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.

 

CHAPTER 1: NUMBER SYSTEMS

CHAPTER 1: NUMBER SYSTEMS

Rational number: A number ‘r’ is called a rational number, if it can be written in the form of  where p and q are integers and q ≠ 0.

Example 1: Are the following statements true or false? Give reasons for your answers.
(i)    Every whole number is a natural number.
(ii)    Every integer is a rational number.
(iii)    Every rational number is an integer.

Solution:
(i) False, because zero is a whole number but not a natural number.
(ii) True, because every integer m can be expressed in the form , and so it is a rational number.
(iii) False, because is not an integer.

Exercise 1.1

1.    Is zero a rational number? Can you write it in the form , where p and q are integers and q ≠ 0 ?
Answer: Zero is a rational number.
Zero can be written as  , or , or  etc.

2.    Find six rational numbers between 3 and 4.
Solution:

Now,

3.    Find five rational numbers between  and
Solution:


Now,



4.    State whether the following statements are true or false. Give reasons for your answers.
(i)    Every natural number is a whole number.
Answer: True, because whole number contains zero and all the natural numbers.

(ii)    Every integer is a whole number.
Answer: False, because -1 is an integer but not a whole number.

(iii)    Every rational number is a whole number.
Answer: False, because is a rational number but not a whole number.

Additional questions
Q. Look at several examples of rational numbers in the form of  (q ≠ 0), where p and q are integers with no common factor other than 1 and having terminating decimal  representations. Can you guess what property q must satisfy?
Solution: A rational number  is a terminating decimal only, when prime factorization of q must have only powers of 2 or 5 or both.

1. The Fun They Had

1. The Fun They Had

BEFORE YOU READ

• The story we shall read is set in the future, when books and schools as we now know them will perhaps not exist. How will children study then? The diagram below may give you some ideas.

• In pairs, discuss three things that you like best about your school and three things about your school that you would like to change. Write them down.

• Have you ever read words on a television (or computer) screen? Can you imagine a time when all books will be on computers, and there will be no books printed on paper? Would you like such books better?

1. MARGIE even wrote about it that night in her diary. On the page headed 17 May 2157, she wrote, “Today Tommy found a real book!”

It was a very old book. Margie’s grandfather once said that when he was a little boy his grandfather told him that there was a time when all stories were printed on paper.

They turned the pages, which were yellow and crinkly¹, and it was awfully funny to read words that stood still instead of moving the way they were supposed to — on a screen, you know. And then when they turned back to the page before, it had the same words on it that it had had when they read it the first time.

2. “Gee,” said Tommy, “what a waste. When you’re through with the book, you just throw it away, I guess. Our television screen must have had a million books on it and it’s good for plenty more. I wouldn’t throw it away.”

“Same with mine,” said Margie. She was eleven and hadn’t seen as many telebooks as Tommy had. He was thirteen.

She said, “Where did you find it?”

“In my house.” He pointed without looking, because he was busy reading. “In the attic².”

“What’s it about?”

“School.”

3. Margie was scornful³. “School? What’s there to write about school? I hate school.”

Margie always hated school, but now she hated it more than ever. The mechanical teacher had been giving her test after test in geography and she had been doing worse and worse until her mother had shaken her head sorrowfully and sent for the County Inspector.

4. He was a round little man with a red face and a whole box of tools with dials and wires. He smiled at Margie and gave her an apple, then took the teacher apart. Margie had hoped he wouldn’t know how to put it together again, but he knew how all right, and, after an hour or so, there it was again, large and black and ugly, with a big screen on which all the lessons were shown and the questions were asked. That wasn’t so bad. The part Margie hated most was the slot⁴ where she had to put homework and test papers. She always had to write them out in a punch code they made her learn when she was six years old, and the mechanical teacher calculated the marks in no time.

5. The Inspector had smiled after he was finished and patted Margie’s head. He said to her mother, “It’s not the little girl’s fault, Mrs Jones. I think the geography sector was geared ⁵a little too quick. Those things happen sometimes. I’ve slowed it up to an average ten-year level. Actually, the overall pattern of her progress is quite satisfactory.” And he patted Margie’s head again.

Margie was disappointed. She had been hoping they would take the teacher away altogether. They had once taken Tommy’s teacher away for nearly a month because the history sector had blanked out completely.

So she said to Tommy, “Why would anyone write about school?”

6. Tommy looked at her with very superior eyes. “Because it’s not our kind of school, stupid. This is the old kind of school that they had hundreds and hundreds of years ago.” He added loftily⁶, pronouncing the word carefully, “Centuries ago.”

Margie was hurt. “Well, I don’t know what kind of school they had all that time ago.” She read the book over his shoulder for a while, then said, “Anyway, they had a teacher.”

“Sure they had a teacher, but it wasn’t a regular ⁷teacher. It was a man.”

“A man? How could a man be a teacher?”

“Well, he just told the boys and girls things and gave them homework and asked them questions.”

7. “A man isn’t smart enough.”

“Sure he is. My father knows as much as my teacher.”

“He knows almost as much, I betcha⁸.”

Margie wasn’t prepared to dispute⁹ that. She said, “I wouldn’t want a strange man in my house to teach me.”

Tommy screamed with laughter. “You don’t know much, Margie. The teachers didn’t live in the house. They had a special building and all the kids went there.”

“And all the kids learned the same thing?”

“Sure, if they were the same age.”

8. “But my mother says a teacher has to be adjusted to fit the mind of each boy and girl it teaches and that each kid has to be taught differently.”

“Just the same they didn’t do it that way then. If you don’t like it, you don’t have to read the book.”

“I didn’t say I didn’t like it,” Margie said quickly. She wanted to read about those funny schools.

They weren’t even half finished when Margie’s mother called, “Margie! School!”

Margie looked up. “Not yet, Mamma.”

“Now!” said Mrs Jones. “And it’s probably time for Tommy, too.”

Margie said to Tommy, “Can I read the book some more with you after school?”

9. “May be,” he said nonchalantly¹⁰. He walked away whistling, the dusty old book tucked beneath his arm.

Margie went into the schoolroom. It was right next to her bedroom, and the mechanical teacher was on and waiting for her. It was always on at the same time every day except Saturday and Sunday, because her mother said little girls learned better if they learned at regular hours.

The screen was lit up, and it said: “Today’s arithmetic lesson is on the addition of proper fractions. Please insert yesterday’s homework in the proper slot.”

10. Margie did so with a sigh. She was thinking about the old schools they had when her grandfather’s grandfather was a little boy. All the kids from the whole neighborhood came, laughing and shouting in the schoolyard, sitting together in the schoolroom, going home together at the end of the day. They learned the same things, so they could help one another with the homework and talk about it.

And the teachers were people…

The mechanical teacher was flashing on the screen: “When we add fractions ½ and ¼...”

Margie was thinking about how the kids must have loved it in the old days. She was thinking about the fun they had.

ISAAC ASIMOV

Thinking about the Text

Activity

Calculate how many years and months ahead from now Margie’s diary entry is.

I. Answer these questions in a few words or a couple of sentences each.

1. How old are Margie and Tommy?

2. What did Margie write in her diary?

3. Had Margie ever seen a book before?

4. What things about the book did she find strange?

5. What do you think a telebook is?

6. Where was Margie’s school? Did she have any classmates?

7. What subjects did Margie and Tommy learn?

II. Answer the following with reference to the story.

1. “I wouldn’t throw it away.”

(i) Who says these words?

(ii) What does ‘it’ refer to?

(iii) What is it being compared with by the speaker?

2. “Sure they had a teacher, but it wasn’t a regular teacher. It was a man.”

(i) Who does ‘they’ refer to?

(ii) What does ‘regular’ mean here?

(iii) What is it contrasted with?

III. Answer each of these questions in a short paragraph (about 30 words).

1. What kind of teachers did Margie and Tommy have?

2. Why did Margie’s mother send for the County Inspector?

3. What did he do?

4. Why was Margie doing badly in geography? What did the County Inspector do to help her?

5. What had once happened to Tommy’s teacher?

6. Did Margie have regular days and hours for school? If so, why?

7. How does Tommy describe the old kind of school?

8. How does he describe the old kind of teachers?

IV. Answer each of these questions in two or three paragraphs (100 –150 words).

1. What are the main features of the mechanical teachers and the schoolrooms that Margie and Tommy have in the story?

2. Why did Margie hate school? Why did she think the old kind of school must have been fun?

3. Do you agree with Margie that schools today are more fun than the school in the story? Give reasons for your answer.

Thinking about Language

I. Adverbs

Read this sentence taken from the story:

They had once taken Tommy’s teacher away for nearly a month because the history sector had blanked out completely.

The word complete is an adjective. When you add –ly to it, it becomes an adverb.

1. Find the sentences in the lesson which have the adverbs given in the box below.

awfully	sorrowfully	completely	loftily
carefully	differently	quickly	nonchalantly

2. Now use these adverbs to fill in the blanks in the sentences below.

(i) The report must be read____________ so that performance can be improved.

(ii) At the interview, Sameer answered our questions__________ , shrugging his shoulders.

(iii) We all behave_______________ when we are tired or hungry.

(iv) The teacher shook her head _________________when Ravi lied to her.

(v) I______________ forgot about it.

(vi) When I complimented Revathi on her success, she just smiled __________and turned away.

(vii) The President of the Company is________________ busy and will not be able to meet you.

(viii) I finished my work _____________so that I could go out to play.

Remember:

An adverb describes action. You can form adverbs by adding –ly to adjectives.

Spelling Note: When an adjective ends in –y, the y changes to i when you add –ly to form an adverb.

For example: angr-y → angr-i-ly

3. Make adverbs from these adjectives.

(i) angry _________________ (ii) happy_________________

(iii) merry _________________ (iv) sleepy_________________

(v) easy _________________ (vi) noisy_________________

(vii) tidy _________________ (viii) gloomy_________________

II. If Not and Unless

• Imagine that Margie’s mother told her, “You’ll feel awful if you don’t finish your history lesson.”

• She could also say: “You’ll feel awful unless you finish your history lesson.”

Unless means if not. Sentences with unless or if not are negative conditional sentences.

Notice that these sentences have two parts. The part that begins with if not or unless tells us the condition. This part has a verb in the present tense (look at the verbs don’t finish, finish in the sentences above).

The other part of the sentence tells us about a possible result. It tells us what will happen (if something else doesn’t happen). The verb in this part of the sentence is in the future tense (you’ll feel/you will feel ).

Notice these two tenses again in the following examples.

Future Tense		Present Tense
• There won’t be any books	left	unless we preserve them.
• You won’t learn your lessons	if	you don’t study regularly.
• Tommy will have an accident	unless	he drives more slowly.

Complete the following conditional sentences. Use the correct form of the verb.

1. If I don’t go to Anu’s party tonight, _________________

2. If you don’t telephone the hotel to order food, _________________

3. Unless you promise to write back, I_________________

4. If she doesn’t play any games, _________________

5. Unless that little bird flies away quickly, the cat_________________

Writing

A new revised volume of Issac Asimov’s short stories has just been released. Order one set. Write a letter to the publisher, Mindfame Private Limited, 1632 Asaf Ali Road, New Delhi, requesting that a set be sent to you by Value Payable Post (VPP), and giving your address. Your letter will have the following parts.

• Addresses of the sender and receiver

• The salutation

• The body of the letter

• The closing phrases and signature

Your letter might look like this:

Your address

_________________

_________________

_________________

Date _________________ (DD/MM/YY)

The addressee’s address

_________________

_________________

_________________

Dear Sir/Madam,

____________________________________________________________________

____________________________________________________________________

____________________________________________________________________

____________________________________________________________________

Yours sincerely,

Your signature

Remember that the language of a formal letter is different from the colloquial style of personal letters. For example, contracted forms such as ‘I’ve’ or ‘can’t’ are not used.

Speaking

In groups of four discuss the following topic.

‘The Schools of the Future Will Have No Books and No Teachers!’

Your group can decide to speak for or against the motion. After this, each group will select a speaker to present its views to the entire class.

You may find the following phrases useful to present your argument in the debate.

• In my opinion . . .

• I/we fail to understand why . . .

• I wholeheartedly support/oppose the view that . . .

• At the outset let me say . . .

• I’d/we’d like to raise the issue of/argue against . . .

• I should like to draw attention to . . .

• My/our worthy opponent has submitted that . . .

• On the contrary . . .

• I firmly reject . . .

False science creates atheists; true science prostrates Man before divinity.

VOLTAIRE

1 crinkly: with many folds or lines

2 attic: a space just below the roof, used as a storeroom

3 scornful: contemptuous; showing you think something is worthless

4 slot: a given space, time or position

5 geared (to): adjusted to a particular standard or level

6 loftily: in a superior way

7 regular: here, normal; of the usual kind

8 betcha (informal): (I) bet you (in fast speech): I’ m sure

9 dispute: disagree with

10 nonchalantly: not showing much interest or enthusiasm; carelessly