1. ## Abstract Algebra

Mathematicians love abstraction, and the field of abstract algebra is no exception. I'm bringing it out here to make it easier to find than in The Math Thread.

• It is commutative: a + b = b + a
• It is associative: (a + b) + c = a + (b + c)
• It has an identity: a + 0 = a
• It has inverses: a + (-a) = 0

Take those properties and make some binary (two-argument) operator have them or else have only some of them. What possible operators can there be?

Try adding a unary operator or another binary operator, especially one that is intertwined with the original binary operator in some way. What further possible operators can there be?

But why start with a binary operator? Why not a unary (one-argument) or a ternary (three-argument) one? Unary ones are almost too simple, while ternary ones has not been researched very much.  Reply With Quote

2. I thought you would pick up on it Ipetrich.

Abstract Algerbra, or Boolean Algebra as a subset, is the foundation of digital electronics. There are devices called programable logic arrays. You write a Boolean expression and softare configures the chip to implement the logic.

http://electronics-course.com/boolean-algebra  Reply With Quote

3. Originally Posted by lpetrich • It is commutative: a + b = b + a
• It is associative: (a + b) + c = a + (b + c)
• It has an identity: a + 0 = a
• It has inverses: a + (-a) = 0

Take those properties and make some binary (two-argument) operator have them or else have only some of them. What possible operators can there be?
Can't you make up an infinite amount of bullshit *(non set) operators that satisfy those conditions?

Are you only looking for operators on certain well defined sets?

Try adding a unary operator or another binary operator, especially one that is intertwined with the original binary operator in some way. What further possible operators can there be?
Anticpmmutator?

But why start with a binary operator? Why not a unary (one-argument) or a ternary (three-argument) one? Unary ones are almost too simple, while ternary ones has not been researched very much.
There are tons of unary, binary, ternary, polyary operators in modern compute languages... All of which can be broken-down to binary operators designed by polyary operators. (brains).

Off topic? I remember trying to come up with an easy (papertime) unary division algorith!.  Reply With Quote

4. There is no simple binary division operator. It is done in hardware using what is called a barrel shifter. In binary shift one way is division, the other multiplication.

Floating point fast multiplication and division in binary assigns weights to each bit. Algorithms have been established since the 50s.

https://en.wikipedia.org/wiki/Barrel_shifter

1000 = 8 decimal in binary
Shift right 1 bit is 0100 or 4 decimal, divide by 2.

Floating point numbers are represented by exponent, the decimal point, and the mantissa which is 1 maximum. It is called fractional arithmetic. Each bit in the mantissa is binary weighted. The lsb or resolution is 1./2^n mantissa bits.  Reply With Quote

5. Originally Posted by Kharakov Anticpmmutator?
Do you mean a commutator? An anticommutator is a symmetric binary operator, while a commutator is an antisymmetric one. One can certainly do abstract algebra with antisymmetric operators. With a few other constraints, that's how one gets Lie algebras.

There are tons of unary, binary, ternary, polyary operators in modern compute languages... All of which can be broken-down to binary operators designed by polyary operators. (brains).
True, but I'm talking about their abstract mathematical properties. Especially properties that can be deduced from constraints like being commutative or associative or whatever.  Reply With Quote

6. Let's see about unary functions. Consider a function from a set S to itself. The function's range or set of possible output values I will call R. It will always be a subset of S.

The function will be invertible only if (1) R = S and (2) for all a, b in S with a != b, f(a) != f(b). In short, f is a bijection.

If S is finite and there are some a, b in S such that a != b but f(a) = f(b), then R must be a proper subset of S. That is not necessarily the case if S is infinite, and it should be easy to come up with counterexamples.

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Let us consider repeated application of f: f(n)(a) = f (f(...f(a))) -- n f's. f(0)(a) = a, and if f is invertible, then n can be negative.

It is easy to prove that f(m)(f(n)(a)) = f(m+n)(a)

For S being finite, it can be shown that for any a, there is some m and n such that f(m+n)(a) = f(m)(a). It is easy to prove with a counting argument, so it fails for S infinite. It is also easy to find counterexamples if S is infinite.

So for S finite, repeated applications of f on every element will converge on some "limit cycle", a set of elements where applying f goes through all of them until doing so repeats the element that one started with.

A limit cycle can have length 1: f(a) = a
Length 2: f(a) = b, f(b) = a
Length 3: f(a) = b, f(b) = c, f(c) = a
...

When S is infinite, limit cycles may exist, but they do not have to. Infinite strips may also exist.

If f is a bijection, then if S is finite, then all of its elements are in cycles, while if S is infinite, then all of its elements are in cycles and/or strips.

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I hope that this post is a good example of what one does in abstract algebra.  Reply With Quote

7. Algebraic structure has a big list of them. No mention of ternary operators, however.

Abstract algebra with ternary operators has been asked about a few times in the math Stack Exchange:

abstract algebra - Is anybody researching "ternary" groups? - Mathematics Stack Exchange -- its responders mentioned Heap (mathematics) and a now dead link on work on ternary-operator algebras.

Heaps are defined with:
Para-associativity: f(f(a,b,c),d,e) = f(a,f(d,c,b),e) = f(a,b,f(c,d,e))
Identity: f(x,a,a) = f(a,a,x) = x

Group to heap: f(a,b,c) = a*b-1*c
Heap to group: a*b = f(a,e,b), identity e, inverse a-1 = f(e,a,e)

abstract algebra - algebraic ternary operators - Mathematics Stack Exchange -- no responses

abstract algebra - how many "pure ternary" operators are there? - Mathematics Stack Exchange

One that cannot be turned into a composition of binary operators: for args a,b,c no possible f(g(a,b),c) and similar.  Reply With Quote

8. Just for the heck of it, I decided to find all the pure ternary operators over {0,1}.
• A unary operator has 2^1 = 2 possible arg values, and thus 2^2 = 4 possibilities.
• A binary operator has 2^2 = 4 possible arg values, and thus 2^4 = 16 possibilities.
• A ternary operator has 2^3 = 8 possible arg values, and thus 2^8 = 256 possibilities.

I found that only 152 of the possible ternary operators can be reduced to compositions of binary ones. Leaving 104 pure ternary ones.  Reply With Quote

9. 1,2,4,8,16,32...2^n
I do not understand what you mean by ternary.  Reply With Quote

10. Originally Posted by steve_bank 1,2,4,8,16,32...2^n
I do not understand what you mean by ternary.
Ternary operator: operator with three arguments (input values)  Reply With Quote

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