SELINA Solutions for Class 9 Maths Chapter 1 - Rational and Irrational Numbers
Chapter 1 - Rational and Irrational Numbers Exercise Ex. 1(A)
Question 1
Is zero a rational number? Can it
be written in the form 
, where p and q are integers and q≠0?
Solution 1
Yes, zero is a rational number.
As it can be
written in the form of , where p and q are integers and q≠0?
⇒ 0 = 
Question 2
Are the following statements true or false? Give reasons for your answers.
i. Every whole number is a natural number.
ii. Every whole number is a rational number.
iii. Every integer is a rational number.
iv. Every rational number is a whole number.
Solution 2
i. False, zero is a whole number but not a natural number.
ii. True, Every
whole can be written in the form of , where p and q are integers and q≠0.
iii. True, Every integer can be written in
the form of , where p and q are integers and q≠0.
iv. False.
Example: 
is a rational
number, but not a whole number.
Question 3
Arrange 
in ascending order of their magnitudes.
Also, find the difference between the largest and smallest of these fractions. Express this difference as a decimal fraction correct to one decimal place.
Solution 3
Question 4
Arrange 
in descending order of their
magnitudes.
Also, find the sum of the lowest and largest of these fractions. Express the result obtained as a decimal fraction correct to two decimal places.
Solution 4
Question 5(i)
Solution 5(i)
Question 5(ii)
Solution 5(ii)
Question 5(iii)
Solution 5(iii)
Question 5(iv)
Solution 5(iv)
Question 5(v)
Solution 5(v)
Question 5(vi)
Solution 5(vi)
Question 5(vii)
Solution 5(vii)
Question 5(viii)
Solution 5(viii)
Chapter 1 - Rational and Irrational Numbers Exercise Ex. 1(B)
Question 1
State, whether the following numbers are rational or not:
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
Solution 1
(i) 
Irrational
(ii) 
Irrational
(iii) 
Rational
(iv) 
Irrational
(v) 
Rational
(vi) 
Rational
Question 2
Find the square of:
Solution 2
(i)
(ii)
(iii)
(iv)
Question 3
State, in each case, whether true or false:
(i) 
(ii) 
(iii) 
(iv) 
is an irrational number
(v) 
is a rational number.
(vi) All rational numbers are real numbers.
(vii) All real numbers are rational numbers.
(viii) Some real numbers are rational numbers.
Solution 3
(i) False
(ii) 
which is true
(iii) True.
(iv) False because
which is recurring and non-terminating and hence it is rational
(v) True because 
which is recurring and non-terminating
(vi) True
(vii) False
(viii) True.
Question 4
Given universal set =
From the given set, find :
(i) set of rational numbers
(ii) set of irrational numbers
(iii) set of integers
(iv) set of non-negative integers
Solution 4
(i)
(ii)
(iii)
(iv)
Question 5
Use method of contradiction to show that 
and 
are irrational numbers.
Solution 5
Let us suppose that and are rational numbers
and 
(Where a, b 
7 and b, y 
0 x , y)
Squaring both sides
a2 and x2 are odd as 3b2 and 5y2 are odd .
a and x are odd....(1)
Let a = 3c, x = 5z
a2 = 9c2, x2 = 25z2
3b2 = 9c2, 5y2 = 25z2(From equation 
)
b2 =3c2, y2 = 5z2
b2 and y2 are odd as 3c2 and 5z2 are odd .
b and y are odd...(2)
From equation (1) and (2) we get a, b, x, y are odd integers.
i.e., a, b, and x, y have common factors 3 and 5 this contradicts our assumption that 
are rational i.e, a, b and x, y do not have any common factors other than.
is not rational
and are irrational.
Question 6
Prove that each of the following numbers is irrational:
Solution 6
Let 
be a rational number.
⇒ = x
Squaring on both the sides, we get
Here, x is a rational number.
⇒ x2 is a rational number.
⇒ x2 - 5 is a rational number.
⇒ 
is also a rational number.
is a rational number.
But 
is an irrational number.
is an irrational number.
⇒ x2- 5 is an irrational number.
⇒ x2 is an irrational number.
⇒ x is an irrational number.
But we have assume that x is a rational number.
∴ we arrive at a contradiction.
So, our assumption that is a rational number is wrong.
∴ is an irrational number.
Let 
be a rational number.
⇒ = x
Squaring on both the sides, we get
Here, x is a rational number.
⇒ x2 is a rational number.
⇒ 11 - x2 is a rational number.
⇒ 
is also a rational number.
is a rational number.
But 
is an irrational number.
is an irrational number.
⇒ 11 - x2 is an irrational number.
⇒ x2 is an irrational number.
⇒ x is an irrational number.
But we have assume that x is a rational number.
∴ we arrive at a contradiction.
So, our assumption that 
is a rational number is wrong.
∴ is an irrational number.
Let 
be a rational number.
⇒ = x
Squaring on both the sides, we get
Here, x is a rational number.
⇒ x2 is a rational number.
⇒ 9 - x2 is a rational number.
⇒ 
is also a rational number.
is a rational number.
But 
is an irrational number.
is an irrational number.
⇒ 9 - x2 is an irrational number.
⇒ x2 is an irrational number.
⇒ x is an irrational number.
But we have assume that x is a rational number.
∴ we arrive at a contradiction.
So, our assumption that 
is a rational number is wrong.
∴ is an irrational number.
Question 7
Write a pair of irrational numbers whose sum is irrational.
Solution 7
are irrational numbers whose sum is irrational.
which is irrational.
Question 8
Write a pair of irrational numbers whose sum is rational.
Solution 8
and 
are two irrational numbers whose sum is rational.
Question 9
Write a pair of irrational numbers whose difference is irrational.
Solution 9
and 
are two irrational numbers whose difference is irrational.
which is irrational.
Question 10
Write a pair of irrational numbers whose difference is rational.
Solution 10
and 
are irrational numbers whose difference is rational.
which is rational.
Question 11
Write a pair of irrational numbers whose product is irrational.
Solution 11
Question 13
Write in ascending order:
(i) 
(ii) 
(iii) 
Solution 13
(i) 
and 45 < 48 
(ii) 
and40 < 54 
(iii) 
and 128 < 147 < 180
Question 14
Write in descending order:
(i) 
(ii) 
Solution 14
(i) 
Since 162 > 96
(ii) 
141 > 63 
Question 15
Compare.
Solution 15
(i) 
and 
Make powers 
and 
same
L.C.M. of 6,4 is 12
and 
(ii) 
and 
L.C.M. of 2 and 3 is 6.
, 
Question 16
Insert two irrational numbers between 5 and 6.
Solution 16
Question 17
Insert five irrational numbers between 
and 
.
Solution 17
Question 18
Write two rational numbers between 
Solution 18
We want rational numbers a/b and c/d such that: 
< a/b < c/d < 
Consider any two rational numbers between 2 and 3 such that they are perfect squares.
Let us take 2.25 and 2.56 as 
Thus we have,
Question 19
Write three rational numbers between 
Solution 19
Consider some rational numbers between 3 and 5 such that they are perfect squares.
Let us take, 3.24, 3.61, 4, 4.41 and 4.84 as
Question 20(i)
Solution 20(i)
Question 20(ii)
Solution 20(ii)
Question 20(iii)
Solution 20(iii)
Question 20(iv)
Solution 20(iv)
Chapter 1 - Rational and Irrational Numbers Exercise Ex. 1(C)
Question 1
State, with reasons, which of the following are surds and which are not:
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
(vii) 
(viii) 
Solution 1
(i) 
Which is irrational
is a surd
(ii) 
Which is irrational
is a surd
(iii) 
is a surd
(iv) 
which is rational
is not a surd
(v) 
is not a surd
(vi) 
= -5
is is not a surd
(vii) not a surd as 
is irrational
(viii) is not a surd because 
is irrational.
Question 2
Write the lowest rationalising factor of:
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
(vii) 
(viii) 
(ix) 
Solution 2
(i) 
which is rational
lowest rationalizing factor is 
(ii) 
lowest rationalizing factor is 
(iii) 
lowest rationalizing factor is 
(iv)
Therefore, lowest rationalizing factor is 
(v) 
lowest rationalizing factor is
(vi)
Therefore lowest rationalizing factor is 
(vii)
Its lowest rationalizing factor is 
(viii)
Its lowest rationalizing factor is 
(ix)
its lowest rationalizing factor is 
Question 3
Rationalise the denominators of :
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
(vii) 
(viii) 
(ix) 
Solution 3
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
Question 4
Find the values of 'a' and 'b' in each of the following:
(i) 
(ii) 
(iii) 
(iv) 
Solution 4
(i) 
(ii) 
(iii) 
(iv) 
Question 5
Simplify:
(i) 
(ii) 
Solution 5
(i)
(ii)
Question 6
If x = 
and y = 
; find:
(i) x2 (ii) y2
(iii) xy (iv) x2 + y2 + xy.
Solution 6
(i) 
(ii) 
(iii) xy = 
(iv) x2 + y2 + xy = 161 - 
= 322 + 1 = 323
Question 7
(i) m2
(ii) n2
(iii) mn
Solution 7
Question 8
If x = 2
+ 2
, find:
(i) 
(ii) 
(iii) 
Solution 8
(i) 
(ii) 
(iii) 
Question 9
Solution 9
Question 10
Solution 10
Question 11
Solution 11
Question 12
Solution 12
Question 13(i)
Solution 13(i)
Question 13(ii)
Solution 13(ii)
Question 13(iii)
If 
= 1.4 and 
= 1.7, find the
value of each of the following, correct to one decimal place:
Solution 13(iii)
Question 14
Solution 14
Question 15
Solution 15
Chapter 1 - Rational and Irrational Numbers Exercise Ex. 1(D)
Question 1
Simplify: 
Solution 1
Question 2
Simplify: 
Solution 2
Question 3
Evaluate, correct to one place of decimal, the expression
, if 
= 2.2 and 
= 3.2.
Solution 3
Question 4
If x = 
, find the value of:
Solution 4
Question 5
Show that:
(i) Negative of an irrational number is irrational number.
(ii) The product of a non - zero rational number and an irrational number is a rational number.
Solution 5
Proof:
(i)
The given statement is TRUE.
Let us assume that negative of an irrational number is a rational number.
Let p be an irrational number,
→ - p is a rational number.
→ - (-p) = p is a rational number.
But p is an irrational number.
Therefore our assumption was wrong.
So, Negative of an irrational number is irrational number.
(ii)
The given statement is FALSE.
We know that 3 is a
non - zero rational number and 
is an irrational number.
So, 3 ×= 3 is an irrational.
Therefore, the product of a non - zero rational number and an irrational number is an irrational number.
Question 6
Draw a line segment of length 
cm.
Solution 6
Since 
So, first we need
to find 
and mark it on number line and then find 
Steps to draw 
on the number line are:
- Draw a number line and mark point O.
- Mark point A on it such that OA = 1 unit.
- Draw right triangle OAB such that ∠A = 90° and AB = 1 unit.
- Join OB.
- By Pythagoras Theorem,
. - Draw a line BC perpendicular to OB such that BC = 1.
- Join OC. Thus, OBC is a right
triangle. Again by Pythagoras Theorem, we have
- With centre as O and radius OC draw a circle which meets the number line at a point E.
units. Thus,
OE represents 
on the number line.
Question 7
Draw a line segment of length 
cm.
Solution 7
Since, 
We need to construct a right - angled triangle OAB, in which
∠A = 90°, OA = 2 units and AB = 2 units.
By Pythagoras theorem, we get
OB2 = OA2 + AB2
∴
OB = 
units
STEPS:
- Draw a number line.
- Mark point A on it which is two points (units) from an initial point say O in the right/positive direction.
- Now, draw a line AB = 2 units which is perpendicular to A.
- Join OB to represent the hypotenuse of a triangle OAB, right angled at A.
- This hypotenuse of triangle OAB is showing a length of OB.
- With centre as O and radius as OB, draw an arc on the number line which cuts it at the point C.
- Here, C represents
.
Question 8
Show that: 
Solution 8
Question 9
Show that:
Solution 9
Question 10
Show that x is irrational, if:
(i) x2 = 6 (ii) x2 = 0.009 (iii) x2 = 27
Solution 10
(i) x2 = 6
(ii) x2 = 0.009
(iii) x2 = 27
Question 11
Show that x is rational, if:
(i) x2 = 16 (ii) x2 = 0.0004 (iii) x2 = 
Solution 11
(i) x2 = 16
Taking square root on both the sides, we get
x = ± 4
As we know that 4 =
and -4 =
are rational numbers.
⇒ x is rational , if: x2 = 16 .
(ii) x2 = 0.0004 (iii) x2 =
x2 = 0.0004 = 
Taking square root on both the sides, we get
…. is a rational number.
⇒ x is rational , if: x2 =0.0004.
(iii) x2 =
Taking square root on both the sides, we get
…. is a rational number.
⇒ x is rational , if: x2 =.
Question 12
Using the following figure, show that BD =
.
Solution 12
From the image, it is cleared that
AB = x, BC = 1
⇒ AC = AB + BC = x + 1
O is a centre of the semi - circle with AC as diameter.
OA, OC and OD are the radii.
⇒ OA = OC = OD = 
OB = OC - BC = 
m∠OBD = 90° … given in the image
∴By Pythagoras theorem, we get
OD2 = OB2 + BD2
⇒ BD2 = OD2 - OB2
⇒
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Sign Up
Verify mobile number
Enter the OTP sent to your number
Change