1
$\begingroup$

I have come up with this definition: n is prime iff

$$(n>1)\ \wedge \ \left[ \forall x.\ \ (\exists k. n=kx) \implies (x=1 \vee x=n) \right] $$

Is this definition correct?

$\endgroup$
  • 1
    $\begingroup$ Yes .. assuming your domain is the natural numbers $\endgroup$ – Bram28 May 28 at 17:06
  • $\begingroup$ Yes it is. I should put it in there $\endgroup$ – FizzleDizzle May 28 at 17:06
  • 1
    $\begingroup$ Fun historical fact: The old definition of prime was just the right conjunct. That is, 1 used to be a prime. Put differently: under the old definition, a prime number was one with at most two divisors. Now, it is a number with exactly two divisors. Which itself you can formalize as: $\forall n (Prime(n) \leftrightarrow \exists x \exists y (x \not = y \land \forall z (\exists k \ n = kz \leftrightarrow (z = x \lor z = y))))$ $\endgroup$ – Bram28 May 28 at 17:09

Your Answer

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

Browse other questions tagged or ask your own question.