Course Content
Class 9th Science
0/30
Class 9th Math
0/81
Class 9 Social Science History: India and the Contemporary World – I
0/16
Class 9 Social Science Geography: Contemporary India – I
0/12
Class 9 Social Science Civics (Political Science): Democratic Politics – I
0/12
Class 9 Social Science Economics: Understanding Economic Development – I
0/8
Class 9 English Beehive
0/22
Class 9 English Beehive Poem
0/19
Class 9 English Moments
0/20
Online Class For 9th Standard Students (CBSE) (English Medium)

## Exercise 1.2 Page: 8

1. State whether the following statements are true or false. Justify your answers.

(i) Every irrational number is a real number.

Solution:

True

Irrational Numbers – A number is said to be irrational, if it cannot be written in the p/q, where p and q are integers and q ≠ 0.

i.e., Irrational numbers = 0, 19/30, 2, 9/-3, -12/7, √2, √5, , 0.102….

Real numbers – The collection of both rational and irrational numbers are known as real numbers.

i.e., Real numbers = √2, √5, , 0.102…

Every irrational number is a real number, however, every real numbers are not irrational numbers.

(ii) Every point on the number line is of the form √m where m is a natural number.

Solution:

False

The statement is false since as per the rule, a negative number cannot be expressed as square roots.

E.g., √9 =3 is a natural number.

But √2 = 1.414 is not a natural number.

Similarly, we know that there are negative numbers on the number line but when we take the root of a negative number it becomes a complex number and not a natural number.

E.g., √-7 = 7i, where i = √-1

The statement that every point on the number line is of the form √m, where m is a natural number is false.

(iii) Every real number is an irrational number.

Solution:

False

The statement is false, the real numbers include both irrational and rational numbers. Therefore, every real number cannot be an irrational number.

Real numbers – The collection of both rational and irrational numbers are known as real numbers.

i.e., Real numbers = √2, √5, , 0.102…

Irrational Numbers – A number is said to be irrational, if it cannot be written in the p/q, where p and q are integers and q ≠ 0.

i.e., Irrational numbers = 0, 19/30, 2, 9/-3, -12/7, √2, √5, , 0.102….

Every irrational number is a real number, however, every real number is not irrational.

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, the square roots of all positive integers are not irrational.

For example,

√4 = 2 is rational.

√9 = 3 is rational.

Hence, the square roots of positive integers 4 and 9 are not irrational. ( 2 and 3, respectively).

3. Show how √5 can be represented on the number line.

Solution:

Step 1: Let line AB be of 2 unit on a number line.

Step 2: At B, draw a perpendicular line BC of length 1 unit.

Step 3: Join CA

Step 4: Now, ABC is a right angled triangle. Applying Pythagoras theorem,

AB2+BC2 = CA2

22+12 = CA2 = 5

⇒ CA = √5 . Thus, CA is a line of length √5 unit.

Step 4: Taking CA as a radius and A as a center draw an arc touching

the number line. The point at which number line get intersected by

arc is at √5 distance from 0 because it is a radius of the circle

whose center was A.

Thus, √5 is represented on the number line as shown in the figure.

4. Classroom activity (Constructing the ‘square root spiral’) : 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 OP1 of unit length (see Fig. 1.9). Now draw a line segment P2P3 perpendicular to OP2. Then draw a line segment P3P4 perpendicular to OP3. Continuing in Fig. 1.9 :

Constructing this manner, you can get the line segment Pn-1Pn by square root spiral drawing a line segment of unit length perpendicular to OPn-1. In this manner, you will have created the points P2, P3,….,Pn,… ., and joined them to create a beautiful spiral depicting √2, √3, √4, …

Solution:

Step 1: Mark a point O on the paper. Here, O will be the center of the square root spiral.

Step 2: From O, draw a straight line, OA, of 1cm horizontally.

Step 3: From A, draw a perpendicular line, AB, of 1 cm.

Step 4: Join OB. Here, OB will be of √2

Step 5: Now, from B, draw a perpendicular line of 1 cm and mark the end point C.

Step 6: Join OC. Here, OC will be of √3

Step 7: Repeat the steps to draw √4, √5, √6….