2. Examples of irrational numbers

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 $q \neq 0$.

 

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.

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 $q \neq 0$.

 

Therefore, our assumption is wrong.

 

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

 

Hence, we proved.