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Let us define primes in $\mathbb Z$ as follows:

Def$1$ $p>1$ is said to be a prime if only positive divisors of $p$ are $1$ and $p$.

Def$2$ A natural number $p>1$ is said to be a prime if $p|ab\implies p|a $ or $p|b$.

I want to show these definitions are equivalent.But I have no clue how to proceed.Can someone show me a way out?$(1)\implies (2)$ is easy but what about the converse?

Kishalay Sarkar
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    What is expected is that you assume one as a definition, and proceed to derive the other as a property. Then switch which one is assumed and do the other proof. [details on steps are in many number theory books.] – coffeemath Sep 24 '21 at 05:01
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    @KishalaySarkar: Have you checked Burton's elementary number theory book? I remember seeing something along these lines there before. – Jose Arnaldo Bebita Dris Sep 24 '21 at 05:04
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    Suppose that $p = r \times s ~: ~r,s \in \Bbb{Z_{\geq 2}}.$ Can you think of specific values for $a,b$ such that $p ~| ~(a \times b)~$ and yet $p$ does not divide either $a$ or $b$? – user2661923 Sep 24 '21 at 05:16
  • @user2661923 if we consider $r,s$ to be prime, and let's consider $a$ has $r$ in prime factorisation but not $s$ and b has $s$ in prime factorisation but not $r$,so $p|a \times b$ and $p\nmid a$ ,$p\nmid b$.am I correct? – user69608 Sep 24 '21 at 07:14
  • @user69608 Re-examine my last comment. Irrelevant whether $r,s$ prime. The comment, as is, suggests specific values for $a,b$ directly in terms of $r,s$ such that $p ~| ~(a\times b)$ and $p$ does not divide either $a$ or $b$. – user2661923 Sep 24 '21 at 07:17
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    Related The relation between irreducible and primes. – Conifold Sep 24 '21 at 08:01

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