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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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    $\begingroup$ look at the first sentence of en.wikipedia.org/wiki/Fundamental_Theorem_of_Arithmetic $\endgroup$
    – miracle173
    Commented Jun 23, 2011 at 14:32
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    $\begingroup$ @miracle173: That is just one way of stating the theorem. $\endgroup$
    – TonyK
    Commented Jun 23, 2011 at 14:44
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    $\begingroup$ From a wider point of view, $1,-1$ are units in $\Bbb Z$. More generally, a ring $A$ is called a UFD or said to be factorial if every nonzero nonunit element admits a unique factorization into primes save order and unit multipliers. Since $1,-1$ are units, we don't care about them. =) $\endgroup$
    – Pedro
    Commented Apr 19, 2014 at 21:24

5 Answers 5

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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 unique product of primes.

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    $\begingroup$ Or, as exponent vectors, e.g. $\:2^3\cdot 3^0\cdot 5^1\mapsto\: <3,0,1,0,0,\cdots>,\ \ 1\mapsto\: <0,0,0,\cdots>\:$. $\endgroup$ Commented Jun 23, 2011 at 15:37
  • $\begingroup$ what about prime factorization of 0? $\endgroup$
    – Yukulélé
    Commented Jan 24, 2020 at 6:59
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    $\begingroup$ Re: my prior comment: see also here for more general remarks on empty products and the utility of adjoining neutral elements to semigroups to get a monoid. $\endgroup$ Commented Dec 14, 2021 at 9:14
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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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    $\begingroup$ 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. $\endgroup$
    – TonyK
    Commented Jun 23, 2011 at 14:42
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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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    $\begingroup$ How fascinating. $\endgroup$ Commented Oct 15, 2011 at 2:34
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    $\begingroup$ 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. $\endgroup$
    – t.b.
    Commented Oct 15, 2011 at 4:46
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You need to change the theorem because anything that works for this contradicts the theorem. Any natural number greater than 1 can be written as a product of prime factors.

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