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. $\:2^3\cdot 3^0\cdot 5^1\mapsto\: <3,0,1,0,0,\cdots>,\ \ 1\mapsto\: <0,0,0,\cdots>\:$. – Gone 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

The OP hasn't misinterpreted the theorem. Every nonzero integer can be written as a product of primes.(GTM84 P.3) Just the exponents are all zeros...

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How fascinating. – The Chaz 2.0 Oct 15 '11 at 2:34
This is rather lousy referencing style. OP will very likely be unaware of what GTM means and thus be unable to track it down. Why not just link to it? For the record: you're referring to Ireland and Rosen, A classical introduction to modern number theory, Springer Graduate Texts in Mathematics, volume 84. Since you have that book in front of you, what does it cost you to give that information? While it costs the reader some minutes to figure out and locate what you're referring to. – t.b. Oct 15 '11 at 4:46