Prime factorization of 1

Question

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?

look at the first sentence of http://en.wikipedia.org/wiki/Fundamental_Theorem_of_Arithmetic — miracle173, Jun 23 '11 at 14:32
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. =) — Pedro, Apr 19 '14 at 21:24

score 44 · Accepted Answer · edited Apr 19 '14 at 21:22

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 unique product of primes.

edited Apr 19 '14 at 21:22

Travis Bemrose

1,098

answered Jun 23 '11 at 14:36

ncmathsadist

49,383

13

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
what about prime factorization of 0? – Yukulélé Jan 24 '20 at 06:59
1

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. – Bill Dubuque Dec 14 '21 at 09:14

score 20 · Answer 2 · answered Jun 23 '11 at 14:31

20

It has (uniquely!) zero prime factors.

answered Jun 23 '11 at 14:31

TonyK

64,559

score 2 · Answer 3 · answered Jun 23 '11 at 14:31

2

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

answered Jun 23 '11 at 14:31

cobaltduck

1,483

8

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

score 2 · Answer 4 · edited Oct 15 '11 at 04:49

2

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...

edited Oct 15 '11 at 04:49

t.b.

78,116

answered Oct 15 '11 at 02:32

Vladimir

2,879

6

How fascinating. – The Chaz 2.0 Oct 15 '11 at 02:34
9

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 04:46

score 1 · Answer 5 · answered May 28 '14 at 18:38

1

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.

answered May 28 '14 at 18:38

user153905

21

Prime factorization of 1

5 Answers5

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