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?

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

1 Answer 1


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.

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.