**Real Numbers**

**Question 1.**Prove that √5 is irrational.

**Solution:**

Let √5 = be a rational number, where p and q are co-primes and q ≠ 0.

Then, √5q = p => 5q

^{2}=p

^{2}⇒ p

^{2}– Sq

^{2}… (i)

Since 5 divides p

^{2}, so it will divide p also.

Let p = 5r

Then p

^{2}– 25r

^{2}[Squaring both sides]

⇒ 5q

^{2}= 25r

^{2}[From(i)]

⇒ q

^{2}= 5r

^{2}Since 5 divides q

^{2}, so it will divide q also. Thus, 5 is a common factor of both p and q.

This contradicts our assumption that √5 is rational.

Hence, √5 is irrational. Hence, proved.

**Question 2.****Show that 3 + √5 is irrational.**

**Solution:**Let 3 + 2√5 = be a rational number, where p and q are co-prime and q ≠ 0.

Then, 2√5 = – 3 =

⇒ √5 =

since is a rational number,

therefore, √5 is a rational number. But, it is a contradiction.

Hence, 3 + √5 is irrational. Hence, proved.

**Question 3.Prove that the following are irrational.**

**Solution:****(i)** Let = be a rational number,

where p and q are co-prime and q ≠ 0.

Then, √2 =

Since is rational, therefore, √2 is rational.

But, it is a contradiction that √2 is rational, rather it is irrational.

Hence, is irrational.

Hence, proved.

**(ii)** Let 7√5 = be a rational number, where p, q are co-primes and q ≠ 0.

Then, √5 =

Since is rational therefore, √5 is rational.

But, it is a contradiction that √5 is rational rather it is irrational.

Hence, 7√5 s is irrational.

Hence proved.

**(iii)** Let 6 + √2 = be a rational number, where p, q are co-primes and q ≠ 0.

Then, √2 = – 6 =

Since is rational therefore, √2 is rational.

But, it is a contradiction that √2 is rational, rather it is irrational.

Hence, 6 + √2 is irrational.