Talk:Divisor/Archive 1

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Archive 1

Old comments

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Is the definition of divisor truly strictly as an integer in all math systems and practical applications of math?

I would have thought the term in more general use would be almost synonimous with denominator.

In other words it would not bother me unduly if an engineer referred to the denominator in the expressions 3/3.14 or 3/pi or (3x+1)/(x+1) as a divisor.

Would this be completely wrong? user:mirwin

I wrote this generalization:

"Let C, D and B be natural numbers, being B>D, Let C digits be written in a B base. For any D that complies with (B mod D = 1), if the D modulus of the sumatorium of C digits is 0, this implies that C mod D = 0."

...but I am not a professional mathematician. This theorem is likely to be hundreds of years old. ¿Can some expert tell me who discovered this, or if this is considered "common knowledge" or something. I would be most grateful.

Sincerely

Vicente Aceituno

I don't know who discovered it, but I'm sure it's as old as modular arithmetic. In your statement, you can safely drop the assumption B>D, and you can sharpen it a bit by stating: (sum of digits of C) mod D = C mod D. In particular, the sum of digits of C is divisible by D if and only if C is divisible by D. The proof goes like this: if B ≡ 1 (mod D), then B^j ≡ 1^j = 1 (mod D), and so

\sum a_{j}B^{j}\equiv \sum a_{j}\qquad ({\mbox{mod }}D)

AxelBoldt (talk) 18:58, 17 September 2004 (UTC)

Thanks a lot. I am not sure if the sharpening breaks something, as what I wanted to do was to explain why divisibility rules work. Particularly the division by 3 rule, that tantalized me since I was a kid. vaceituno (talk) 12:0, 19 September 2004 (UTC)

iff?

Being a Dane, I wonder if(f) this is a word missing in my vocabulary :-) [[1]]

Shall we say iff. It is math lingo. Oleg Alexandrov 15:09, 22 April 2005 (UTC)

Change of wording

In the rule to find n mod 1001 in order to test wheter n is a multiple of 7, or 13, I had written "divide the number in groups of three digits from right to left". It was changed to "left to right", which isn't what I meant. Dividing 1048576 in groups of three digits from left to right would yield 104'857'6, wouldn't it? Army1987 14:38, 2 May 2005 (UTC)

Problems

Looks like some technical problem in red text...I have no idea what to do about it. Gert2 03:35, 4 January 2007 (UTC)

Nevermind...I fixed it. Gert2 03:39, 4 January 2007 (UTC)

Divisibility rule

should I think be reinstated as a separate article - it's not central to the concept of "divisor" and there is enough material for a separate article. — ciphergoth 20:13, 24 July 2005 (UTC)

I'd certainly vote for that. --Jay (Histrion) 19:33, 4 August 2005 (UTC)
my vote goes for that too -- Yannick Gingras 17:36, 6 August 2005 (UTC)
Yes please. The way it stands now, this takes way too much space in the article, and now with the proofs thrown in. By the way, according to Wikipedia:WikiProject_Mathematics/Proofs proofs should not be that proeminent in articles, especially if they are long, as the recently added proof in the divisor article. Oleg Alexandrov 17:52, 6 August 2005 (UTC)

I agree - a new article "Divisibilty rules", and link to that. Bubba73 18:14, 6 August 2005 (UTC)

For some reason, though, I feel "rules" is the wrong term. Would "checks," "shortcuts," or "tricks" be better? Or maybe "criteria," which has a meaning closer to "rules" but seems a bit more appropriate? --Jay (Histrion) 20:08, 6 August 2005 (UTC)

OK, I moved the material to Divisibility rule, although I'd still like to talk about a name change for the "new" topic. --Jay (Histrion) 20:23, 6 August 2005 (UTC)

I just found this page. I read a book, Vedic Mathematics, 1965, 1978. In chapters 29 and 30, pages 273-295, the author gives an algorithm for divisibility called osculation. It gets complicated as there are several methods: digit-wise, whole number, positive and negative, complex, and multiplex osculation. It can be done mentally or with some notation of the results of an addition and multiplication. It does not require division! The positive osculator plus the negative osculator equals the divisor, P+Q=D. Example: is the number 5,293,240,096 divisible by 139? Round up to fourteen tens. The osculator (multiplier) is 14. Start on the right, moving column-wise to the left. 14x6=86. Then adding to the tens column gives 86+9=93. Osculate 93. 3x14+9+0=51. Next column, osculating 51 gives 1x14+5+0=19. Osculating 19 gives 14x9+1+4=131. Then osculate 131, 14x1+13+2=29. Next osculate 29, 14x9+2+3=131. Next osculate 131, 1x14+13+9=36. Osculate 36, 6x14+3+2=89. Osculate 89, 9x14+8+5=139. Yes, the number is divisible by 139. When we end on either zero or the divisor 139 (or a multiple of 139), then, yes, the number is divisible by 139.

5    2   9   3    2   4    0   0   9   6 
139  89  36  131  29  131  19  51  93
YES.

More later. Larry R. Holmgren 03:08, 1 March 2007 (UTC)

Dividing by 13

Article currently says, "A number is divisible by 13 iff the result of adding 4 times the last digit to the original number" ... which doesn't look finished. How should it end?

ZagrebMike 23:15, 22 November 2004 (UTC)

Yes, i agree. Who can explain?

...Nobody know

I know, I think. 4 is the multiplicative inverse of 10 mod 13. So (10a + b) | 13 iff (40a + 4b) | 13 iff (a + 4b) | 13. So for example you can reduce determining if 12056629 divides 13 to determining if 1205662 + 4*9 = 1205698 divides 13. — ciphergoth 17:40, 24 July 2005 (UTC)

However, -9 == 4 mod 13 so the existing rule also works, and multiplying by 9 is probably easier than multiplying by 4 - just calculate 1205662 - 90 + 9. — ciphergoth 17:47, 24 July 2005 (UTC)

A Vedic algorithm says that since a divisor of 13 times 3 gives 39. 39 is in the nines family so we add one and drop the units digit. Hence, the Ekādhika or multiplier is 4.

Example: Is 12,056,629 divisible by 13? Start on the right, move column-wise to the left. Multiply the first digit by 4, add the product to the next digit to the left, repeat the process on this result and add that result to the next digit to the left. If you end with either 13 or zero, then, yes, the number is divisible by 13, otherwise, no, not divisible. This method is called digit-wise osculation. It follows the vedic ideal of one-line notation. The lower line is the whole necessary notation. Multipling the units digit by 4 and adding the tens and the next digit is mental math.

1   2   0   5   6   6   2   9 
13  3   10  22  14  41  38   
YES.

Larry R. Holmgren 03:28, 1 March 2007 (UTC)

Possible error regarding abundant / deficient numbers

In the section "Further notions and facts" is this text:

"If a number equals the sum of its proper divisors, it is said to be a perfect number. Numbers less than the sum of their proper divisors are said to be abundant; while numbers greater than that sum are said to be deficient."

Is that right? I thought it was the other way around. Spadewarrior (talk) 00:39, 6 August 2008 (UTC)

I dont know the definition!!!!! sry!!!!! —Preceding unsigned comment added by 24.93.163.108 (talk) 21:51, 21 January 2009 (UTC)

Grammar

In the article, there are several times when the phrase "we say..." is used. I think it needs to be changed. Anyone disagree? Alan16 ^talk 00:25, 29 March 2009 (UTC)

Two questions re factors

Is a factor the same thing as a proper divisor?

Is n a factor of itself? The first part of the article seems to allow that possibility.

Thanks, CBHA (talk) 07:33, 30 May 2009 (UTC)

A software

I create a software for this. Please, add it in external links —Preceding unsigned comment added by Thekiing (talk • contribs) 16:25, 31 October 2009

Replied on user's talk page. JamesBWatson (talk) 16:54, 31 October 2009 (UTC)

wow

I came here for a quick reference to see if it's the one on the top, or the one on the bottom, and it doesn't tell you straight out at the top of the article. I think maybe at the top of the page there could be an image showing "dividend/divisor". Anyway, so the divisor is the one on the bottom. That's all I was making sure of. --Neptunerover (talk) 08:21, 25 January 2010 (UTC)

Two definitions of gcd

There seem to be two different definitions of gcd. According to the definition at Greatest common divisor, namely "gcd(x,y) = greatest n such that m*n = x for some m and m*n = y for some m", gcd(0,0) is undefined: setting m = 0 the equations hold for all integers n, and since the ordering on the integers has no greatest element, there is no greatest such n.

However, according to the definition in this article (infimum of the lattice of integers ordered by divisibility), gcd(0,0) is defined to be 0, the top element.

So while it is true, as this article says, that the meet is gcd, it is a different gcd than the one defined at Greatest common divisor, namely one extended to be total. Hairy Dude (talk) 21:55, 30 October 2010 (UTC)

Wrong link to Dutch Wikipedia

This is the current link: http://nl.wikipedia.org/wiki/Deelbaar This should be the link: http://nl.wikipedia.org/wiki/Deler I would correct it if I knew how to do that, since I don't I hope that someone else does. The first word "deelbaar" is the translation of divisible, the second word "deler" is the translation of divisor. The Dutch Wikipedia has both entries while the Englisch Wikipedia doesn't have an entry for divisible. 82.170.40.166 (talk) 15:20, 24 April 2011 (UTC)

d(n) is a multiplicative function??

It's written that:
The total number of positive divisors of n is a multiplicative function d(n) (e.g. d(42) = 8 = 2 x 2 x 2 = d(2) x d(3) x d(7))

But I don't think this is always true.

Let's take 12 whose divisors are 1, 2, 3, 4, 6 and 12. So d(12) = d(2 x 6) = 6.
But d(2) = 2 and d(6) = 4, so d(2) x d(6) = 8 !! 81.64.224.95 (talk) 11:24, 2 August 2011 (UTC)

You didn't read the link to find the actual definition of multiplicative function, did you? f(ab)=f(a)f(b) is only required to be true when a and b are coprime. In your example, 2 and 6 are not coprime. —David Eppstein (talk) 15:53, 2 August 2011 (UTC)

One correct way to express the number of divisors of a number n is to express n in terms of its integer factorization. For example:

d(12)=d\left(2^{2}\times 3^{1}\right)=d\left(2^{2}\right)\times d\left(3^{1}\right)=(2+1)\times (1+1)=3\times 2=6

. — Glenn L (talk) 16:48, 2 August 2011 (UTC)

0

From article:

...and every integer is a divisor of 0.

Does 0 divide itself?

no it doesn't, because any division by 0 is undefined Xrchz

No, 0 does divide itself. a|b <==> exists k: ka = b. Obviously for any k, k0 = 0, so the condition is satisfied. — ciphergoth 08:11, 2 August 2005 (UTC)

Yeah, but the problem is the "any k" part. If 0|0, then what does 0 ÷ 0 come out to? You can't make a case for any particular answer, so it's considered "indeterminate." (Division by 0 is indeterminate for 0, and undefined for any other number -- either way, it's not allowed.) Perhaps the definition should properly read: "We say m|n for any integers m and n iff there exists a unique integer k such that n = km." Just to clear it up. --Jay (Histrion) 19:53, 4 August 2005 (UTC)

Sadly, I can't find my copy of "the theory of numbers" at the moment, but I'm pretty sure that my definition of divisibility is the one they give, not yours. Using yours would make proofs much longer, because you'd have to prove that k was unique all the time. And it's sad to have an exception to the rule "every integer divides zero". It's better to allow an anomaly in the relationship between a|b and b &divides; a. — ciphergoth 06:57, 5 August 2005 (UTC)

I see your point. --Jay (Histrion) 18:28, 6 August 2005 (UTC)

I made a survey of mathematics textbooks in my college library. Including Knuth's Concrete Mathematics and Hardy, G. H. and Wright, E. M. An Introduction to the Theory of Numbers. The only one I could find that permitted 0 divides 0 was Ross and Wright's Discrete Mathematics. However, they then added a note in every proof and theorem that m could not be zero. In short the convention is that 0 does not divide 0 and that is what the article should state.

This issue is similar to wheather 1 is prime. In many imprecise definitions of primality it is--and historically it was considered to be--but to define 1 to be prime is not useful and force you to add an exception for it to almost every theorem involving prime numbers.

If no one has further information on this topic I will edit the definition accordingly. --Elfan 02:26, 7 March 2006 (UTC)

I shall have to find my copy of Niven and Zuckerman's "The Theory of Numbers" to contradict you on this. Suffice to say at this stage that the article is now inconsistent with itself, since the definition has not been changed. You'll need to update the definition to reflect this convention... — ciphergoth 16:26, 24 March 2006 (UTC)

Fixed consistency. Do Niven and Zuckerman actually use 0|0 anywhere? --Elfan 18:19, 26 March 2006 (UTC)

It needs to be mentioned that although 0 is a factor of itsef, it is NOT a divisor of itself. —Preceding unsigned comment added by 35.11.50.242 (talk) 10:34, 7 October 2009 (UTC)

It seems strange to me to agree that 0 is no divisor of 0 (as to achieve this 0 must be explicitly excluded in the definition), and in my study of mathematics in Belgium 0 was always a divisor of 0. But of course other agreements can be made, however there are some problems here. Namely the set $N$ of non-negative integers with the relation of divisibility, mentioned in this article, is now no partially ordered set (and no Lattice (order). Also as already mentioned in the paragraph 'Two definitions of gcd', the definition of greatest common divisor doesn't fit with this 'other' concept of gcd. In my country we also use 'a greatest common divisor' in a more general way where this is non-unique (so 'a' greatest common divisor in stead of 'the' greatest common divisor). For integers, -gcd(a,b) is then also 'a greatest common divisor' of a and b. When talking about 'the greatest common divisor' of course the unique positive greatest common divisor is meant then. Samaritaan (talk) 22:14, 21 October 2011 (UTC)

Straight-line "divides" symbol & absolute value symbol, aliquot, plus more on "wow" or why does my head hurt...worse?

Granted, my pre-frontal processing is partially impaired because of my sleepiness, but the language (not the terminology, but the explanations thereof) used in Wikipedia math articles hurts my head, even in this article. Like the writer of the earlier "wow" section, I find it hard to find and process information in this math article and others. I don't expect to understand even half the math in Wikipedia, but even the first half of this article made my brain backtrack.

I know this is not the Simple English Wikipedia, but I assume we would like to create articles or sections that open with information aimed at the novice but that grows in complexity until an expert would be satisfied. I don't know if I can help with this due to chronic illness, but I wanted to point it out.

I came to this article from the Aliquot disambiguation page. Why is aliquot used here? Its etymology or definition might help. As the text is, my mind searches for meaning and only finds, "Alice in Wonderland's kumquat?"

Also, my mind hit speed-bumps at the straight-line division symbol. That space in my head was already taken up by the absolute value symbol. A brief comparison might help. Oh, and there's two absolute value articles that have to do with math. I don't know if they are both needed. But one of them used a smaller in-line math font that required less of a transition in my mind for single-line equations.

For a lot of us, unfortunately, math eats up a lot of energy to process. Smoother transitions between levels of complexity, conceptual connections, and even fonts can encourage us to process a little further. It's tough going when you're already at a deficit in energy when you haven't even hit the core of the topic.

--Geekdiva (talk) 12:07, 17 March 2012 (UTC)