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Title: Dozenal benefits - 2s and 3s, continued


icarus - April 19, 2006 12:00 PM (GMT)
oops,

CODE

threes
1      3        0.4
2      9        0.14
3     23        0.054
4     69        0.00194
5    183        0.000714


That's a lot better than decimal, and even octal or hex. Because it handles both three and two, dozenal manages sixes. Dozenal handles fours also, and doesn't do badly handling eights and nines, because these are factors of the square of 12.

JDozen - August 17, 2007 12:38 PM (GMT)
The most obvious reason that three is better than five, is that THERE ARE MORE NUMBERS THAT DIVIDE BY THREE.

I haven't noticed anyone even mentioning that, please correct me if I'm wrong...

Ruthe - August 17, 2007 09:08 PM (GMT)
QUOTE (JDozen @ Aug 17 2007, 12:38 PM)
The most obvious reason that three is better than five, is that THERE ARE MORE NUMBERS THAT DIVIDE BY THREE.

I haven't noticed anyone even mentioning that, please correct me if I'm wrong...

Well, yes and no.

I seem to recall that there are as many numbers divisible by 3 as by 5, and the same is true for any number. We get into the realm of infinities, of which there are several.

But for any given finite upper limit, your statement is correct. So for general day to day usage, 3 is better than 5.

Dan - August 18, 2007 05:37 AM (GMT)
QUOTE (Ruthe @ Aug 17 2007, 03:08 PM)
I seem to recall that there are as many numbers divisible by 3 as by 5

Right. In general, any infinite set of the form {f(n): n in Z} has a cardinality of aleph-null. Therefore, | {3n: n in Z} | = | {5n: n in Z} |.

uaxuctum - August 19, 2007 02:25 AM (GMT)
Yes, but, as you say, that's just abstract set theory which assumes infinity as something real or even possible. But keeping ourselves within the realm of reality, i.e. the realm of finite sets, it's mathematically and intuitively clear that those divisible by 3 constitute a larger subset than those divisible by 5. The theory of infinite sets may be good as a hypothetical construct to entertain mathematicians, but when it produces results such as that the set of integers and its subset of even numbers turn out to have the same cardinality, which contradicts all common sense (supposedly, if you have a infinite set, and then another set which you obtain from the previous one by discarding every other of its elements, both sets end up with the same number of elements and none is any larger than the other because they are both infinite and of the same infinite cardinality), then the concept of infinite sets and transfinite numbers doesn't seem very helpful to talk about reality, and in fact Cantor's postulates about infinite sets have been contested by some mathematicians (called "finitists"). It's like that paradox about movement attributed to Zeno, who argued that since to reach from one point to another you first have to reach their middle point, and since mathematically it is assumed that there are infinite points between any two, then by applying the middle-point argument recursively ad infinitum one would never be able to move an inch, to which Diogenes replied that movement is proven by walking.

uaxuctum - August 19, 2007 09:26 PM (GMT)
Rephrasing it in mathematical terms of finite sets:

[DOHTML]<s>V</s> S[/DOHTML] = {n | n [DOHTML]<s>C</s>[/DOHTML] N, n[DOHTML]<sub>m</sub>[/DOHTML]= n[DOHTML]<sub>m−1</sub> + 1[/DOHTML] = n[DOHTML]<sub>m+1</sub> − 1[/DOHTML]} | |S| [DOHTML]<s>C</s>[/DOHTML] N, |S| > 5, S[DOHTML]<sub>3</sub>[/DOHTML]= {n [DOHTML]<s>C</s>[/DOHTML] S | n mod 3 = 0}, S[DOHTML]<sub>5</sub>[/DOHTML]= {n [DOHTML]<s>C</s>[/DOHTML] S | n mod 5 = 0} : |S[DOHTML]<sub>3</sub>[/DOHTML]| > |S[DOHTML]<sub>5</sub>[/DOHTML]|

For every finite set of contiguous natural numbers, of cardinality strictly greater than 5, the cardinality of its subset of multiples of 3 is strictly greater than the cardinality of its subset of multiples of 5. This is proven easily: there are two multiples of 3 and one multiple of 5 in every six contiguous naturals; to get the second multiple of 5, the set has to be increased to at least ten contiguous naturals, but then a third multiple of 3 is also added; and so on, so the cardinality of the multiples of 3 remains strictly larger than the cardinality of the multiples of 5. Except when the cardinality of the set "reaches" infinity and crazy things start to happen, that is. :rolleyes:

JDozen - August 23, 2007 12:18 PM (GMT)
QUOTE
For every finite set of contiguous natural numbers, of cardinality strictly greater than 5, the cardinality of its subset of multiples of 3 is strictly greater than the cardinality of its subset of multiples of 5. This is proven easily: there are two multiples of 3 and one multiple of 5 in every six contiguous naturals; to get the second multiple of 5, the set has to be increased to at least ten contiguous naturals, but then a third multiple of 3 is also added; and so on, so the cardinality of the multiples of 3 remains strictly larger than the cardinality of the multiples of 5. Except when the cardinality of the set "reaches" infinity and crazy things start to happen, that is.


That is exactly what I thought. There isn't even one set where there are more multiples of five than multiples of three. Hurrah for three! (just felt like saying that, donno why)

icarus - July 2, 2008 12:43 PM (GMT)
Amusing:

The US Government's Census Bureau requires my business to fill out a form. One of the passages on this form states:

"D. For the person(s) owning the largest percentage(s) in this business in 2007, please list the percentage owned by each person and his or her position title.

"Do not report percentages owned by parent companies, estates, trusts, etc.

"If more than 4 persons owned this business equally, select any 4.

"Round percentages to whole numbers. for example, report 1/3 ownership (33.3%) as: 33.0%"

So they're asking us to report up to four owners, but if there are three, to call it 33.0%. This, by the rules of accounting, has to total 99.0% technically. Very interesting accounting...hmmm.

Dan - August 24, 2008 04:25 AM (GMT)
You should have some kind of contest to determine which person gets to claim 34% ownership.




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