Question

Prove that √3-√5 is irrational​

Answer 1

To Prove :

$\sf{ \sqrt{3} - \sqrt{5} \: \: is \: irrational}$

Let us assume that $\sqrt{3} - \sqrt{5}$ is rational, which let us take as a.

Now,

$\sf{ \sqrt{3} - \sqrt{5} = a}$

Squaring on both sides,

⟶ $\sf{( \sqrt{3} - \sqrt{5} )^{2} = {a}^{2} }$

⟶ $\sf{8 - 2 \sqrt{15} = {a}^{2} }$

⟶ $\sf{ - 2 \sqrt{15} = {a}^{2} - 8}$

⟶ $\sf{ \sqrt{15} = \frac{ {a}^{2} - 8}{ - 2} }$

$\sf{ \frac{ {a}^{2} - 8}{ - 2}}$ is a rational number, whereas $\sqrt{15}$ is irrational.

We know that, One rational number cannot be equal to an irrational number.

Which means that,

Our assumption was wrong.

⤐ $\sqrt{3} - \sqrt{5}$ is irrational.

$\sf{ \red{Hence \: Proved!}}$

___________________

Answer 2

To Prove :

$\sf{ \sqrt{3} - \sqrt{5} \: \: is \: irrational}$

Let us assume that $\sqrt{3} - \sqrt{5}$ is rational, which let us take as a.

Now,

$\sf{ \sqrt{3} - \sqrt{5} = a}$

Squaring on both sides,

⟶ $\sf{( \sqrt{3} - \sqrt{5} )^{2} = {a}^{2} }$

⟶ $\sf{8 - 2 \sqrt{15} = {a}^{2} }$

⟶ $\sf{ - 2 \sqrt{15} = {a}^{2} - 8}$

⟶ $\sf{ \sqrt{15} = \frac{ {a}^{2} - 8}{ - 2} }$

$\sf{ \frac{ {a}^{2} - 8}{ - 2}}$ is a rational number, whereas $\sqrt{15}$ is irrational.

⤐ $\sqrt{3} - \sqrt{5}$ is irrational.

$\sf{ \red{Hence \: Proved!}}$