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Binary Operations : Equivalence Classes

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Note. I don't how to make a letter with a line overtop of it so the equivalent notation here is *. ex) a* = a bar (a with a line overtop of it)

Let M be a commutative monoid. Define a relation ~ on M by a ~ b if a = bu for some unit u.
(a) Show that ~ is an equivalence on M and if a* deontes the equivalence class of a, let M* = {a*| a belongs to M} denote the set of all equivalence classes. Show that a*b* = (ab)* is a well-defined operation on M* deontes.
(b) If M* is as in (a), show that M* is a commutative monoid in which the identity 1* is the only unit.

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Solution Summary

Equivalence of a monoid is investigated. The commutative monoid identified is provided.

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(a) We want to show that a*b*=(ab)*. By the definition, a* is the equivalence class of a, and b* is the equivalence class of b. So,
for any element a' in a*, it can be written as a'=au for some unit u. Similarly, for any element b' in b*, ...

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  • BSc , Wuhan Univ. China
  • MA, Shandong Univ.
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  • "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
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