Contents
Real Numbers
Question 1:
Exercise 1A
The numbers of the form
Ten examples of rational numbers are:
, , , , , , , , 1,
Question 2:
 5
 3
(v) 1.3
(vii)
Question 3:
Therefore, we have
Therefore, three rational numbers between
, and
Question 5:
Let
Then, x < y because
Or we can say that,
We know that, 8 < 9 < 10 < 11 < 12 < 13 < 14 < 15.
Therefore, we have,
, , , and
Question 6:
Let x = 3 and y = 4
Then, x < y, because 3 < 4 We can say that,
We know that, 21 < 22 < 23 < 24 < 25 < 26 < 27 < 28.
Therefore, we have,
, , , and
Question 7:
Let x = 2.1 and y = 2.2
Then, x < y because 2.1 < 2.2 Or we can say that,
That is, we have,
We know that, 2100 < 2105 < 2110 < 2115 < 2120 < 2125 < 2130 < 2135 < 2140 <
2145 < 2150 < 2155 < 2160 < 2165 < 2170 < 2175 < 2180 < 2185 < 2190 < 2195 <
2200
Therefore, 16 rational numbers between, 2.1 and 2.2 are:
So, 16 rational numbers between 2.1 and 2.2 are:
2.105, 2.11, 2.115, 2.12, 2.125, 2.13, 2.135, 2.14, 2.145, 2.15, 2.155, 2.16, 2.165, 2.17,
2.175, 2.18
Exercise 1B
Question 1:
If the prime factors of the denominator are 2 and/or 5 then the rational number is a terminating decimal.
Since, 80 has prime factors 2 and 5,
If the prime factors of the denominators of the fraction are other than 2 and 5, then the rational number is not a terminating decimal.
Since, 24 has prime factors 2 and 3 and 3 is different from 2 and 5,
If the prime factors of the denominators of the fraction are other than 2 and 5, then the rational number is not a terminating decimal.
Since 12 has prime factors 2 and 3 and 3 is different from 2 and 5,
If the prime factors of the denominators of the fraction are other than 2 and 5, then the rational number is not a terminating decimal.
Since 35 has prime factors 5 and 7, and 7 is different from 2 and 5,
If the prime factors of the denominator are 2 and/or 5 then the rational number is a terminating decimal.
Since 125 has prime factor 5 only
Question 2:
Question 3:

Let x =
i.e x = 0.333 …. (i)
⇒ 10x = 3.333 …. (ii) Subtracting (i) from (ii), we get 9x = 3

Let x = 1.
i.e x = 1.333 …. (i)
⇒10x = 13.333 …. (ii)
Subtracting (i) from (ii) we get; 9x = 12
⇒ x =

Let x = 0.
i.e x = 0.3434 …. (i)
⇒ 100x = 34.3434 …. (ii)
Subtracting (i) from (ii), we get 99x = 34
⇒ x =

Let x = 3.
i.e x = 3.1414 …. (i)
⇒ 100x = 314.1414 …. (ii)
Subtracting (i) from (ii), we get 99x = 311
⇒ x =

Let x = 0.
i.e. x = 0.324324 ….(i)
⇒ 1000x = 324.324324….(ii)
Subtracting (i) from (ii), we get 999x = 324
⇒ x =

Let x = 0.
i.e. x = 0.177 …. (i)
⇒ 10x = 1.777 …. (ii)
and 100x = 17.777…. (iii) Subtracting (ii) from (iii), we get 90x = 16
⇒ x =

Let x = 0.
i.e. x = 0.544 …. (i)
⇒ 10 x = 5.44 …. (ii) and 100x = 54.44 ….(iii)
Subtracting (ii) from (iii), we get 90x = 49
⇒ x =
⇒ 10x = 1.6363 …. (ii)
and 1000 x = 163.6363 …. (iii) Subtracting (ii) from (iii), we get
990x = 162
Question 4:
 True. Since the collection of natural number is a sub collection of whole numbers, and every element of natural numbers is an element of whole numbers
 False. Since 0 is whole number but it is not a natural number.
 True. Every integer can be represented in a fraction form with denominator 1.

False. Since division of whole numbers is not closed under division, the value of
, p and q are integers and q ≠ 0, may not be a whole number.  True. The prime factors of the denominator of the fraction form of terminating
decimal contains 2 and/or 5, which are integers and are not equal to zero.
 True. The prime factors of the denominator of the fraction form of repeating decimal contains integers, which are not equal to zero.

True. 0 can considered as a fraction
, which is a rational number.
Question 1:
Exercise 1C
Irrational number: A number which cannot be expressed either as a terminating decimal or a repeating decimal is known as irrational number. Rather irrational numbers cannot
be expressed in the fraction form,
For example, 0.101001000100001 is neither a terminating nor a repeating decimal and so is an irrational number.
Also,
Question 2:
(i)
We know that, if n is a perfect square, then
(ii)
We know that, if n is a perfect square, then
(iii)
We know that, if n is a not a perfect square, then
(iv)
We know that, if n is a not a perfect square, then
(v)
Theorem: The sum of a rational number and an irrational number is an irrational number.
So by the above theorem, the sum,
(vi)
Theorem: The sum of a rational number and an irrational number is an irrational number.
So by the above theorem, the sum,
So,
(vii)
Theorem: The product of a nonzero rational number and an irrational number is an irrational number.
Thus, by the above theorem,
So,

0.
Every rational number can be expressed either in the terminating form or in the non terminating, recurring decimal form.
Therefore, 0.
Question 3:
Let X’OX be a horizontal line, taken as the xaxis and let O be the origin. Let O represent 0.
Take OA = 1 unit and draw BA ⊥ OA such that AB = 1 unit, join OB. Then,
With O as centre and OB as radius, drawn an arc, meeting OX at P. Then, OP = OB =
Thus the point P represents
Now draw BC ⊥ OB such that BC = 1 units Join OC. Then,
With O as centre and OC as radius, draw an arc, meeting OX at Q. The, OQ = OC =
Thus, the point Q represents
Now draw DE ⊥ OD such that DE = 1 units Join OE. Then,
With O as centre and OE as radius draw an arc, meeting OX at R. Then, OR = OE =
Thus, the point R represents
Question 4:
Draw horizontal line X’OX taken as the xaxis Take O as the origin to represent 0.
Let OA = 2 units and let AB ⊥ OA such that AB = 1 units
With O as centre and OB as radius draw an arc meeting OX at P. Then, OP = OB =
Now draw BC ⊥ OB and set off BC = 1 unit Join OC. Then,
With O as centre and OC as radius, draw an arc, meeting OX at Q. Then, OQ = OC =
Thus, Q represents
With O as centre and OD as radius, draw an arc, meeting OX at R. Then OR = OD =
Thus, R represents
Question 5:
(i)
Since 4 is a rational number and
So,
(ii)
Since – 3 is a rational number and
So,
(iii)
Since 5 is a rational number and
So,
(iv)
Since 3 is a rational number and
So,
Question 6:
 True
 False
 True
 False
 True
 False
 False
 True
 True
Exercise 1D
Question 1:
We have:
We have:
Question 2:
Question 3:
(ii)
Question 4:
Draw a line segment AB = 3.2 units and extend it to C such that BC = 1 units. Find the midpoint O of AC.
With O as centre and OA as radius, draw a semicircle. Now, draw BD AC, intersecting the semicircle at D. Then, BD =
With B as centre and BD as radius, draw an arc meeting AC produced at E. Then, BE = BD =
Question 6:
Draw a line segment AB = 7.28 units and extend it to C such that BC = 1 unit. Find the midpoint O of AC.
With O as centre and OA as radius, draw a semicircle.
Now, draw BD AC, intersecting the semicircle at D. Then, BD =
With D as centre and BD as radius, draw an arc, meeting AC produced at E. Then, BE = BD =
Question 7:
Closure Property: The sum of two real numbers is always a real number. Associative Law: (a + b) + c = a + (b + c) for al real numbers a, b, c.
Commutative Law: a + b = b + a, for all real numbers a and b.
Existence of identity: 0 is a real number such that 0 + a = a + 0, for every real number a. Existence of inverse of addition: For each real number a, there exists a real number (a) such that
a + (a) = (a) + a= 0
a and (a) are called the additive inverse of each other. Existence of inverse of multiplication:
a and
Question 1:
Exercise 1E
Question 2:
Question 3:
Question 5:
Question 9:
Question 12:
Question 14:
Question 15:
Question 17:
Exercise 1F
Question 1:
(ii)
Question 2:
(iii)
Question 3:
(ii)
Question 4:
(ii)
Question 5:
(ii)
Question 6:
(ii)
(iii)
Question 7:
(ii)