December 21 Information

Why is there no Roman numeral to represent zero? 2602:252:D13:6D70:6533:6D2D:ACBE:8031 ( talk) 05:22, 21 December 2015 (UTC) reply

I don't know why, but see Roman_numerals#Zero. They had the concept of zero, but zero really aids in decimal calculations, so it isn't really applicable to Roman Numerals. Bubba73 ^{You talkin' to me?} 05:33, 21 December 2015 (UTC) reply

Thanks. Yes, I had read that (which is why a placed the Wikipedia link -- Roman numeral -- in my original question). But what do you mean it is not "applicable" to Roman numerals? How -- for example -- is the number "7" (or VII) applicable? And how is zero not applicable? I don't follow what you mean. Thanks. 2602:252:D13:6D70:F5CC:81BD:3808:6FD5 ( talk) 05:42, 21 December 2015 (UTC) reply

There is the number zero and the numeral for zero. The numeral zero is essential in a Positional number system but Roman Numberals aren't a positional number system. I don't think having a Roman numeral for zero would really help. Bubba73 ^{You talkin' to me?} 06:45, 21 December 2015 (UTC) reply

Help for what? The OP didn't mention any application of zero; he/she just asked how you represent zero (or more precisely, why you can't). The answer seems to be, as 76.69.45.64 explains below, because the Romans didn't know about zero and therefore didn't feel the need to represent it. The same reason, in other words, that you can't represent −1 or i or the smallest infinite ordinal. -- Trovatore ( talk) 07:07, 21 December 2015 (UTC) reply

Help with arithmetic. The Romans has no numeral for zero, but they knew about the number zero and used "nulla" for it, accouring to the article. Bubba73 ^{You talkin' to me?} 07:13, 21 December 2015 (UTC) reply

I think you read that part too fast. It doesn't say that the Romans at the time of the Roman Empire knew about the number zero. It says that "medieval computists" did, several centuries later. -- Trovatore ( talk) 07:16, 21 December 2015 (UTC) reply

Bubba is saying, I think, that in our system of numerals you might have need a 0 in any position so you can tell the number 23 from 203 or 2300. In Roman numerals they did not have the concept of positional notation where the same symbol might mean 2 or 20 or 200 or 2000 depending on where it occurs. They wrote 2 as II, 20 as XX, 200 as CC, and so on. So their system did not need zero within a numeral the way ours does.

This is a separate issue from having the concept that zero is a number. In our time someone writing up a report on a train accident might say there were "0 killed, 2 seriously injured, 4 slightly injured, 214 uninjured". The 0 here is just as much a number as the 2, the 4, or the 214. The ancient Romans did not have this concept; it developed later. They would do the equivalent of writing "none were killed" in words. Note that the linked article says what word was used in medieval times, not in ancient times when Roman numerals were invented. Medieval writers might have invented a symbol for zero as a number to be used with Roman numerals, but they didn't, or if some of them did, it didn't catch on. -- 76.69.45.64 ( talk) 06:08, 21 December 2015 (UTC) reply

OK. That makes some sense. But, I think of two examples that might need further clarification. One: What would they do in a case such as this? In our way, we would write 7 - 7 = 0. They would write VII - VII = _____ (what)? And Two: What would they do for a decimal number that requires a "zero" position? For example, 5.907 (where the zero is positioned between the 9 and 7). Thanks. 2602:252:D13:6D70:B04B:668:A3D0:7138 ( talk) 07:10, 21 December 2015 (UTC) reply

One: Take seven away from seven, and you have nothing left. According to Plus and minus signs, the - sign wasn't used until much later.

Two: 5907 = MMMMMCMVII

Rojomoke ( talk) 16:45, 21 December 2015 (UTC) reply

• Actually they didn't need five of one symbol; there was an alternate system used for large numbers in ancient Rome where 1,000 looked like a large O with a small I in the center, and then marks like () were added around the central bar to multiply by 10, and one half of this symbol represented half the value. This is where D for 500 comes from, then you had ↀ = M = 1,000; ↁ = 5,000; ↂ = 10,000; ↇ = 50,000; and ↈ = 100,000. They didn't continue beyond that; the largest number represented in an ancient inscription is 2,300,000 (this information from memory, sorry) and it consists of 23 ↈ signs in a row. The Unicode font on my screen shows the ↀ with the bar joined to the outer loop at top and bottom, but the way I've seen it in books, it was not joined. -- 76.69.45.64 ( talk) 01:43, 22 December 2015 (UTC) reply

Thanks. But I don't understand your responses. One: Whether or not they used the "minus" sign, they still must have had the concept of subtraction. So, the question is: VII subtracting out VII equals ______ (what)? Two: My question was about the decimal number 5.907 and not the number 5,907 (in other words, there is a decimal point -- not a comma -- between the 5 and the 9). Thanks. 2602:252:D13:6D70:9D55:2706:817F:DC85 ( talk) 18:10, 21 December 2015 (UTC) reply

Sorry, I completely missed the decimal point. I don't know how they represented fractions. Maybe some variation of our numerator/divisor format? I doubt they ever needed to represent more than a couple of decimal places. And as was said above, they didn't comprehend zero as a number. So they simply said "there's nothing left". Rojomoke ( talk) 00:21, 22 December 2015 (UTC) reply

Yes, they would have said that asking what is 7 minus 7 made no more sense than asking what is 7 minus 8. You can't do that and get a number, that's all. -- 76.69.45.64 ( talk) 01:43, 22 December 2015 (UTC) reply

You state that, in their view, talking about 7 minus 7 makes no sense. I can't imagine that to be true. Perhaps they would not say "well, 7 minus 7 yields zero". And, as suggested above, they might conclude "well, there is nothing left over". But I doubt that they would assert that the question itself makes no sense and cannot be answered. Certainly they could see that if they had 7 apples and ate 7 apples, there are no more apples remaining to eat. The concept of 7 minus 7 would not "stump them". 2602:252:D13:6D70:9D55:2706:817F:DC85 ( talk) 02:57, 22 December 2015 (UTC) reply

Exactly. How can you be asking how many apples there are, when there aren't any? It doesn't make sense. -- 76.69.45.64 ( talk) 22:55, 22 December 2015 (UTC) reply

Well, it's all semantics, I guess. There aren't any apples. Or there is a quantity, and that quantity of apples is zero/none. Thanks. 2602:252:D13:6D70:DD24:7700:74C1:76C6 ( talk) 15:33, 23 December 2015 (UTC) reply

Here's a page of tables from al-Khwarizmi from the 12th century, primarily using Roman numerals and using the Hellenistic zero for zero. This symbol goes back at least to Ptolemy and looked like a circle with a bar over it. -- Amble ( talk) 07:48, 21 December 2015 (UTC) reply

I think the problem here is that the Romans only used Roman numerals for saying how many of something there was. Their written work doesn't really show what they were doing in everyday life. They used an abacus for any actual computation, these were quite sophisticated device with fractions of 1/12 and 1/144. Maths or astronomy used Greek letters. Long before then the Egyptians had a symbol for zero in accounting and for the origin in height or distance - and could represent negative numbers by writing the numbers backwards. Dmcq ( talk) 22:02, 23 December 2015 (UTC) reply

Thanks, all. 2602:252:D13:6D70:5509:A93B:8297:3B2F ( talk) 19:40, 26 December 2015 (UTC) reply

A zero in the modern sense is part of a place-value system of numbers. To those who have not used such a system since childhood, a zero is actually a counter-intuitive concept. It is a symbol that doesn't represent anything, because it represents nothing. Place-value systems have been invented three times independently in history, by the Babylonians, by the native Central American Mayans, and in India. (Therefore to refer to Indian math is ambiguous, at least if we call native Americans what Columbus incorrectly called them.) Interestingly, these three inventions used different bases. Babylonian mathematics used base sixty. We still have a few vestiges of it in geography and timekeeping. The Mayans used base twenty. What was invented in India was then adopted by the Arabs in the Middle Ages, and we know them as Arabic numerals, using base ten. Base sixty Babylonian mathematics was known in Greco-Roman times to a priestly learned class, the magi, and was used by Greek astronomers, but was difficult to learn because of the need to master a massive multiplication table. Roman numerals don't have a zero, as mentioned, because they are not positional or place-value. A zero is a counter-intuitive concept except to those who have learned it in childhood, because it is a symbol that doesn't represent anything, because it represents nothing. Robert McClenon ( talk) 19:49, 26 December 2015 (UTC) reply

It's not true that it doesn't represent anything, and "represents nothing" is just a pun. The numeral zero is part of a place-value system, but that has little to do with the number zero.

The numeral 0 represents a certain abstract object; namely, the number zero. The numeral 2 also represents an abstract object, the number two. The number two is just as abstract as the number zero. Once you have fully accepted the number two as an abstract object and not as either a mark on a piece of paper or any specific two physical objects, I don't think accepting zero is a big leap. -- Trovatore ( talk) 20:03, 26 December 2015 (UTC) reply

Either zero or place-value evidently was a big leap because for most of intellectual history most mathematical systems did not have place value or zero. Anyway, in referring to abstract objects, and in calling 0 and 2 abstract objects, you are using modern terminology. Robert McClenon ( talk) 20:22, 26 December 2015 (UTC) reply

Using modern terminology? Plato would have understood. -- Trovatore ( talk) 20:30, 26 December 2015 (UTC) reply

Plato would have understood the concept, but the terminology used to explain the concept is late-nineteenth-century. Also, to say that the underlying mathematics is the same is to use modern terminology as to what mathematics is. In the past, the art of calculation was considered to be mathemtical (as it still is), and the art of calculation is base-dependent. The underlying foundational mathematics is the same, but the foundations were put in under existing mathematics in the early twentieth century. Sometimes math is like that, in that the underlying logic is put in place after the art is already developed, just as the Weierstrass foundations of calculus were used to give logical rigor to the techniques that Newton and Leibniz had used successfully. Newton didn't need Weierstrass's concepts to determine that an inverse-square law would explain Kepler ellipses (and you can still jettison Weierstrass's approach by using Robinson's approach). Robert McClenon ( talk) 22:22, 26 December 2015 (UTC) reply

It's irrelevant whether I'm using modern terminology. Much the same way that the representation of numbers is irrelevant to the numbers themselves, the terminology used is irrelevant to the concepts. -- Trovatore ( talk) 22:36, 26 December 2015 (UTC) reply

By the way, in your 19:49 contribution, you're conflating numbers with numerals. Numbers are the underlying thing; numerals are the symbols we use to represent them. Numeral representation is accidental rather than essential. In particular there is no such thing as, say, base-sixty or base-ten mathematics. The underlying mathematics is the same. Only the representation differs. -- Trovatore ( talk) 20:45, 26 December 2015 (UTC) reply

Something I've been puzzling over since the recent Rugby Union world cup:

Points are scored in Rugby in the following ways:

• 3 points for a drop goal or score from a penalty,
• 5 points for a try,
• An additional 2 points for a successful conversion (which is only taken after a try).

For simplicity, I'll define "unconverted try" as a try without conversion (5 points) and "converted try" (7 points) as a try plus conversion. Therefore a team's score at the end of the match is 3p+5u+7c, where p=number of penalties/drop goals scored, t=number of unconverted tries scored and c=number of converted tries. Obviously c, p and t are non-negative integers.

My question is this - if a team ends with a given final score, how many combinations of penalties/drop goals, tries and converted tries are there that could lead to that result? For example, a score of 8 must have been an unconverted try plus a penalty/drop goal. A score of 10 could have been two unconverted tries,or a converted try plus a penalty/drop goal.

To put it mathematically then, for a given value of f, how many combinations of p, u and c are there so that f=3p+5u+7c, so that f, p, u and c are non-negative integers? Optimist on the run ( talk) 10:57, 21 December 2015 (UTC) reply

The equation 3x+5y+7z=a where the unknowns are non-negative integers is a linear diophantine equation. Geometrically the solutions (x,y,z) are points lying in the triangle having corners at (a/3,0,0) and (0,a/5,0) and (0,0,a/7). Computing the area of that triangle may be a step towards answering your question. Bo Jacoby ( talk) 12:18, 21 December 2015 (UTC). reply

(ec) Not exactly an answer, but A261155 from OEIS gives "The highest rugby (union) score that can be made in n or fewer ways." The question is related to the Frobenius coin problem (in fact I see that the article discusses rugby scores). See also Change-making problem. AndrewWTaylor ( talk) 12:22, 21 December 2015 (UTC) reply

For a from 1 to 100 the number of solutions are

 0  0  1  0  1  1  1  1  1  2
 1  2  2  2  3  2  3  3  3  4
 4  4  4  5  5  5  6  6  6  7
 7  7  8  8  9  9  9 10 10 11
11 12 12 12 14 13 14 15 15 16
16 17 17 18 19 19 20 20 21 22
22 23 24 24 25 26 26 27 28 29
29 30 31 31 33 33 34 35 35 37
37 38 39 40 41 41 43 43 44 46
46 47 48 49 50 51 52 53 54 55

This is counted by this J program

  10 10$+/(,(3*i.>.>:3%~a)+/(5*i.>.>:5%~a)+/(7*i.>.>:7%~a))=/1+i.a=.100

For bigger values of a the number of solutions is approximately proportionate to a², say 55⋅(a/100)². Bo Jacoby ( talk) 14:02, 21 December 2015 (UTC). reply

This is OEIS A008677. You will find more information and references there. Note particularly how this type of problems can be expresses as a series for a rational function.

Note that this is a type of partition problem, with the restriction that there are 3 summands which are multiples of 3, 5 and 7 respectively. Perhaps there is something in that article that will be useful.

Of course, brute forcing is also possible. Here is a program like Bo's, but in Mathematica:

maxV = 100; Map[Counts[Sort[Flatten[Table[3 p + 5 u + 7 c, {p, 0, maxV/3}, {u, 0, maxV/5}, {c, 0, maxV/7}]]]], Range[maxV]] /. {Missing[__] -> 0}

Fitting a quadratic to the values up to 1000, we get the following approximation:${\displaystyle (0.2314+0.07144a+0.0047618972a^{2})}$. The difference between the real and approximate value seems to have a roughly repeating pattern. -- Meni Rosenfeld ( talk) 15:54, 21 December 2015 (UTC) reply

Meni's link says that the generating function f(x) for the number of combinations A_N of penalties/drop goals (3 points) tries (5 points) and converted tries (7 points), for a given value of the total number of points N, is

${\displaystyle f(x)=\sum _{N=0}^{\infty }A_{N}x^{N}=((1-x^{3})(1-x^{5})(1-x^{7}))^{-1}}$

Tell me why! Bo Jacoby ( talk) 20:25, 21 December 2015 (UTC). reply

One of the more beautiful pieces of Generating functions. 1/(1-x^3) = 1+x^3+x^6+x^9+x^12..., and similarly for 1/(1-x^5) and 1/(1-x^7). When all of these are expanded the factor of x^10 is equal to 2, one from the x^10 factor in the 1/(1-x^5) and the other from the x^3 factor from 1-x^3 times the x^7 factor from 1-x^7. Naraht ( talk) 20:54, 21 December 2015 (UTC) reply

Thanks! Bo Jacoby ( talk) 21:57, 21 December 2015 (UTC). reply

Of course, this generalizes to other situations. For example, if I wanted to know the number of ways to represent a number as a sum of five multiples of 6, I would look at the coefficients of the series for ${\displaystyle \left({\frac {1}{1-x^{6}}}\right)^{5}}$.

This also suggest an efficient way to calculate these values. The naive methods took cubic time (since they iterate over each one of the variables), but with fast polynomial multiplication algorithms, it can be only slightly more than linear time. Here is an implementation in Mathematica:

CoefficientList[Series[1/((1 - x^3)(1 - x^5)(1 - x^7)), {x, 0, 10000}], x]

-- Meni Rosenfeld ( talk) 23:27, 21 December 2015 (UTC) reply

That is amazing! Bo Jacoby ( talk) 07:05, 22 December 2015 (UTC). reply