Question

i)Prove that sq.root 7 + sq.root2 is an irrational number.
$\sqrt{7} + \sqrt{2}$
​

Answer 1

Answer~

To prove:- √7 + √2 is an irrational number.

________________

➭Let us first consider √2 + √7 is a rational number.

$\longrightarrow \bold{ \sqrt{2} + \sqrt{7} = \frac{a}{b} } \red \rightarrow \: eq.1$

$\longrightarrow \bold{ \sqrt{2} = \frac{a}{b} - \sqrt{7} } \red \rightarrow \: eq.2$

➦ Now square eq.2

$\longrightarrow \bold{( \sqrt{2})^2 = ( \frac{a}{b} - \sqrt{7})^2 }$

$\longrightarrow \bold{2 = \frac{a^2}{b^2} + 7 - \frac{2 \sqrt{7}a }{b} }$

$\longrightarrow \bold{ \frac{2 \sqrt{7a} }{b} = \frac{a^2}{b^2} + 5}$

$\longrightarrow \bold { \sqrt{7} = \frac{a^2 + 5b^2}{2ab} }$

$\longrightarrow \bold \green{So \: here \sqrt{7} \: is \: irrational \: and \: \frac{a^2 + 5b^2}{2ab}is \: rational }$

_____________________

➦ From these we can conclude that our contradiction is wrong as rational number cant be equal to irrational number.

So √2+√7 is irrational number.

Answer 2

${\huge{\red{αղsωєя-:}}}$

To prove:- √7 + √2 is an irrational number.

________________

➭Let us first consider √2 + √7 is a rational number.

$\implies \bold{ \sqrt{2} + \sqrt{7} = \frac{a}{b} } \red \rightarrow \: eq.1$

$\implies \bold{ \sqrt{2} = \frac{a}{b} - \sqrt{7} } \red \rightarrow \: eq.2$

➦ Now square eq.2

$\implies \bold{( \sqrt{2})^2 = ( \frac{a}{b} - \sqrt{7})^2 }$

$\implies \bold{2 = \frac{a^2}{b^2} + 7 - \frac{2 \sqrt{7}a }{b} }$

$\implies \bold{ \frac{2 \sqrt{7a} }{b} = \frac{a^2}{b^2} + 5}$

$\implies \bold { \sqrt{7} = \frac{a^2 + 5b^2}{2ab} }$

$\bold \pink{So \: here \sqrt{7} \: is \: irrational \: and \: \frac{a^2 + 5b^2}{2ab}is \: rational }$