Question

Prove that √2 + √3 is an irrational number​

Answer 1

Step-by-step explanation:

Let √2 + √3 = (a/b) is a rational no. On squaring both sides , we get 2 + 3  + 2√6 = (a2/b2) So,5 + 2√6 = (a2/b2) a rational no. So, 2√6 = (a2/b2) – 5 Since, 2√6 is an irrational no. and (a2/b2) – 5 is a rational no. So, my contradiction is wrong. So, (√2 + √3) is an irrational no.

thanks pls mark me

........

Answer 2

Answer:

$\sqrt{2} + \sqrt{3} = \frac{a}{b} (taking \: them \: rational) \\ squaring \: on \: both \: sides \\ 2 + 3 + 2 \sqrt{6} = \frac{ {a}^{2} }{ {b}^{2} } \\ \sqrt{6} = \frac{ {a}^{2} - 5 {b}^{2} }{2 {b}^{2} } \\ but \: now \: at \: lhs \: we \: have \: irrational \: no \\ and \: at \: rhs \: we \: have \: rational \\ this \: is \: due \: to \: our \: wrong \: assumption \\ \\ \\ the \: given \: number \: is \: irrational$

Step-by-step explanation:

hope it helps you friend.