**Download NCERT Solutions For Class 9 Maths Chapter 1 Exercise 1.5 – Number System. This Exercise contains 5 questions, for which detailed answers have been provided in this note. In case you are looking at studying the remaining Exercise for Class 9 for Maths NCERT solutions for Chapter 1 or other Chapters, you can click the link at the end of this Note.**

**1. Classify the following numbers as rational or irrational:**

**(i) 2 – √5**

**(ii) ( 3 + √23 ) – √23**

**(iii) **

**(iv) **

**(v) 2π**

**Solution:**

(i) 2 – √5 : It is irrational.

(ii) ( 3 + √23 ) – √23 = 3 + √23 – √23 = 3 : It is rational.

(iii) = : It is rational

(iv) = : It is irrrational.

(v) 2π : It is irrational.

### Simplification of Irrational Numbers – Video Explanation

**(2). Simplify each of the following expressions:**

**(i)( 3 + √3 )( 2 + √2 )**

**(ii)( 3 + √3 )( 3 – √3 )**

**(iii)**

**(iv)( √5 – √2 )( √5 + √2 )**

**Solution:**

(i) ( 3 + √3 )( 2 + √2 )

( 3 + √3 )( 2 + √2 )

= ( 3 + √3 )2 +( 3 + √3 )√2

= 6 + 2√3 + 3√2 + √6

(ii)( 3 + √3 )( 3 – √3 )

Using the identity

( a + b )( a – b) = a^{2} – b^{2}

We get

( 3 + √3 )( 3 – √3 ) = – √3^{2} = 9 – 3 = 6

(iii)

Using the identity

= a^{2} + 2ab + b^{2}

We get

= 5 + 2 √10 + 2 = 7 + 2 √10

(iv)( √5 – √2 )( √5 + √2 )

Using the identity

( a – b )( a + b) = a^{2} – b^{2}

We get

( √5 – √2 )( √5 + √2 ) = 5 – 2 = 3

**(3) Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π = . This seems to contradict the fact that π is irrational. How will you resolve this contradiction?**

**Solution:**

As we measure c and d using any physical device we do not get the exact value manually , instead we get an approximate rational value and this is why this contradiction arises.

**4. Represent √9.3 on the number line.**

**Solution:**

Follow the steps below to locate √9.3 on the number line as shown above:

Step (I) Draw a line segment of length 9.3 units EA and extend it to C by 1 unit .

Step (II) Now with O as mid point of EC draw a semicircle with O as center and OC as radius.

Step (III) Draw a line perpendicular to EC at A and intersecting the semicircle at B.

Step (IV) Taking A as center and AB as radius draw an arc from B to intersect the extended line EAC at D.

If we assign A as origin i.e. 0

Then D represents √9.3 on the number line as shown in the figure.

**NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.5 – Number System, has been designed by the NCERT to test the knowledge of the student on the topic – Operations on Real Numbers**

**The next Exercise for** **NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.6 – Number System can be accessed by clicking here.**

** Maths – NCERT Solutions Class 9**

**Download NCERT Solutions For Class 9 Maths Chapter 1 Exercise 1.5 – Number System**

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