Question

If √2 is an irrational number then Prove that 5−√2 is an irrational number.​

Answer 1

let 5 - √2 is an irrational number which is in the form of p/q where p and q are co-prime numbers and q ≠ 0

then,

$5 - \sqrt{2} = \frac{p}{q} \\ \\ \bold{: \implies \: 5 = \frac{p}{q} + \sqrt{2} } \\ \\ \bold{: \implies \: \sqrt{2} = 5 - \frac{p}{q} } \\ \\ \bold{: \implies \: \sqrt{2} = \frac{5q - p}{q} }$

here, 5q , -p and q are Integers

$\bold{ ➪ \: \frac{5q - p}{q} \: is \: a \: rational \: number }$

but √2 is an irrational number (given)

so, our assumption is wrong

hence, 5 - √2 is an irrational number

Answer 2

$\underline{\underline{\huge{\gray{\tt{\textbf Answer :-}}}}}$

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$\sf\large\bold{\orange{\underline{\blue{ Given :-}}}}$

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• √2 is an irrational number

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$\sf\large\bold{\orange{\underline{\blue{ To \: Prove :-}}}}$

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• 5 - √2 is irrational number

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$\sf\large\bold{\orange{\underline{\blue{ Solution :-}}}}$

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• Let us assume 5 - √2 to be rational number

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We know that rational number could be expressed in the form of $\sf \dfrac { p } { q }$ where , q ≠ 0 and p & q are integers

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Thus, 5 - √2 could be expressed in the form of $\sf \dfrac { p } { q }$ where , q ≠ 0

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So,

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➜ $\sf 5 - \sqrt 2 = \dfrac { p } { q }$

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➜ $\sf 5 - \dfrac { p } { q } = \sqrt 2$

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➜ $\sf \dfrac { 5q - p } { q } = \sqrt 2$

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Here in LHS 5 , q , p are integers thus LHS is a rational number So RHS need to be a rational number i.e √2 is a rational number

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But it is given that √2 is an irrational number

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This contradiction has been arisen due to our wrong assumption as 5 - √2 is a rational number

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• Thus 5 - √2 is a irrational number