Question

prove that rootP+rootQ is irrarational, where P,Q are primes

Answer 1

Step-by-step explanation:

assume tha

$\sqrt{p} + \sqrt{q}$

is rational

then,there exist co-prime a and b such that

$\sqrt{p} + \sqrt{q} = \frac{a}{b} \\ \implies\sqrt{p} = \frac{a}{b} - \sqrt{q}$

squaring on both side

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

since a,b,p,q are integers so,

$\frac{ {a}^{2}b + {b}^{3} (p - q) }{2a}$

is rational

thus

$\sqrt{q}$

is also rational

but q being prime

$\sqrt{q}$

is irrational

since , a contradiction arises so our assumptions is incorrect

hence

$( \sqrt{p} + \sqrt{q} )$

is irrational no