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We can say that $p$ is prime if and only if :

$$p\ne 1$$ $$p\space is\space prime \leftrightarrow \left\{(1|p)\wedge(p|p)\wedge\forall i\space2\leq i\leq p-1 (p\equiv r(mod\space i),r\neq0\right),(p,r)\in N\}$$

$$Edit: $$

$$p\ne 1$$ $$p\space is\space prime \leftrightarrow \left\{(\forall i\space2\leq i\leq p-1 (i\not\mid p)\right),(p,i)\in \mathbb{Z^{+}}\}$$

Noodle
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    The statements $1|p$ and $p|p$ hold for every $p\in Z^+$ so you can ignore them. – Yanko Sep 11 '18 at 14:24
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    Also another problem with your definition is that $1$ happens to be a prime number – Yanko Sep 11 '18 at 14:25
  • Can this be corrected if I say $1$ is not a prime number, and $p,r \in N$? – Noodle Sep 11 '18 at 14:32
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    yes if you add $p\not=1$ then it's fine. – Yanko Sep 11 '18 at 14:32
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    Indicating that $p$ has a non-zero remainder $\mod{i}$, i.e. upon division by $i$, is equivalently stated in the more traditional manner, $i\not\mid p$. – Keith Backman Sep 11 '18 at 14:41
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    Also, for natural numbers $p$, the statements $1\mid p$ and $p\mid p$ are superfluous, as such is true of all natural numbers. – Keith Backman Sep 11 '18 at 14:44
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    @Yanko 1 is not prime. – Did Sep 11 '18 at 14:48
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    @Did That's what I meant. I mentioned that $p=1$ satisfies the conditions in the OP. – Yanko Sep 11 '18 at 14:50
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    Another characterization of primes that generalizes well to rings other than $\mathbb{Z}$ is that $p$ is prime if whenever $p$ divides $ab$, then $p$ divides $a$ or $p$ divides $b$. – Daniel Mroz Sep 11 '18 at 15:02
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    What is your real question? Why would you ever want to define primes in this way when perfectly good, and well-understood, definitions abound? – David G. Stork Sep 11 '18 at 15:16
  • I wanted to define them formally and concisely using quantifiers , because every time I don't like writing a whole sentence just to say that a number is prime , it's just for fun. – Noodle Sep 11 '18 at 15:21
  • Why do you keep $p\ne1$ outside of the RHS of the definition? – Did Sep 11 '18 at 20:14

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