Question

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Answer 1

Answer:

(i) 1/√2

Let 1/√2 = a/b , be a rational number where a and b are coprimes. ( b is not equal to 0)

1/√2 = a/b

√2 = a/b

√2b = a

Squaring both the sides,

(√2b)² = a².................(1)

2b² = a²

√2 is a factor of a², so √2 is a factor of a

Let a = 2c

squaring both the sides,

a² = (2c)²

putting the value of a² from equation (1)

2b² = 4c²

b² = 2c²

2 is a factor of b², so 2 is a factor of b

But a and b are co primes.

This Contradiction occurs due to our wrong assumption.

Hence, 1/√2 is an irrational number.

(ii) 7√5

Let 7√5 = a/b, be a rational number, where a and b are integers, co primes and b is not equal to zero.

7 √5 = a/b

√5 = a/b*7

a, b and 7 are integers so a/b*7 is a rational number.

So √5 is also a rational number but √5 is irrational number.

This Contradiction occurs due to our wrong assumption.

Hence, 7√5 is an irrational number.

(iii) 6+√2

Let 6+√2 = a/b, be a rational number where a and b are integers, coprimes. (b is not equal to 0)

6+√2 = a/b

√2 = a-6/b

a,b and -6 are integers , so a-6/b is a rational number.

So √2 is also a rational number but √2 is a irrational number.

This Contradiction occurs due to our wrong assumption.

Hence, 6+√2 is a irrational number.

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