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Okay, so I was thinking about my first semester of logic. I know that there are 2 "0-ary" functions: True or False. There are 4 "binary" functions:

1) f(A) = True always

2) f(A) = True when A is true, f(A) = False when A is false (ie the identity function)

3) f(A) = True when A is false, f(A) = Fales when A is true (ie NOT A)

4) f(A) = False always

Notice that 2 and 3 are the only "meaningful" functions that depend on A...1 and 4 are basically 0-ary functions.

There are 16 binary functions that take the form A@B. Here's their standard list:

1) True

2) False

3) A

4) not A

5) B

6) not B

7) A AND B

8) A NAND B

9) If A then B

10) Not (If A then B)

11) If B then A

12) Not (If B then A)

13) A OR B

14) A NOR B

15) A iff B

16) A XOR B

Now, I'm lacking some vocabulary here....but there seems to be a little redudancy. The unary operators I labled "True" and "False" dont' depend on A...so in fact they're really equivalent to 0-ary operators. Same for the binary operators 1 and 2. In addition, the binary operators 3, 4, 5, and 6 only depend on ONE of the statements....not both. So they're effectively unary operators, as they don't "use" the second input.

Two simple questions for my own edification (this is NOT homework!).

1) Is there some name for these "redundant" or "useless" operators? I know that strictly speaking, all binary operators are useless if you just have NAND, but I'm talking about something different. I mean specifically about an n-ary operator where you KNOW the output even if you know less than n of the inputs.

2) There are 2^(2^3) = 256 ternary operators. What's the redundancy schema of these operators? How many of these operators REALLY require information of all three inputs?

So I was randomly thinking of two patterns in math that somehow seem to "go together":

Patterns with a period of 1:

If arrows represent multiplication by 1: 1 => 1 => 1 => 1

If arrows represent differentiation: e^x => e^x => e^x => e^x

Patterns with a period of 2:

If arrows represent multiplication by -1: -1 => 1 => -1 => 1

If arrows represent differentiation: e^-x => -e^-x => e^-x => -e^-x

Patterns with a period of 4:

If arrows represent multiplicatoin by i: i ==> -1 ==> -i ==> 1 ==> i

If arrows represent differentiation: sin x ==> cos x ==> -sin x ==> -cos x ==> sin x

It's been awhile since a formal math class for me, but I'm 100% competent in anything from high school or first year calc, and remember most of multivar. Never did diff-eqs, although I recognise that sin x, cos x, e^x, and e^-x are all solutions to the equation d^4y/dx^4 = y. Is there some tie-in between the roots of x^4 = 1 and the solution to that differential equation?

So, I've never taken real analysis, but I've taken multivariable calculus (aka vector analysis), linear algebra, and a couple of semesters of logic. Never taken any real analysis, as I was a chem major in college. All this got me thinking of an interesting question.

When discussing limits and continuity, my high school calc teacher put up this function:

f(x) = 0 if x is not rational

f(x) = 1 if x is rational

It made sense that this function was defined for all real x, but continuous for NO real x, since there is a rational number between any two irrational numbers, and an irrational number between any two real numbers.

I want to integrate this function! In logic I learned that irrationals aren't countable, so it would seem that the integral of f(x)dx from 0 to 1 would equal 0, looking at the integral from an "average value of a function" perspective. By that logic, the integral of the function EVERYWHERE must be zero. So if the integral of the function is zero, the original function must have to be an unvarying constant...which my function isn't! Someone help me with this paradox...it's been burning up my brain as a casual mather...

Here's a question for you. I am going to answer every question on the board, if and only if that person is unable to answer his or her own question. Do I answer this question?