Question

Prove
$2 - 3 \sqrt{5}$

as
irrational.​

Answer 1

$\mathtt{ \huge{ \fbox{Solution :)}}}$

Let , 2 - 3√5 is an rational number

So , it can be written in the form of A/B

$\sf \mapsto 2 - 3 \sqrt{5} = \frac{a}{b} \\ \\ \sf \mapsto - 3 \sqrt{5} = \frac{a}{b} - 2 \\ \\ \sf \mapsto \sqrt{5} = \frac{ - a + 2b}{3b}$

Here , √5 is an irrational number but (-a + 2b)/3b is rational number

Since , Irrational ≠ rational

Thus , our assumptions is incorrect

Hence , 2 - 3√5 is an irrational

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Answer 2

$\huge{\boxed{\boxed{\sf{\red{Solution}}}}}$

To prove : $\sf 2-3\sqrt{5}\:is\:irrational$

Proof:

$\large\sf Let\:the\:2-3\sqrt{5}\:be\:rational$

$\large\sf Let\:its\:simplest\:form\:be\:\frac{p}{q}$

$\large\sf Then\:p\:and\:q\:are\:integers\:having\:no$

$\large\sf factor\:other\:than\:1$

$\implies\large\sf 2-3\sqrt{5}=\frac{p}{q}$

$\implies\large\sf -3\sqrt{5}=\frac{p}{q}-2$

$\implies\large\sf -3\sqrt{5}=\frac{p-2q}{q}$

$\implies\large\sf \sqrt{5}=\frac{-p+2q}{3q}$

$\large\sf Since\:p\:and\:q\:are\:integers$

$\large\sf so,\frac{-p+2q}{3q}$

$\large\sf Thus,\sqrt{5}\:is\:also\:rational$

$\large\sf But\:the\:contradiction\:the\:fact\:that$

$\large\sf \sqrt{5}\:is\:irrational\:So\:our\:assumption\:is\:wrong$

$\sf Hence(2-3\sqrt{5})\:is\:irrational$