Fill In The Blanks[1 Marks ]

Question 11 Mark

$\Big(\frac{2+\sqrt{5}}{3}\Big)$is _________ number.

Answer

$\Big(\frac{2+\sqrt{5}}{3}\Big)$is irrational number.Solution:
$\therefore\Big(\frac{2+\sqrt{5}}{3}\Big)$is not be written in the form of $\frac{\text{n}}{\text{d}}$
Where n amd d is non zero number. So, $\Big(\frac{2+\sqrt{5}}{3}\Big)$is irrational number.

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Question 21 Mark

Prove that $\sqrt{5}$ is an irrational number.

Answer

$\sqrt{5}=\frac{\text{p}}{\text{q}}, p q$ are coprimes $q \neq 0$
$5q^2 = p^2$
$\Rightarrow 5$ divides $p^2$
$\Rightarrow 5$ divides $p$ also
Let $p = 5a,$ for some integer $a$
$5q^2 = 25a^2$
$\Rightarrow q^2 = 5a^2$
$\Rightarrow 5$ divides $q^2$
$\Rightarrow 5$ divides $q$ also
$\therefore 5$ is a common factor of $p, q,$ which is not possible as $1\ p, q$ are coprimes.
Hence assumption is wrong $\sqrt{5}$ is irrational no.

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Question 31 Mark

Prove that $\sqrt{5}$ is an irrational number.

Answer

$\sqrt{5}=\frac{\text{p}}{\text{q}}, p q $are coprimes $q \neq 0$
$5q^2 = p^2$
$\Rightarrow 5$ divides $p^2$
$\Rightarrow 5$ divides $p$ also
Let $p = 5a,$ for some integer $a$
$5q^2 = 25a^2$
$\Rightarrow q^2 = 5a^2$
$\Rightarrow 5$ divides $q^2$
$\Rightarrow 5$ divides $q$ also
$\therefore 5$ is a common factor of $p, q,$ which is not possible as $1\ p, q$ are coprimes.
Hence assumption is wrong $\sqrt{5}$ is irrational no.

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