Talk:Divisor

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Revision as of 09:52, 3 April 2007 by imported>Larry Sanger
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Here's another perfect example of a topic that could benefit from a plainer-language, if inexact, definition given first (and billed as "rough" or "inexact")--followed by the more precise, but harder-to-understand, definition. --Larry Sanger 17:47, 31 March 2007 (CDT)

For example (I'm not going to actually edit the article--this no doubt needs worthsmithing):

A divisor of a number, roughly speaking, "goes into" the number evenly, with nothing left over (no remainder). For example, 2 is a divisor of 4, because 2 goes into 4 two times, with nothing left over. But 2 is a not a divisor of 5, because 2 goes in 5 2.5 times.
More exactly, given two numbers d and a, d is a divisor of a if, and only if, a divided by d equals an integer, that is, a number without fractions. So if d = 5 and a = 15, then d/a = 3, and so d is a divisor of a.
Even more exactly and formally, given two integers d and a, where d is nonzero, d is said to divide a, or d is said to be a divisor of a, if and only if there is an integer k such that dk = a. For example, 3 divides 6 because 3*2 = 6. Here 3 and 6 play the roles of d and a, while 2 plays the role of k. Though any number divides itself (as does its negative), it is said not to be a proper divisor. The number 0 is not considered to be a divisor of any integer.

"proper divisors" comment

1 and -1 might be proper divisors, contrary to the current version. I think they're called trivial divisors instead. My evidence: The statement "6 is perfect because it is the sum of its proper divisors 1, 2, and 3" is everywhere.Rich 20:09, 31 March 2007 (CDT)

I'll fix it. Thanks. Greg Woodhouse 20:21, 31 March 2007 (CDT)
Man I'm over the hill! By my quote about 6 above negative numbers like -1 or -3 can't be proper divisors of 6. Sorry.Rich 12:37, 2 April 2007 (CDT)

Further Reading

"Fearless Symmetry" is certainly a fascinating read, but really out of place in this article. (It's an introduction to the ideas behind the proof of Fermat's last theorem for non-specialists.) I was a bit, well, ebulient, in placing it here. I'll probably use it elsewhere, such as in an article on reciprocity laws. Greg Woodhouse 13:22, 2 April 2007 (CDT)