Question

Prove that √p+√q is irrational where p and q are primes in long method​

Answer 1

Step-by-step explanation:

here's the answer friend

Answer 2

$\huge{\green{\underline{\underline{\bf{\red{Solution}}}}}}$

$\large{\red{\underline{\underline{\bf{\blue{Given}}}}}}$

• √p+√q is irrational where p and q are primes in long method.

$\large{\red{\underline{\underline{\bf{\blue{To \: find}}}}}}$

• Prove that √p+√q is irrational.

$\huge{\green{\underline{\underline{\bf{\red{Explanation}}}}}}$

$\large{\red{\underline{\blue{A.T.Q}}}}$

$\rightarrow \consider \ \sqrt{p} + \ \sqrt{q} \is \rational.$

$\sqrt{p} \ \sqrt{q} \ = \x$

$x \is \also \rational \number.$

$squaring \both \sides$

$( \sqrt{p} + \sqrt{q} ^{2} ) \ = \x ^{2}$

$( \sqrt{p ^{2} } ) + ( \sqrt{q} ^{2} ) + \sqrt[2]{pq} \ = \x ^{2}$

$p \+ \ q \ + \sqrt[2]{pq} \ = \ {x}^{2}$

$\sqrt[2]{?pq} \ = \x \ - \( \p + \q )$

$\sqrt{pq} \ = \ \frac{x - (p + q)}{2}$

$\sqrt{pq} \is \irrational \and \ \frac{x - (p + q)}{2} \is \rational . \\ \\ \sqrt{p} + \sqrt{q} \is \irrational.$

$\rule{200}{2}$