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Let $x$ and $y$ denote variables over the set $\{a, b, c\}=S$, and $t$ some constant of $S$. Let "X" denote an unknown binary operation on $S$. Suppose we have the following set of properties:

  1. For all x, xxX=t
  2. There exists a y, such that for all x, xyX=y=t (the same constant as in 1.)
  3. There exists an x, such that for all y, xyX=t
  4. There exists an x, such that for all y, xyX=y
  5. There exists an x, such that there exists a y not equal to x, and xyX=x.

There exist at least 6 structures (all automorphic to each other, which indicates how I obtained them in the first place, THEN I wrote the properties) which satisfy this set of properties as follows:

A  a  b  c
a  c  c  c
b  b  c  c
c  a  b  c

B  a  b  c
a  b  b  b
b  a  b  c
c  c  b  b

C  a  b  c
a  c  a  c
b  c  c  c
c  a  b  c

D  a  b  c
a  a  b  c
b  a  a  a
c  a  c  a

E  a  b  c
a  b  b  a
b  a  b  c
c  b  b  b

F  a  b  c
a  a  b  c
b  a  a  b
c  a  a  a

Do any other binary operations on {a, b, c} satisfy properties 1.-5. above? I would think "yes", since, if I've gotten things right, property 1. specifies something about the diagonal, property 2. specifies a particular column, for which all of its values equal that of the diagonal, 3. specifies a row that always equals the valued of the diagonal, 4. specifies a row which always equals the values of the second coordinate, and I think 5. specifies the only value left undetermined by 1.-4. Do any other structures exist here?

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    +1 for insistence on polish postfix notation and adherence to denote operations by integers, -1 for not getting Mariano's and others' points. -1 for the oxymoronic "automorphic to each other". Total: -1.2011-06-21
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    @Theo I'm not sure Mariano had any point. All I saw indicated a complaint about my use of notation. It's not standard. So what? Denoting operations by integers immediately starts a quantification and doesn't require a map from say (1, 2, 3, 4) to the operations (A, B, C, D). Substitute a*b for x, and c for y, in x*y. You obtain a*b*c an ambiguous expression in infix notation (none of these operations associate). Now substitute ab* for x, and c for y, in xy*. You obtain ab*c* an *unambiguous* expression. (x*y) needs more symbols than xy*. Now, do you see why infix notation has a problem?2011-06-21
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    @Doug: Your complaint on "needs more symbols" simply means that you value *your* time (the insignificant amount that it would take to type extra prentheses), while at the same time finding other people's time not worth your concern (the time it takes them to figure out what you mean); in short, contempt for others while you ask for their help. As they say in Mexico, "limosnero con garrote." Given that my time is so worthless to you, this was my last effort on your behalf.2011-06-21
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    @Arturo The point about symbols doesn't just apply to me, but anyone typing symbols. So, it requires less time for them to type such expressions also. And it requires fewer symbols to get seen and processed by any brain. So, once the new patterns get learned, it requires less time for anyone.2011-06-21
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    @Doug: In short, if people want to talk to you, they need to do it in whatever way *you* prefer. Like I said, limosnero con garrote. Enjoy your monologues, Doug; please don't address any more comments to me.2011-06-21
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    @Doug: There's a reason why we name lists of variables things such as $x_0,x_1,\ldots$, and not $0,1,\ldots$. It's because $0,1,\ldots$ **already have a meaning**. It's absolutely true, we **could** decide to swap the numerals $0,1,\ldots,9$ with the first 10 letters $a,b,\ldots,j$, replace $+$ with $\&$, and write $a$ to the $b$th power as $b^a$ instead of $a^b$, and it would be logically equivalent. However, human beings will (rightly) hate it, and unless you plan on talking about math only to computers, you should use the traditional notations and names for things.2011-06-21
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    -1 for purposefully bad notation. How could you expect anyone to be able to read this question?2011-06-21
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    @Zev I don't see how 0, 1, ... already have a meaning, since, for example, I can't tell if one means them as finite ordinals or cardinals without a context. I can see them as suggesting meaning, but the reader ultimately determines if any symbol has any meaning whatsoever. This human being wrote those symbols.2011-06-21
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    @Jim Whatever I may have expected it happened. Second, I find such notation much better than writing x*y=t, because the rule of substitution can't legitimately get used in that form, since a*b*c is ambiguous. Not writing operations also leaves open the possibility of ambiguity, though it doesn't work out as bad in this respect since it doesn't imply ambiguity necessarily. (x*y)=t takes more symbols than xy*=t or *xy=t.2011-06-21
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    An old saw about honey and vinegar comes to mind...2011-06-21
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    @Doug, the purpose of notation is not to minimize the number of symbols used. The purpose is to *communicate* with other mathematicians, using whatever notation would be the most clear to others.2011-06-21
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    @Jim It's not an either-or, but not both. Notation preferably will use relatively few symbols AND make things very clear. I see no reason to optimize one and totally ignore the other.2011-06-21
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    @Doug, minimizing symbols isn't a goal at all. For example, general linear groups are typically denoted something like $\text{GL}(3,\mathbb{C})$. There is no reason why this notation should be replaced by $\text{G}3\mathbb{C}$ or any other shorter form, and indeed the latter notation would be worse. The only reason to make notation shorter would be to make it easier for a human being to read it, and $\text{G}3\mathbb{C}$ is considerably less readable than $\text{GL}(3,\mathbb{C})$.2011-06-21
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    @Jim Look at Arturo's comment. It is NOT clear that any operation exists for what a string like "xy=t" means from just the symbols given. So, it's not easier for a human being to read what he wrote as if there existed an operation there implicitly in comparison to reading his writing as if he made several mistakes by using several meaningless strings such as "xy=t". So, why did he omit the operations, since it made things harder to read? Because it saved space in reading and time in writing.2011-06-21
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    @Jim Additionally, the real question isn't whether you or I find prefix, infix, or suffix notation easier to read. The real question comes as "Which notation would a person with exactly equal experience reading all three notations find easiest? Hardest?"2011-06-21
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    @Doug: as a passerby to this conversation, let me say that your most recent comment strikes me as simply obnoxious. You have decided that you wish to communicate with a hypothetical audience rather than the audience that you actually have (most of the world -- or more! -- learns mathematics using "infix notation"). Arturo, Jim and Mariano are all experienced mathematicians and talented expositors and communicators of mathematics: they know what they are talking about. Let me also say that I was one of the ones who stopped thinking about your question because of the unfriendly notation.2011-06-21
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    @Doug: Do not presume to tell others why I did or did not do anything; you have little or no clue about my reasons, and your conclusions are unwarranted (and as it happens, incorrect; I did not do it "because it saved space in reading and time in writing"). Yet again, you take other people's actions and interpret them to suit your prejudices, bringing them forth as unwilling witnesses on your behalf. It's at best disingenuous, at worst dishonest, and at the very least I find it obnoxious.2011-06-21
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    When in grad school, I sat in on a course where the professor spent three lectures defining polynomials in $n$ variables, because he defined them as the image of the adjoint to the forgetful functor. Later in the course, he spent 10 minutes defining Ext as the right derived functor of Hom. Here was someone who figured people didn't really know what polynomials were, but were quite comfortable with derived functors. The idea that someone will wonder whether "$xy=t$" is the result of several mistakes but be comfortable with reverse polish notation strikes me as in the same realm.2011-06-21
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    @Arturo I'm not wondering here Arturo. I know you didn't intend xy=t as a mistake. However, such an interpretation of what you actually wrote does come as quite consistent with the definition of a wff. Is "xy" *as stated* a term, constant, or variable? No. Is it an expression involving a binary operation? No, and we don't have any other operations or relations here. So, *without* any implicitly understood operation there, it qualifies as nonsense and impossible to formalize. Note, I don't assume that, I deduced it. So why did you write xx=t? To just follow convention?2011-06-21
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    @Doug: Your reading comprehension approaches your communication skills down in the nadir. I did not assert that *you* were wondering, so your entire response starts up with an incorrect premise. But *you* brought forth a *hypothetical* individual who would be confused by "$xy=t$", and implied such an individual would be perfectly comfortable with reverse polish notation, and it was that silliness that I addressed. We don't communicate in a vacuum, and people with valid points don't engage in strawmen. You might want to try it sometime, but not with me.2011-06-21
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    @Arturo I never indicated a hypothetical individual who would feel confused. I said "So, it's not easier for a human being to read what he wrote as if there existed an operation there implicitly in comparison to reading his writing as if he made several mistakes by using several meaningless strings such as "xy=t"." Is such an interpretation of "xy=t" as meaningless possible? Yes. Which is easier, interpreting "xy=t" as not having an operation *as it looks like it does not* or interpreting it as *implicitly* having an operation? The former requires fewer assumptions, so the former.2011-06-21
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    @Arturo And the latter implies that you have to believe something *other than* what you actually see, which requires more cognitive effort than believing what you see.2011-06-21
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    @Doug: Sigh. Puesto que hay mas hispanohablantes como lengua nativa que angloparlantes, obviamente debemos comunicarnos en español, pues eso requiere de menos suposiciones. O quizas en Mandarin. En cualquier caso, Sr Maderacuchara, que tenga usted buen dia, buena semana, buena decada, y buena vida. Estoy harto de sus argumentos de espantapajaro y su malrepresentacion de sus propias palabras en el servicio de pretender que usted no dijo las absurdeces que ya dijo, una y otra vez, por no decir cuando trata de meter palabras en mi boca o pensamientos en mi cerebro. Que se divierta, pero solito.2011-06-22
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    Since someone voted on this, and I looked at this again, I'll add this comment. I don't think I had realized it when I wrote this question, but you can interpret each of these structures as representing the conditional for Lukasiewicz 3-valued logic (also Reichenbach 3-valued logic). So, the above indicates one way to write them in terms of first-order logic. Also, if you look at each of the tables, you can either cover the element which appears most with 1 discrete equilateral triangle, or 1 discrete equilateral triangle and a discrete right triangle.2013-07-19

1 Answers 1

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So, suppose you have a 3 element magma $M$, and an element $t$ such that:

  1. For all $x\in M$, $xx=t$.
  2. There exists $y\in M$ such that for all $x\in M$, $xy=t$.
  3. There exists $w\in M$ such that for all $x\in M$, $wx=t$.
  4. There exists $z\in M$ such that for all $x\in M$, $zx=x$.
  5. There exist $x,y\in M$, $x\neq y$, such that $xy=x$.

Note that $z\neq w$: letting $x\neq t$ (possible since $M$ has three elements), then $wx = t \neq x = zx$ by 3 and 4, so $z\neq w$.

Note that $z=t$: for $t=zz=z$, by 1 and 4.

Likewise, $y=t$, since $t = zy = y$ (by 2 and 4). Thus, $y=z=t$, and $w$ is different from $t$.

Thus, $t$ is a right zero and a left identity. So we have:

$M$ is a 3-element magma; there are distinguished elements $t$ and $w$ such that:

  1. $t\neq w$.
  2. $xx = t$ for all $x\in M$.
  3. $xt = t$ for all $x\in M$.
  4. $tx = x$ for all $x\in M$.
  5. $wx = t$ for all $x\in M$.
  6. There exists $z,r\in M$, $z\neq r$, such that $zr=z$.

From 4 and the clause $z\neq r$ in 6, we have that $z\neq t$. From this and 5, we have that $z\neq w$. Thus, the three elements of $M$ are $t$, $w$, and $z$. From 3, we know that $r\neq t$, and since $r\neq z$, hence $r=w$. Thus:

$M$ is a three element magma. The three elements are $t$, $w$, and $z$. They satisfy:

  • $xx = t$ for all $x$.
  • $xt = t$ for all $x$.
  • $tx = x$ for all $x$.
  • $wx = t$ for all $x$.
  • $zw = z$.

Thus, the multiplication table of $M$ is: $$\begin{array}{c|ccc} &t&w&z\\ \hline t&t&w&z\\ w&t&t&t\\ z&t&z&t \end{array}$$ The first row is forced by the third listed property in the last quote box. The first column is forced the second listed property; the second row by the fourth listed property; the last diagonal entry by the first listed property; and the $zw$ entry by the last listed property. This gives all the entries of the table.

Thus, the properties completely determines $M$ up to the names of the elements. If you want to call the elements $a$, $b$, and $c$, then there are exactly $3!=6$ ways in which you can assign the names $a$, $b$, and $c$ to $t$, $w$, and $z$, so that gives exactly six different tables in terms of $a$, $b$, and $c$, though they are all isomorphic to $M$.

(And it would have taken me about half as long to figure out what you are asking if you were using standard notations and nomneclature instead of insisting on your own personal language; I wonder if your favorite movie is Nell.)