Question

Prove that
$\frac{3}{2 \sqrt{5} }$
is a irrational.
​

Answer 1

In Order to solve it , first we have to prove that root 5 is irrational.

Let Root 5 a rational number

Root5= p/q where p and q are co-primes

$\sqrt{5} = \frac{p}{q} \\ \\ \implies 5 = \frac{ {p}^{2} }{ {q}^{2} }(squaring \: both \: sides) \\ \\ \implies {q}^{2} = \frac{ {p}^{2} }{5} \\ \\ \implies {p}^{2} \: is \: divisible \: by \: 5 \\ \\ \implies p \: is \: divisible \: by \: 5 \: (eq.01)$

Let P = 5S ,

Now again we will square both the sides,

${p}^{2} = 25 {s}^{2} \\ \\ \implies 5 {q}^{2} = 25 {s}^{2} \\ \\ \implies {q}^{2} = 5 {s}^{2} \\ \\ \implies \frac{{q}^{2} }{5} = {s}^{2} \\ \\ \implies {q}^{2} \: is \: divisible \: by \: 5 \\ \\ \implies q \: is \: divisible \: by \: 5$

Hence, both p and q are divisible by 5,they arenot co-primes.

It contradicts our assumption.Therefore, root 5 is an irrational number.

Now, let us prove the question.

$\frac{3}{2 \sqrt{5} } = \frac{a}{b} where \: a \: and \: b \: are \: rational \: numbers \\ \\ \implies 3b = 2 \sqrt{5} a \\ \\ \implies \sqrt{5} = \frac{3b}{2a}$

Now, if a and b are rational then 3a and 2b is also rational which means the RHS is rational but LHS is irrational (proved above) .

Hence, the given number is irrational.