Wikipedia:Reference desk/Archives/Mathematics/2024 May 15

= May 15 =

How many such binary operations exist in a set with n elements?
There are four possible properties for a binary operation:


 * 1) Idempotence
 * 2) Commutative property
 * 3) Associative property
 * 4) Cancellation property

So, in a set with n elements, how many such binary operations (which are closed) exist?

2402:7500:92D:FD81:F115:AC09:9228:B1A8 (talk) 10:58, 15 May 2024 (UTC)
 * 1) Satisfy property 1
 * 2) Satisfy property 2
 * 3) Satisfy property 3
 * 4) Satisfy property 4
 * 5) Satisfy properties 1 and 2 simultaneously
 * 6) Satisfy properties 1 and 3 simultaneously
 * 7) Satisfy properties 1 and 4 simultaneously
 * 8) Satisfy properties 2 and 3 simultaneously
 * 9) Satisfy properties 2 and 4 simultaneously
 * 10) Satisfy properties 3 and 4 simultaneously
 * 11) Satisfy properties 1, 2, and 3 simultaneously
 * 12) Satisfy properties 1, 2, and 4 simultaneously
 * 13) Satisfy properties 1, 3, and 4 simultaneously
 * 14) Satisfy properties 2, 3, and 4 simultaneously
 * 15) Satisfy all four properties simultaneously


 * Case 1, just idempotence, is easy. A binary operation on a set of $$n$$ elements (a finite magma) can be completely described by the $$n^2$$ entries of the $$n\times n$$ operation table. Idempotence fixes the $$n$$ entries on the diagonal. For each of the remaining $$n(n-1)$$ entries there are $$n$$ choices, so there are $$n^{n(n-1)}$$ distinct tables.
 * Case 2, just commutativity, is also easy. The $$n$$ entries on the diagonal can be chosen freely, as can the $$\tfrac 12n(n-1)$$ entries of the triangle below the diagonal; the upper triangle is thereby fixed. So the number equals $$n^{\frac 12n(n+1)}.$$
 * For case 4, just cancellation, there is a one-to-one correspondence with the Latin squares of order $$n$$. See the section Number of $n\times n$ Latin squares.
 * There is no simple formula for this case, and I suppose also not for most, if not all, other cases. Some have been tabulated; for case 3, the number of finite semigroups, see . --Lambiam 13:51, 15 May 2024 (UTC)
 * How about the cases 5 to 15? I found the OEIS sequences:
 * All binary operations (without any condition): (labeled),  (isomorphism classes)
 * Case 1: (labeled),  (isomorphism classes)
 * Case 2: (labeled),  (isomorphism classes)
 * Case 3: (labeled),  (isomorphism classes)
 * Case 4: (labeled),  (isomorphism classes)
 * Case 8: (labeled),  (isomorphism classes)
 * How about other cases? 49.217.196.102 (talk) 10:12, 18 May 2024 (UTC)
 * Case 5 is . Using the analysis method given above for cases 1 and 2, you should have been able to derive the formula yourself. --Lambiam 10:48, 18 May 2024 (UTC)
 * 220.132.230.56 (talk) 17:20, 18 May 2024 (UTC)
 * Case 10 seems to be exactly the groups, i.e. a binary operation is a group if and only if it satisfies condition 10 (i.e. both associative property and cancellation property), the only exception is the empty set, which satisfies all these four properties but is not group since it does not have the identity element, and I found these OEIS sequences: (note: a(0) = 1 in every case, although not all these sequences have a(0) = 1)


 * But I cannot find OEIS sequences for other cases, for case 7 (isomorphism classes), I found a table, the last element in each row is exactly the answer, but when I search this sequence "1, 1, 0, 1, 1, 4, 18", I found no sequence in OEIS, and for case 12 (isomorphism classes), I found a table , the last element in each row is exactly the answer, but the last row of this sequence is currently "1, 1, 0, 1, 0, 1, 0", which may have too many sequences in OEIS, the answer for case 12 seems to be with offset 0, i.e. the answer is 0 for all even n >= 2 and 1 for n = 0 and all odd n >= 1, also for cases 13 and 15, the answer is 0 for all n >= 2 and 1 for n = 0 and n = 1, i.e.  with offset 0, since there is no idempotent group with >= 2 elements (since all groups have an identity element and have the cancellation property). 61.224.168.169 (talk) 12:10, 24 May 2024 (UTC)

Measuring Coefficient of Variation
I have a group of 39 subjects who evaluated 20 different answers to questions on a scale of 1 to 5. The mean is 2.13 and the Standard Deviation is 1.077. I want to say that there is a lot of variation in the answers. I asked ChatGPT and it said to compute the Coefficient of Variation which is: CV= 2.131.077 ×100% ≈ 50.47%  It said that: "The interpretation of the coefficient of variation (CV) can vary depending on the context and the field of study. However, as a general guideline: Low variability: CV less than 15% Moderate variability: CV between 15% and 30% High variability: CV greater than 30%" Which seems to support my hypothesis which is that there is significant variation in the answers. That's also what a quick look at the data indicates. Haven't done statistics in decades so wanted to check with a human as well. Does this all sound reasonable? I want to say in an academic paper that there was considerable variation in our subjects as indicated by the CV being greater than 50% does that seem reasonable? MadScientistX11 (talk) 22:49, 15 May 2024 (UTC)
 * Computing 1.077/2.13 yields 0.5056, not 0.5047 – but a precision of four digits, also for the SD, is excessive. If only one of the subjects, vacillating between 2 and 3, had picked the other choice, the SD would almost certainly have diverged from 1.077 already in the second digit after the decimal point.
 * Reporting the CV may be common practice, but is too often meaningless. Your subjects scored on a scale of 1 to 5, which is an arbitrary convention for Likert scales (see ). If the scale had been labeled 0 to 4, the SD would remain the same, but the mean would have been 1 less, only 1.13. So computing the value of the CV would in that case have resulted in 1.077/1.13 = 0.9531.
 * Another commonly used measure is the index of dispersion, which is even more problematic for the distribution of scores using an arbitrary scale.
 * I'm not a social scientist, but, assuming that each of the five possible responses is a reasonable one, the dispersion does not appear that considerable to me. It is definitely less than the expected 1.41 if respondents had given uniformly random answers. If you show a histogram, readers can form their own assessments about how considerable the dispersion is.
 * A final word of advice. In reporting the statistics of research findings, avoid the terms "significance" and "significant" unless you definitely mean to refer to the technical notion of statistical significance. --Lambiam 05:31, 16 May 2024 (UTC)
 * Excellent. I've come to realize that these LLMs often sound very convincing but when you look into the details they often are incorrect. FYI: the survey is really a fairly minor part of the work we're doing so we didn't spend as much time as (with hindsight) we should have to set it up appropriately and think about the statistical analysis before hand. Based on what you said, I don't think it makes sense for us to talk about any statistics because the sample size was small and in this phase of the work it was just a trivial part. Thanks again. MadScientistX11 (talk) 17:05, 16 May 2024 (UTC)