$\frac{64}{455}$ =$\frac{64}{5\times 7\times 13}$
Since, the denominator has factors other than 2 and 5; it has a non-terminating decimal expansion.