Question

Prove that 3/√5 is irrational, given that √5 is irrational

Answer 1

$\huge\sf\blue{Given}$

✭ √5 is irrational

✭ $\sf\dfrac{3}{\sqrt{5}}$

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$\huge\sf\gray{To \:Find}$

◈ The above statement

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$\huge\sf\purple{Steps}$

We know that,

Rational Numbers will be of the form $\sf \dfrac{p}{q}$

Where p and q are co primes and q ≠ 0

Let's assume that $\sf\dfrac{3}{\sqrt{5}}$ is a rational number

Equating the given number with $\sf\dfrac{p}{q}$

➝ $\sf\dfrac{3}{\sqrt{5}} = \dfrac{p}{q}$

➝ $\sf3\times q = p\times \sqrt{5}$

➝ $\sf\quad\bigg\lgroup \because On \ Cross \ Multiplying\bigg\rgroup$

➝ $\sf 3q = p\sqrt{5}$

➝ $\sf\dfrac{3q}{p} = \sqrt{5}$

But here this is a $\sf\longrightarrow \ \longleftarrow$(contradiction) as LHS is rational but RHS is irrational (√5 is Irrational as per the Question)

$\sf \therefore$ Our assumption is wrong and $\sf\dfrac{3}{\sqrt{5}}$ is irrational

$\sf Hence \ Proved !!$

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Answer 2

Given ,

• √5 is an irrational number

Let , 3/√5 is an rational number

Thus ,

$\sf \mapsto \frac{3}{ \sqrt{5} } = \frac{a}{b} \\ \\ \sf \mapsto \sqrt{5} = \frac{3b}{a}$

Here , √5 is an irrational number but 3b/a is rational number

Since , irrational ≠ rational

Thus , our assumptions is wrong

$\sf \therefore{ \underline{ \frac{3}{ \sqrt{5} } \: is \: irrational \: number}}$