Question

Prove that (3 + 2 √5) is irrational.​

Answer 1

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

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

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• (3 + 2 √5) is irrational

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

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• Let us assume (3 + 2 √5) to be a rational number

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Rational number are the numbers that can be expressed in the form of $\sf \dfrac { p } { q }$ Where p & q are co primes and q ≠ 0

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So as per our assumptions (3 + 2 √5) could be expressed in the form of $\sf \dfrac { p } { q }$ where , p & q are co primes and q ≠ 0

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So,

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➜ (3 + 2 √5) = $\sf \dfrac { p } { q }$

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➜ 2 √5 = $\sf \dfrac { p } { q } - 3$

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➜ 2 √5 = $\sf \dfrac { p - 3q} { q }$

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➜ √5 = $\sf \dfrac { p - 3q} { 2q }$

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Here in the RHS p and q are rational number also 2 & 3 are rational numbers hence RHS is rational thus LHS must be a rational number

But this contradicts the fact that √5 is a irrational number , this contradiction has been arisen due to our wrong assumption that (3 + 2 √5) is rational

Hence (3 + 2 √5) is irrational

Answer 2

Step-by-step explanation:

hope that above answer will help you

tao left first

and tap right side after tapping left