n-N-n-is-not-a-square-of-any-integer-Show-that-n-Q-

Question Number 157098 by mathocean1 last updated on 19/Oct/21

n \in N^{*}; n i s n o t a s q u a r e o f a n y

i n t e g e r . S h o w t h a t \sqrt{n} \notin Q .

Answered by mindispower last updated on 19/Oct/21

\sqrt{n} = \frac{p}{q}, p q = 1 \Rightarrow

\Rightarrow q^{2} n = p^{2}

\Rightarrow q^{2} ∣ p^{2}, p q = 1 \Rightarrow p^{2} q^{2} = 1

q^{2} ∣ p^{2}, q^{2} ∣ p^{2} \Rightarrow q^{2} ∣ p^{2} q^{2} = 1 \Rightarrow q = 1

\sqrt{n} = p \in N \Rightarrow n = p^{2} b u t n i s n o t s q u a r e \Rightarrow \sqrt{n} \notin Q

Commented by mathocean1 last updated on 22/Oct/21

t h a n k s m i n d i s p o w e r