NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.2

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.2

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.2 are the part of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for Class 9 Maths Chapter 1 Number System Ex 1.2.

• NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.1
• NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.2
• NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3
• NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.4
• NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.5
• NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6

Ex 1.2 Class 9 Maths Question 1.
State whether the following statements are true or false. Justify your answers.
(i) Every irrational number is a real number.
(ii) Every point on the number line is of the form √m, where m is a natural number.
(iii) Every real number is an irrational number.

Solution:
(i) True
Because the collection of all rational numbers and all irrational numbers is called a set of real numbers.
(ii) False
Because negative numbers cannot be the square root of any natural number.
(iii) False
Because real numbers have rational and irrational both types of numbers. For example, 5, ½, 12, 2/3, etc. are real numbers but they are not irrational.

 

Ex 1.2 Class 9 Maths Question 2.
Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.

Solution:
No, if we take a positive integer, say 16, its square root is 4, which is a rational number. Again, the square root of 25 is 5 which is a rational number.

 

Ex 1.2 Class 9 Maths Question 3.
Show how √5 can be represented on the number line.

Solution:
Method 1:

Draw a number line and mark a point O on it. Take a point A on it such that OA = 1 unit. Draw BA ⊥ OA as BA = 1 unit. Join OB = √2 units.
Now draw BB₁ ⊥ OB such that BB₁ =1 unit. Join OB₁ = √3 units.
Next, draw B₁B₂ ⊥ OB₁ such that B₁B₂ = 1 unit. Join OB₂ = √4 = 2 units.
Again, draw B₂B₃ ⊥ OB₂ such that B₂B₃ = 1 unit. Join OB₃ = √5 units.

Take O as centre and OB₃ as radius, draw an arc which cuts the number line at D. Point D represents √5 on the number line and OD = √5 units.

Method 2: We have √5 = √(4 + 1) = √(2² + 1²)

Draw a number line and mark a point O on it. Mark … , -2, -1, 0, 1, 2, … as shown in the figure below. Take a point Q such that OQ = 2 units. Draw PQ ⊥ OQ. With point Q as centre and radius as 1 unit, cut an arc at P. Join OP.

Now, O as centre and OP as radius, draw an arc which cuts the number line at R. Point R represents √5 and OR = √5.

We can verify the result using Pythagoras theorem,

OP² = OQ² + PQ²

OP² = 2² + 1²

OP² = 4 + 1

OP² = 5

OP = √5

OR = OP = √5 units

Ex 1.2 Class 9 Maths Question 4.

A classroom activity (constructing the ‘square root spiral’).

Solution:
Take a large sheet of paper and construct the ‘square root spiral’ in the following fashion. Start with a point O and draw a line segment OP₁ of unit length. Draw a line segment P₁P₂ perpendicular to OP, of unit length (see figure).

Now, draw a line segment P₂P₃ perpendicular to OP₂. Then draw a line segment P₃P₄ perpendicular to OP₃. Continuing in this manner, you can get the line segment P_n-1P_n by drawing a line segment of unit length perpendicular to OP_n-1. In this manner, you will have created the points P₂, P₃, …., P_n …, and joined them to create a beautiful spiral depicting √2, √3, √4, …

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