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

$$n\in {\mathbb{N}}^{\ast };nisnotasquareofany$$$$integer.Showthat\sqrt{n}\notin Q.$$

Answered by mindispower last updated on 19/Oct/21

$$\sqrt{n}=\frac{p}{q},pq=1\Rightarrow$$$$\Rightarrow q^{2}n=p^2$$$$\Rightarrow q^2\mid p^2,pq=1\Rightarrow p^2{q}^2=1$$$$q^2\mid p^2,q^2\mid p^2\Rightarrow q^2\mid p^2{q}^2=1\Rightarrow q=1$$$$\sqrt{n}=p\in \mathbb{N}\Rightarrow n=p^{2}butnisnotsquare\Rightarrow \sqrt{n}\notin Q$$

Commented by mathocean1 last updated on 22/Oct/21

$$thanksmindispower$$

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