Question

Prove that 2✓3 + 5✓7 is an irrational number.


Solve it

Answer 1

we know that a,b,2 and 5 are integers and they are also rational
-2 root 3 = a/b - 5
2 root 3 = 5 - ab 
2 root 3 = 5b/b - a/b
root 3 = 5b - a / 2b
therefore,root3 will be rational

but we know that root3 is irrational

there is a contradiction

so,5-2root3 is irrational

Answer 2

➡HERE IS YOUR ANSWER⬇

Let us consider
$(2 \sqrt{3} + 5 \sqrt{7} ) \: is \: a \: rational \: number \: \frac{a}{b} \\ where \: b \: is \: non \: zero. \\ \\ then \\ 2 \sqrt{3} + 5 \sqrt{7} = \frac{a}{b} \\ \\ or \: \: 2 \sqrt{3} = \frac{a}{b} - 5 \sqrt{7} \\ \\ or \: \: ( {2 \sqrt{3} })^{2} = {( \frac{a}{b} - 5 \sqrt{7} })^{2} \\ \\ or \: \: 12 = \frac{ {a}^{2} }{ {b}^{2} } - 10 \sqrt{7} \frac{a}{b} + 175 \\ \\ or \: \: 10 \sqrt{7} \frac{a}{b} = \frac{ {a}^{2} }{ {b}^{2} } + 163 \\ \\ or \: \: \sqrt{7} = \frac{a}{10b} + \frac{163b}{10a} \\ \\ which \: \: implies \: \: that \: \: \sqrt{7} \: \: is \\ a \: rational \: number \\ a \: \: contradiction. \\ \\ thus \: \: (2 \sqrt{3} + 5 \sqrt{7} ) \: \: is \: \: an \: \: \\ irrational \: \: number.$

(Proved)

⬆HOPE THIS HELPS YOU⬅