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

Question

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

As @N74 says, $p,q \in \Bbb N$! – Error 404 Feb 19 '17 at 09:09 — Error 404, Feb 19 '17 at 09:09

score 2 · Accepted Answer · answered Feb 19 '17 at 07:21

2

$r={p^2\over q^2}$ so it is rational contradiction.

answered Feb 19 '17 at 07:21

Tsemo Aristide

score 1 · Answer 2 · answered Feb 19 '17 at 09:53

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.

score 0 · Answer 3 · answered Feb 19 '17 at 07:20

0

Then $r = p^2/q^2$, which is rational, a contradiction.

answered Feb 19 '17 at 07:20

user49640

3 Answers3