### Theory:

Let us prove the theorem by assuming its contrary.
Example:
1. Prove that $$\sqrt{7}$$ is an irrational number.

Proof: Let us assume that $$\sqrt{7}$$ is a rational number.

Therefore, it can be written as $$\sqrt{7} = \frac{p}{q}$$ where $$p, q$$ are integers and $$q \neq 0$$.

Here, $$p$$ and $$q$$ are coprime(no common factors between $$p$$ and $$q$$).

Thus, $$\sqrt{7} = \frac{p}{q}$$

$$q \sqrt{7} = p$$

Squaring on both sides, we have:

$$7q^2 = p^2$$ ---- ($$1$$)

Therefore, $$p^2$$ is divisible by $$7$$.

And, $$p$$ is also divisible by $$7$$ [If $$p$$ be a prime number. If $$p$$ divides $$a^2$$, then $$p$$ divides $$a$$, where $$a$$ is a positive integer.]

So, we can write $$p = 7a$$ for some integer $$a$$.

Substituting the value of $$p$$ in equation ($$1$$), we have:

$$7q^2 = 49a^2$$

$$q^2 = 7a^2$$

This implies that $$q^2$$ is divisible by $$7$$ and $$q$$ is also divisible by $$7$$. [If $$p$$ be a prime number. If $$p$$ divides $$a^2$$, then $$p$$ divides $$a$$, where $$a$$ is a positive integer.]

Therefore, $$p$$ and $$q$$ have atleast $$7$$ as a common factor.

This contradicts the fact that $$p$$ and $$q$$ are coprime.

Therefore, our assumption is wrong.

Thus, $$\sqrt{7}$$ is an irrational number.

Hence, we proved.

2. Prove that $$8 - \sqrt{3}$$ is an irrational number.

Proof: Let us assume that $$8 - \sqrt{3}$$ is a rational number.

Therefore, it can be written as $$8 - \sqrt{3} = \frac{p}{q}$$ where $$p, q$$ are integers and $$q \neq 0$$.

Here, $$p$$ and $$q$$ are coprime(no common factors between $$p$$ and $$q$$).

Thus, $$8 - \sqrt{3} = \frac{p}{q}$$

$$8 - \frac{p}{q} = \sqrt{3}$$

$$\frac{8q - p}{q} = \sqrt{3}$$

Since $$8$$, $$p$$ and $$q$$ are integers, then $$\frac{8q - p}{q}$$ is rational. This implies that $$\sqrt{3}$$ is a rational number.

But, we know that $$\sqrt{3}$$ is an irrational number.

Therefore, our assumption is wrong.

Thus, $$8 - \sqrt{3}$$ is an irrational number.

Hence, we proved.
Important!
In grade $$9$$, we have learnt that

1. The sum or difference of a rational and an irrational number is irrational.

2. The product and quotient of a non-zero rational and irrational number is irrational.
Example:
Show that $$\frac{3}{\sqrt{5}}$$ is an irrational number.

Proof:

Let us assume that $$\frac{3}{\sqrt{5}}$$ is a rational number.

Therefore, it can be written as $$\frac{3}{\sqrt{5}} = \frac{p}{q}$$ where $$p, q$$ are integers and $$q \neq 0$$.

Here, $$p$$ and $$q$$ are coprime(no common factors between $$p$$ and $$q$$).

Thus, we have:

$$\frac{3}{\sqrt{5}} = \frac{p}{q}$$

$$3q = \sqrt{5}p$$

$$\frac{3q}{p} = \sqrt{5}$$

Since we know that $$3$$, $$q$$ and $$p$$ are integers and $$\frac{3q}{p}$$ is a rational number. This implies that $$\sqrt{5}$$ is a rational number.

But, we know that $$\sqrt{5}$$ is an irrational number.

Therefore, our assumption is wrong.

Thus, $$\frac{3}{\sqrt{5}}$$ is an irrational number.

Hence, we proved.