# 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)`

**Answer:**

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

Thus, it is a rational number.

(iv) `1/(sqrt2)`

**Answer:**

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.