Real Numbers Ex-1.3 Ch-1 [Solutions] NCERT Maths Class 10th

Exercise 1.3

Q1. Prove that √5 is irrational.

Answer: Let take √5 as rational number If a and b are two co prime number and b is not equal to 0. We can write √5 = a/b Multiply by b both side we get b√5 = a To remove root, Squaring on both sides, we get 5b2 = a2 … (i)

Therefore, 5 divides a2 and according to theorem of rational number, for any prime number p which is divides a2 then it will divide a also. That means 5 will divide a. So we can write a = 5c Putting value of a in equation (i) we get 5b2 = (5c)2 5b2 = 25c2 Divide by 25 we get b2/5 = c2 Similarly, we get that b will divide by 5 and we have already get that a is divide by 5 but a and b are co prime number. so it contradicts. Hence √5 is not a rational number, it is irrational.

Q2. Prove that 3 + 2√5 is irrational.

Answer:

Let take that 3 + 2√5 is a rational number. So we can write this number as 3 + 2√5 = a/b Here a and b are two co prime number and b is not equal to 0 Subtract 3 both sides we get 2√5 = a/b – 3 2√5 = (a-3b)/b Now divide by 2, we get √5 = (a-3b)/2b Here a and b are integer so (a-3b)/2b is a rational number so √5 should be a rational number But √5 is a irrational number so it contradicts. Hence, 3 + 2√5 is a irrational number.

3. Prove that the following are irrationals: (i) 1/√2 (ii) 7√5 (iii) 6 + √2

Answer

(i) Let take that 1/√2 is a rational number. So we can write this number as 1/√2 = a/b Here a and b are two co prime number and b is not equal to 0 Multiply by √2 both sides we get 1 = (a√2)/b Now multiply by b b = a√2 divide by a we get b/a = √2 Here a and b are integer so b/a is a rational number so √2 should be a rational number But √2 is a irrational number so it contradicts. Hence, 1/√2 is a irrational number

(ii) Let take that 7√5 is a rational number. So we can write this number as 7√5 = a/b Here a and b are two co prime number and b is not equal to 0 Divide by 7 we get √5 = a/(7b) Here a and b are integer so a/7b is a rational number so √5 should be a rational number but √5 is a irrational number so it contradicts. Hence, 7√5 is a irrational number.

(iii) Let take that 6 + √2 is a rational number. So we can write this number as 6 + √2 = a/b Here a and b are two co prime number and b is not equal to 0 Subtract 6 both side we get √2 = a/b – 6 √2 = (a-6b)/b Here a and b are integer so (a-6b)/b is a rational number so √2 should be a rational number. But √2 is a irrational number so it contradicts. Hence, 6 + √2 is a irrational number.