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• ##### More arrant pedantry . . .(1+ / 0-)
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Orinoco

. . . from ends with space, who makes the following comments with trepidation, but with no doubt of their correctness.

The "proof" that commutativity implies associativity is invalid.

Using letters for colors, here's the trouble: It starts by considering r+y+g.

But what does this expression mean? Addition is defined on two things at a time. There's no table where we can look up the sum of three things taken together! As an expression, a three-way sum without parentheses is not defined. We can have only (r+y)+g or r+(y+g).

The "proof" makes it clear that the former is intended, because it first exchanges r and y via commutativity. It starts by saying that (r+y)+g = (y+r)+g, perfectly true. But the next step is not permissible---you can't now exchange r and g without using the fact that (y+r)+g = y+(r+g). That is, this proof makes use of precisely the thing that it purports to prove, an error classically called "begging the question". (Use of this phrase to mean "raising the question", as is done nowadays, is still deplored by us arrant pedants.)

In fact, the only justification for expressions like "a+b+c" is associativity. Given that property, it doesn't matter how we insert parentheses to make the expression well-formed, so we can get away with being a little lazy.

One last point, for those who haven't studied it formally: In algebra, associativity is more fundamental. There are many algebraic structures without commutativity and even without identity elements, but if you don't have associativity things aren't very interesting.

• ##### You're right, of course(0+ / 0-)

Hmmm
(r+y)+g = (y+r)+g
also r+y=orange (intermediate sum)
and y+r=o
so (r+y)+g=o+g
then
o+g=g+o
(r+y)+g=g+o
(r+y)+g=g+(y+r)
I suppose now I'm stuck again.

Looks like I'm going to have to re-write this one. :)

"The problems of incompetent, corrupt, corporatist government are incompetence, corruption and corporatism, not government." Jerome a Paris

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• ##### Not sure where you're going with this, (1+ / 0-)
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Orinoco

. . . but if you're trying to repair the proof, it can't be done.

Consider the following (completely artificial) operator ? defined on these same three elements:

?   y   r   g
y   y   r   g
r   r    g   g
g  g    g   r

This operator is not only commutative, but it also has an identity (y) and lacks zero divisors. And yet it's not associative, since (r ? g) ? g != r ? (g ? g).

This shows that associativity is somewhat akin to Buddha-nature: You either have it or you don't, and if you don't, there's nothing at all that can substitute for it.

• ##### So to demonstrate associativity(0+ / 0-)

would I have to do an exhaustive evaluation of all possible combinations arising from my addition chart?

Or: what does Buddha-nature look like on an addition table? Left to right uniform progression? Top to bottom uniform progression? The combination of these?

"The problems of incompetent, corrupt, corporatist government are incompetence, corruption and corporatism, not government." Jerome a Paris

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• ##### I don't myself know a way to prove associativity(1+ / 0-)
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Orinoco

of an arbitrary operator, except by exhaustive checking. Arbitrary means arbitrary. In my ? operator, for example, changing a single result value makes the operator associative.

Of course one never does this in real life, especially when the space is infinite. Associativity is proven via other properties of the operator or the set. For example, the associativity of addition and  multiplication mod N follows from the same properties on the natural numbers, which properties can themselves be proven directly from the Peano axioms (say).

Not sure what you mean by "uniform progression" but I doubt that there's any such characterization. I'm hardly an expert on this topic, though.

• ##### And what about proving commutativity?(1+ / 0-)
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Orinoco

A quibble: The text proves commutativity of + mod 5 by noting that all entries in any given SW-to-NE diagonal are identical. This of course is sufficient but not necessary; all you really need is for the table to be symmetric in the main diagonal. Although there's no incorrect statement in the text as it stands, some people may take away a condition for commutativity that's too strong.

• ##### Ah, yes(0+ / 0-)

I see where I did that. Didn't mean to, though. The condition is symmetry, not identical entries along the entire SW-NE diagonal.

A bit later Awkward Goat says "But if we reverse the order of those numbers, we'll get the sum in the mirror image spot in the other triangle" but the damage was already done when he agrees with Billy Goat Gruff that identical diagonals "...means that...we get two triangles which are mirror images of each other.”

By "uniform progression" I mean there is an order that is preserved on each row (or in each column). b follows a, c follows b, d follows c, e follows d, a follows e.

a b c d e
b c d e a
c d e a b
d e a b c
e a b c d

"The problems of incompetent, corrupt, corporatist government are incompetence, corruption and corporatism, not government." Jerome a Paris

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