​ Prove that the following are irrational, i) 1/ √2

Question

Prove that the following are irrational,

i) \(\frac{1}{\sqrt2}\)

ii) \(\sqrt{3}+\sqrt{5}\)

iii) 6 + \(\sqrt{2}\)

iv) \(\sqrt{5}\)

v) 3 + \(2\sqrt{5}\)

Answer 1

i) \(\frac{1}{\sqrt2}\)

On the contrary, suppose that is a \(\frac{1}{\sqrt2}\) rational number;

then \(\frac{1}{\sqrt2}\) is of the form where \(\frac{p}{q}\) and q are integers.

∴ \(\frac{1}{\sqrt2}\) = \(\frac{p}{q}\)

⇒ \(\frac{\sqrt2}{1}\) = \(\frac{q}{p}\)

(i.e.,) √2 is a rational number and it is a contradiction. This contradiction arised due to our supposition that \(\frac{1}{\sqrt2}\) is a rational number. Hence \(\frac{1}{\sqrt2}\) is an irrational number.

ii) \(\sqrt{3}+\sqrt{5}\)

Suppose √3 + √5 is not an irrational number. Then √3 + √5 must be a rational number. √3 + √5 = \(\frac{p}{q}\) , q ≠ 0 and p, q ∈ Z

Squaring on both sides

but √15 is an irrational number. 

\(\frac{p^2-8q^2}{2q^2}\)is a rational number 

(p²– 8q², 2q²∈ Z, 2q²≠ 0) 

but an irrational number can’t be equal to a rational number, so our supposition that √3 + √5 is not an irrational number is false. 

∴ √3 + √5 is an irrational number.

iii) 6 + √2 

To prove: 6 + √2 is an irrational number. 

Let us suppose that 6 + √2 is a rational number. 

∴ 6 + √2 = \(\frac{p}{q}\) , q ≠ 0 

⇒ √2 = \(\frac{p}{q}\)– 6 

⇒ √2 = Difference of two rational numbers

⇒ √2 = rational number But this contradicts the fact that √2 is an irrational number. 

∴ Our supposition is wrong. 

Hence the given statement is true. 6 + √2 is an irrational number.

iv) √5 

To prove: √5 is an irrational number. 

On the contrary, let us assume that √5 is a rational number. 

∴ √5 = \(\frac{p}{q}\), q ≠ 0 

If p, q have a common factor, on cancelling the common factor let it be reduces to \(\frac{a}{b}\) where a, b are co-primes. 

Now √5 =\(\frac{a}{b}\) , where HCF (a, b) = 1 

Squaring on both sides we get

⇒ (√5)² = (\(\frac{a}{b}\))²

⇒ 5 = a²/b²

⇒ 5b² = a² 

⇒ 5 divides a² and thereby 5 divides 5

Now, take a = 5c 

then, a² = 25c ²

i.e., 5b² = 25c² 

⇒ b² = 5c² 

⇒ 5 divides b² and thereby b. 

⇒ 5 divides both b and a.

This contradicts that a and b are co-primes. 

This contradiction arised due to our assumption that √5 is a rational number. 

Hence our assumption is wrong and the given statement is true, i.e., √5 is an irrational number,

v) 3 + 2√5 

To Prove: 3 + 2√5 is an irrational. 

On the contrary, let us assume that 3 + 2√5 is a rational number.

Here p, q being integers we can say that \(\frac{p-3q}{2q}\) is a rational number. 

This contradicts the fact that √5 is an irrational number. This is due to our assumption “3 + 2√5 is a rational number”. 

Hence our assumption is wrong. 

∴ 3 + 2√5 is an irrational number.