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If $n$ is prime, then $\sqrt{n}$ is irrational. Prove this statement.

If I were to prove this using proof by contradiction, I would do:

Suppose $n$ is prime and $\sqrt{n}$ is rational. Let $\sqrt{n}=\frac{a}{b}$ where $a,b\in \mathbb{R}$ and have no common factor other than 1.

And then I would go on and get a contradiction like $a$ is even and $b$ is even.

However, I was wondering if I could also use proof by counterexample:

($n$ is prime) $\Rightarrow$ ($\sqrt{n}$ is irrational)

The negation of this statement is:

($n$ is prime) $\land$ ($\sqrt{n}$ is rational)

A counter example to this is $n=2$.

Therefore, this statement is disproved and hence, its negation is proven to be true.

Is this also a valid proof?

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  • $\begingroup$ Hope my answer below clarifies it. $\endgroup$
    – Peter
    Mar 5, 2021 at 19:10

1 Answer 1

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The claim is in plain text

"For every prime $n$, $\sqrt{n}$ is irrational."

The negation is in plain text

"There is some prime $n$, such that $\sqrt{n}$ is rational."

This is not a statement about all primes, but only about some prime. Therefore we cannot disprove it by a single counterexample.

To see the flaw : Assume $\sqrt{5}$ would be rational. Then the statement would be false since $5$ is prime, but $\sqrt{5}$ would not be irrational. You see, that the case $n=2$ is not enough.

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    $\begingroup$ The main point is that I forgot about "For every prime" becoming "There is some prime" in its negation. Thank you for pointing that out. $\endgroup$
    – ianc1339
    Mar 5, 2021 at 19:19

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