# Prime factorization of 1

Fundamental Theorem of Arithmetic says every positive number has a unique prime factorisation. Question: If 1 is neither prime nor composite, then how does it fit into this theorem?

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look at the first sentence of en.wikipedia.org/wiki/Fundamental_Theorem_of_Arithmetic –  miracle173 Jun 23 '11 at 14:32
@miracle173: That is just one way of stating the theorem. –  TonyK Jun 23 '11 at 14:44

Let us remember that an empty product is always 1. Hence, 1 has the empty product as its prime factorization. This product is vacuously a product of primes.

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Or, as exponent vectors, e.g. $\:2^3\cdot 3^0\cdot 5^1\mapsto\: <3,0,1,0,0,\cdots>,\ \ 1\mapsto\: <0,0,0,\cdots>\:$. –  Bill Dubuque Jun 23 '11 at 15:37

It has (uniquely!) zero prime factors.

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I think you have simply misinterpreted the theorem. It should be stated as "...every positive number greater than one has a unique prime factor." .c.f. http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic

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This is a question of style, not content. You don't have to exclude 1, if you interpret an empty set of primes as a factorisation; but you might want to for pedagogical purposes, so that it doesn't distract from the important issues. –  TonyK Jun 23 '11 at 14:42