Question

Prove that √2 is an irrational number. Hence, show that 3 - √2 is an irrational
number.​

Answer 1

Answer:

Let us assume that  $\sqrt{2}$  is a rational number

   

             $\sqrt{2} = \frac{a}{b}$ , ( a & b co primers, b ≠ 0 )

Squaring both sides,

   

             $(\sqrt{2} )^2 = \frac{a^2}{b^2}$

     

              $2 = \frac{a^2}{b^2}$ -------(1)

⇒ a² is divided by 2

∴ a is also divided by 2

Let's assume a = 2c  ( c = any integer)

Substitute a in equation (1)

   

            $2 = \frac{(2c)^2}{b^2}$

            $2b^2 = 4c^2$

             $b^2 = 2 c^2$

⇒b² is divided by 2

∴ b is also divided by 2

Form this we understood that 2 is a factor of both a & b.

Therefore a & b are not co primers.

∴ Our assumption is wrong.

ie)  $\sqrt{2}$  is an irrational number.

We also have to show that  $3 -\sqrt{2}$ is an irrational  number.

Now let's assume that $3 -\sqrt{2}$  is a rational number.

   

             $3 - \sqrt{2} = \frac{a}{b}$ , ( a & b co primers, b ≠ 0 )

             $3 - \frac{a}{b} = \sqrt{2}$

             $\frac{3b - a}{b} = \sqrt{2}$

∴ a & b are integers,  $\frac{3b - a}{b}$  is rational.

Which means $\sqrt{2}$ is rational.

But its a contradiction of the fact that $\sqrt{2}$ is irrational.

Therefore again our assumption is wrong.

ie)  $3 -\sqrt{2}$ is an irrational number.

                          Hope it helps you^_^