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Prove the following statement: $\forall r\in \mathbb{R^+}$,if r is irrational then $\sqrt r$ is irrational

My attempt:

if suppose $\sqrt r $ is rational

then there exists $p,q \in \mathbb{R^+}$ such that

$\frac{p}{q}=\sqrt r $ where p and q are prime to each other

$\rightarrow p=q \sqrt r \rightarrow p^2=r q^2$

i can't processed further can any help

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  • $\begingroup$ $p,q \in \mathbb{N}$ $\endgroup$ – N74 Feb 19 '17 at 7:37
  • $\begingroup$ As @N74 says, $p,q \in \Bbb N$! $\endgroup$ – Error 404 Feb 19 '17 at 9:09
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$r={p^2\over q^2}$ so it is rational contradiction.

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  • $\begingroup$ ..how it is irrational can you exaplain $\endgroup$ – user271336 Feb 19 '17 at 7:24
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Then $r = p^2/q^2$, which is rational, a contradiction.

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  • $\begingroup$ @user49640...can u expalin bit more $\endgroup$ – user271336 Feb 19 '17 at 7:31
  • $\begingroup$ $p^2/q^2$ is a quotient of integers. $\endgroup$ – user49640 Feb 19 '17 at 7:32
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easier to do contrapositive, if $\sqrt{r}$ is rational then so is $r$.

If $\sqrt{r} = p/q$ with $p$ and $q$ integers then $$ r = p^2/q^2 $$ so it is a ratio of integers so it is rational.

Taking the contrapositive, we have the statement you wanted.

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