Question

show that √3+√5 is irrational​

Answer 1

ANSWER

Let us assume that 3+5 is a rational number.

Now,

3+5=ba [Here a and b are co-prime numbers]

5=[(ba)−3]

5=[(ba−3b)]

Here, [(ba−3b)] is a rational number.

But we know that 5 is an irrational number.

So, [(ba−3b)] is also a irrational number.

So, our assumption is wrong.

3+5 is an irrational number.

Hence, proved.

Answer 2

$\huge{\blue{\bold{\underline{\underline{Answer :}}}}}$

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$\large{\green{\underline \bold{\tt{Given :-}}}}$

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• √3 + √5

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$\large{\red{\underline \bold{\tt{To \: Prove :-}}}}$

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• √3 + √5 is a irrational number

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$\large{\orange{\underline{\tt{Solution :-}}}}$

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• Let us assume it to be a rational number

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We know that

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Rational numbers are the numbers that can be expressed in the form of $\sf \dfrac { p } { q }$ where p & q are integers and q isn't equal to zero.

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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 on both side ⟯

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➠ (a - b)² = a² + b² - 2ab

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

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

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

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

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As p and q are integers RHS is also rational.

As RHS is rational LHS is also rational

But this contradicts the fact that √5 is irrational

This contradiction arose because of our wrong assumption.

∴ √3 + √5 is an irrational number

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