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.



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