Question

Prove that the following numbers are irrational.
√3+√5​

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 + √5 is an irrational number

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

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

Rational numbers are the ones that can be expressed in the form of $\sf \dfrac { p } { q }$ where p and q are integers and q ≠ 0

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So as per our assumption (√3 + √5) could be expressed in the form of $\sf \dfrac { p } { q }$ where p and q are integers and q ≠ 0

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

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

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⟮ Squaring both sides ⟯

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➜ $\sf (\sqrt 3)^2 = (\dfrac { p } { q } - \sqrt 5)^2$

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➜ $\sf 3 = \dfrac { p ^2 } { q ^2 } + 5 - 2 \times \dfrac { p } { q } \times \sqrt 5$

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➜ $\sf 2 \times \dfrac { p } { q } \times \sqrt 5 = \dfrac { p ^2 } { q ^2 } + 5 - 3$

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

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

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➜ $\sf \sqrt 5= \dfrac { p ^2 + 2q ^2 } { q \times p \times 2}$

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As 2 , p , q are integers thus the RHS is rational hence the LHS need to be rational too

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But √5 is an irrational number , this contradiction has been arisen due to our wrong assumption that √3 + √5 is rational

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∴ √3 + √5 is an irrational number

Answer 2

Answer:

Let √3+√5 be a rational number. A rational number can be written in the form of p/q where p,q are integers. p,q are integers then (p²+2q²)/2pq is a rational number. ... Therefore, √3+√5 is an irrational number.