Monthly Archives: November 2016

Negative numbers, the minus sign, abstract algebra.

I was pondering the reality of negative numbers and after figuring out that a sequence of dots on a line can be extended in each of the two directions, and then arbitrarily selecting one dot as “the zero”. The line can be further labelled as 1, 2, 3, … to one side and -1, -2, -3, … on the other side.
(better to label the 1, 2, 3, … as +1, +2, +3, … and consider the lot as “signed numbers”)

Soon proceeding towards arithmetic I concluded that 7-3 is 4, and also 8-4 is 4, and therefore 13-9 is 4, and then 3-7 is -4, and -2-2 is -4. It was then observed that if a-b=c then a-y-(b-y) is also equal to c, regardless of the signs of the specific numbers involved.
This of course is stunningly obvious when looking at the signed difference of the first and the second number as an extended number line diagram.

The outcome of all this was an arithmetic for 0, 1, 2 modulo 3, and  the signed difference x-y is a binary operation diff(x,y) with table:

…x  … 0     1     2
y
0         0     1     2
1         2      0    1
2         1      2    0

Example: 1-2 is -1, which is 2 modulo 3

So a non abelian, non associative algebra with a not quite identity satisfies the conditions, where A=1, B=2 and C=0
—————————————————–
There are three objects and an operation called “doesn’t have a name”.
Two are similar, and the third is a bit different
They are paired to yield a single object as follows:

AA = BB = CC = C
AB = BC = CA = B
AC = CB = BA = A

Notice that BC and CB are different, so non-abelian.
Worse is that (AC)A = C and A(CA) = B are different, so non-associative.
And consequently A and B and C are different.
—————————————————-

Interestingly, and maybe separately, the minus sign behaves very differently from the plus sign:

a-(-b) is a+b, but there is no way of writing a-b using only addition.

This means that all expressions can be written with “minus” alone.

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Minimal Abstract Algebra

This is a challenge !!!!!

There are three objects, A, B, and C, and an operation called “doesn’t have a name”.
Two are similar, and the third is a bit different
They are paired to yield a single object as follows:

AA = BB = CC = C
AB = BC = CA = B
AC = CB = BA = A

Notice that BC and CB are different, so non-abelian.
Worse is that (AC)A = C and A(CA) = B are different, so non-associative.
And consequently A and B and C are different.

Your job is to identify (model, in current jargon) the objects and the operation.”

The next post will be “the solution”.

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Filed under abstract, algebra, Uncategorized