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Fundamental theorem of arithmetic. Proof:- $P(n)$: $n$ is a prime or can written uniquely as a product of primes. $P(2)$ is true, and assume true for $n=3,4,...,k$. Now if $k+1$ is not a prime it can be written as $k+1= r\times s$ . Now $r

Where did I go wrong? Can I include 'uniquely' in the statement? Because in every book it is proved first that every composite number is product of primes and then uniqueness is proved.

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    Your proof seems correct to me. You should define $r$ and $s$ upon using those variables. – 2017-02-12
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    @kingW3 I know the proof. And that is why I am asking. In the first step when we write the statement why can't we include the uniqueness part in the statement? – 2017-02-12
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    @user398623 I retract my comment,though since both $r,s\leq k$ they can be written as a product of unique primes,now what's left is to conclude that a product of two such numbers must itself be unique. – 2017-02-12
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    @kingW3 Yes, I get you. – 2017-02-12
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    That inductive proof shows only *existence* of prime factorizations, not *uniqueness*. – 2017-02-12

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That $r,\,s$ have unique prime factorisations proves $rs$ has a prime factorisation, but you still need to show it's unique.

The usual proof begins with Bézout's lemma, stating the hcf of two integers is a linear combination of them. Thus if $p|mn$ with $p$ prime and $p\nmid m$ then integers $x,\,y$ exist with $1=px+my$ and $n=pnx+mny$, an obvious multiple of $p$.

Since $p$ divides $m$ or $n$ whenever it divides $n$, if $pM=q_1\cdots q_N$ with $p,\,q_i$ prime then $p$ divides, and hence equals, some $q_i$, say $q_1$ so $M=q_2\cdots q_N$. So if $k+1$ were a natural with multiple prime factorisations, we could cancel one of its prime factors to obtain a smaller such natural.