Question

prove that √2 +√3 is irrational​

Answer 1

Answer:

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

squaring on both sides

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

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

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

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

2✓6=a^2/b^2-5/1

2✓6=a^2-5b^2/b^2

✓6=a^2-5b^2/2b^2

therefore it is a irrational