given that is prime,proof that (√p) is irrational


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Question Number 166910 by rexford last updated on 02/Mar/22

$g i v e n t h a t i s p r i m e, p r o o f t h a t \sqrt{p} i s$ $i r r a t i o n a l$
Commented by rexford last updated on 02/Mar/22

$t h a n k y o u$
	Answered by mr W last updated on 02/Mar/22

	$a s s u m e \sqrt{p} = r a t i o n a l = \frac{a}{b} w i t h g c d (a, b) = 1$ $\Rightarrow p = \frac{a^{2}}{b^{2}} \Rightarrow p i s a p e r f e c t s q u a r e$ $\Rightarrow c o n t r a d i c t i o n$ $a c t u a l l y \sqrt{p} i s a l w a y s i r r a t i o n a l, i f$ $p i s n o t a p e r f e c t s q u a r e . t h e r e f o r e$ $\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{6}, \sqrt{7}, \sqrt{8}, \sqrt{10}, \sqrt{11}, \sqrt{12},$ $\sqrt{13}, \sqrt{14}, \sqrt{15}, \sqrt{17}, \sqrt{18}, \sqrt{19}, . . . a r e a l l$ $i r r a t i o n a l n u m b e r s .$
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