Question

prove that 3 +✓5 is an irrational

Answer 1

Answer:

Let 3+√5 is a rational no.

and its fraction form be 'a/b'

3+√5 = a/b

√5 = a/b -3

since, (a/b) -3 is a rational no.

therefore, √5 is a rational no.

this contradicts the fact that √5 is an irrational no.

therefore , 3+√5 is an irrational no.

Answer 2

$\LARGE{\underline{\mathfrak{\textcolor{blue}{Answer :}}}}$ $\Large{\fcolorbox{pink}{pink}{irrational}}$
$\LARGE{\underline{\mathrm{\textcolor{red}{Step-by-step \: \: explanation :}}}}$

$\LARGE{\underline{\mathtt{\textcolor{violet}{Given :-}}}}$
⚪ 3 + root over 5 [ 3 + √5 ]

$\LARGE{\underline{\mathtt{\textcolor{green}{To \: \: prove :-}}}}$
⚪ $3 + \sqrt{5}$ is an irrational number.

$\LARGE{\underline{\mathtt{\textcolor{teal}{Concept \: \: used :-}}}}$
⚪ Real numbers
⚪ Irrational numbers

$\LARGE{\underline{\mathtt{\textcolor{blue}{Proof :-}}}}$
✒ If possible, $3 + \sqrt{5}$ be a rational number.

let $3 + \sqrt{5} = \Large{ \frac{a}{b} }$, where a and b are co-prime integers and b ≠ 0.

Then,
✒ $3 + \sqrt{5} = \Large{ \frac{a}{b} }$

On squaring both the sides :

$\Rightarrow { \left( 3 + \sqrt{5} \right) }^{2} = \Large{ { \left( \frac{a}{b} \right) }^{2} }$

$\Rightarrow 9 + 5 + 6 \sqrt{5} = \Large{ \frac{ {a}^{2} }{ {b}^{2} } }$

$\Rightarrow 14 + 6 \sqrt{5} = \Large {\frac{ {a}^{2} }{ {b}^{2} } }$

$\Rightarrow \sqrt{5} = \Large{ \frac{ {a}^{2} - 14 {b}^{2} }{6 {b}^{2} } }$

Since a and b are integers.
Therefore, $\tt \frac{ {a}^{2} - 14 {b}^{2} }{6 {b}^{2} }$ is a rational number.

Thus, $\sqrt{5}$ is also an rational number.

This contradicts the fact that $\sqrt{5}$ is an irrational number.
This contradiction arises on assuming $3 + \sqrt{5}$ to be a rational number.
So, our assumption is wrong.
Hence, $3 + \sqrt{5}$ is an $\sf\green{\underline{\blue{irrational \: number}}}$.

$\LARGE{\underline{\mathtt{\textcolor{magenta}{Conclusion :-}}}}$
⚪$3 + \sqrt{5}$ is an $\sf\green{\underline{\blue{irrational \: number}}}$.

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