# Number System

## Exercise 1.5 Part 1

Question 1: Classify the following numbers as rational or irrational:
(i) 2 - sqrt5

Answer: Since this equation contains an irrational number sqrt5 thus, given number is an irrational number.

(ii) (3 + sqrt23) - sqrt23

Answer: Given, (3 + sqrt23) - sqrt23
= 3 + sqrt23 - sqrt23 = 3
Thus, it is rational number.

(iii) (2sqrt7)/(7sqrt7)

Given, (2sqrt7)/(7sqrt7)=2/7

Thus, it is a rational number.

(iv) 1/(sqrt2)

Since given number contains an irrational number as numerator, thus, it is an irrational number.
(v) 2π

Answer: Since, given number an irrational number, π as factor, thus, given number is an irrational number.

Question 2: Simplify each of the following expressions:

(i) (3 + sqrt3)(2 + sqrt2)

Answer: Given, (3 + sqrt3)(2 + sqrt2)
= 3 x 2 + 2sqrt3 + 3sqrt2 + sqrt3 x sqrt2
= 6 + 2sqrt3 + 3sqrt2 + sqrt6

(ii) (3 + sqrt3)(3 - sqrt3)

Asnwer:
Given, (3 + sqrt3)(3 - sqrt3)
= (Since, (a + b)(a – b) = a^2 – b^2
Hence, we get 3^2 – (sqrt3)^2
= 9 – 3 = 6

(iii) (sqrt5 + sqrt2)^2

Answer: Given (sqrt5 + sqrt2)^2
(Since, (a + b)^2 = a^2 + b^2 + 2ab
Hence, we have; (sqrt5)^2 + (sqrt2)^2 + 2 xx sqrt5 xx sqrt2
= 5 + 2 + 2sqrt10
= 7 + 2sqrt10

(iv) (sqrt5 - sqrt2)( sqrt5 + sqrt2)

Answer: Given, (sqrt5 - sqrt2)( sqrt5 + sqrt2)
(Since, (a + b)(a – b) = a^2 – b^2
Hence, we have; (sqrt5)^2 – (sqrt2)^2 = 5 – 2 = 3

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

Answer: There is no contradiction. Remember that when you measure a length with a scale or any other device, you only get an approximate rational value. So, you may not realise that either c or d is irrational.
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