Question

prove that √5 is irrational and hence 7 + 3√5 is irrational

please answwe it I'll mark you brainliest​

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:Let \: assume \: that \: \sqrt{5} \: is \: not \: irrational$

$\rm :\implies\: \sqrt{5} \: is \: rational$

So, Let assume that,

$\rm :\longmapsto\: \sqrt{5} = \dfrac{x}{y}$

where x and y are integers and y is non - zero and HCF of x and y is 1. -------(1)

Now, On squaring both sides, we get

$\rm :\longmapsto\:5 = \dfrac{ {x}^{2} }{ {y}^{2} }$

$\rm :\longmapsto\: {x}^{2} = 5 {y}^{2} - - - (2)$

$\rm :\implies\: {x}^{2} \: is \: divisible \: by \: 5$

$\rm :\implies\: {x} \: is \: divisible \: by \: 5 - - - (3)$

Let assume that,

$\rm :\longmapsto\:x = 5m$

On squaring both sides,

$\rm :\longmapsto\: {x}^{2} = 25 {m}^{2}$

$\rm :\longmapsto\: {5y}^{2} = 25 {m}^{2} \: \: \: \: \: \: \: \{using \: (2) \: \}$

$\rm :\longmapsto\: {y}^{2} = 5 {m}^{2}$

$\rm :\longmapsto\: {y}^{2} \: is \: divisible \: by \: 5$

$\rm :\longmapsto\: {y} \: is \: divisible \: by \: 5 - - - (4)$

From equation (3) and (4),

we concluded that both x and y are divisible by 5, which is contradiction to the fact that HCF ( x, y ) = 1.

Hence, Our assumption is wrong.

$\bf\implies \: \sqrt{5} \: is \: irrational$

Now second Part

$\rm :\longmapsto\:Let \: assume \: that \: 7 + 3\sqrt{5} \: is \: not \: irrational$

$\rm :\implies\: 7 + 3\sqrt{5} \: is \: rational$

Let assume that

$\rm :\longmapsto\: 7 + 3\sqrt{5} = \dfrac{x}{y}$

where x and y are integers and y is non - zero and HCF of x and y is 1

Now,

$\rm :\longmapsto\: 3\sqrt{5} = \dfrac{x}{y} - 7$

$\rm :\longmapsto\: 3\sqrt{5} = \dfrac{x - 7y}{y}$

$\rm :\longmapsto\: \sqrt{5} = \dfrac{x - 7y}{3y}$

As x and y are integers, so x - 7y and 3y are integers

$\rm :\longmapsto\: \dfrac{x - 7y}{3y} \: is \: rational$

$\rm :\implies\: \sqrt{5} \: is \: rational$

which is contradiction as proved above

$\rm :\longmapsto\:\: \sqrt{5} \: is \: irrational$

Hence, our assumption is wrong

$\bf :\implies\: 7 + 3\sqrt{5} \: is \: irrational$

Hence, Proved