Question

Prove that root2+root3 is irrational​

Answer 1

Step-by-step explanation:

question_answer Answers(2)

Let √2 + √3 = (a/b) is a rational no. So,5 + 2√6 = (a2/b2) a rational no. Since, 2√6 is an irrational

Answer 2

Answer:

Suppose root2+root3 is a rational number.

$\therefore \: \sqrt{2} + \sqrt{3} = \frac{a}{b} \:$

(Where a and b are integers)

Squaring both sides

${( \sqrt{2} + \sqrt{3}) }^{2} = {( \frac{a}{b}) }^{2} \\ \\ {( \sqrt{2)} }^{2} + 2 (\sqrt{2})( \sqrt{3}) + {( \sqrt{3}) }^{2} \\ = {\frac{a ^{2} }{b ^{2} } } \\ \\ 2 + 2 \sqrt{6} + 3 = \frac{ {a}^{2} }{ {b}^{2} }$

$5 + 2 \sqrt{6} = \frac{ {a}^{2} }{ {b}^{2} } \\ \\ 2 \sqrt{6} = \frac{ {a}^{2} }{ {b}^{2} } - 5 \\ \\ 2 \sqrt{6} = \frac{ {a}^{2} - {5b}^{2} }{ {b}^{2} } \\ \\ \sqrt{6} = \frac{ {a}^{2} - {5b}^{2} }{ {2b}^{2} }$

Here,

$\frac{ {a}^{2} - {5b}^{2} }{ {2b}^{2} }$

is rational as a and b are integers.

root6 is also rational.

But actually it is irrational.

The construction has arisen due to our incorrect assumption root2+root3 is rational.

Hope it helps you from my side