# 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 part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.2.

• Number Systems Class 9 Ex 1.1
• Number Systems Class 9 Ex 1.2
• Number Systems Class 9 Ex 1.3
• Number Systems Class 9 Ex 1.4
• Number Systems Class 9 Ex 1.5
• Number Systems Class 9 Ex 1.6

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

Ex 1.2 Class 9 Maths Question 1.
State whether the following statements are true or false. Justify your answers.

1. Every irrational number is a real number.
2. Every point on the number line is of the form $\sqrt { m }$, where m is a natural number.
3. Every real number is an irrational number.

Solution:
(1) True (∵ Real numbers = Rational numbers + Irrational numbers.)
(2) False (∵ no negative number can be the square root of any natural number.)
(3) False (∵ rational numbers are also present in the set of real numbers.)

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, the square roots of all positive integers are not irrational.
e.g., $\sqrt { 16 } = 4$
Here, ‘4’ is a rational number.

Ex 1.2 Class 9 Maths Question 3.
Show how $\sqrt { 5 }$ can be represented on the number line.
Solution:
We know that, $\sqrt { 5 }$ $\sqrt {4+1}$ $\sqrt { { 2 }^{ 2 }+{ 1 }^{ 2 } }$ Draw of right angled triangle OQP, such that
OQ = 2 units
PQ= 1 unit
and   ∠OQP = 90°
Now, by using Pythagoras theorem, we have
OP2 =   OQ2 +PQ2
=    22 +12
= op= $\sqrt {4+1}$ = $\sqrt { 5 }$
Now, take O as center OP = 45 as radius, draw an arc, which intersects the line at point R.
Hence, the point R represents $\sqrt { 5 }$.

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 0P3. Continuing in this manner, you can get the line segment Pn-1Pn by 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 $\sqrt { 2 },$ $\sqrt { 3 },$ $\sqrt { 4 }$,………..

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