# Interesting Problem

1. Jan 4, 2005

### Alkatran

Assuming o(1) is +, o(2) is *, o(3) is ^, etc how would you go about solving an equation such as:
y = 3 o(x) 4 (where y is a given number and you must find x)

I can't imagine how you would isolate x, you'd need an inverse o() of sorts.
ie:
x = o'(3, 4, y) (Hey, a 4d graph needed to represent this one!)

Hmmm... come to think of it:
2 o(x) 2 = 4
x can be any positive integer because 2 o(4) 2 = 2^2 = 2*2 = 2+2
2 o(x) 2 = 2 o(x-1) 2 = ... = 2 + 2 = 4 as long as x is > 2 and whole.

2. Jan 4, 2005

### arildno

Hi, I think you should approach this problem as a FUNCTIONAL EQUATION

Basically, "+" on N can be seen as a function $$f:N\times{N}\to{N}$$

so that given arguments and output, you are to deduce which function on $$N\times{N}$$ you're after.

I am somewhat pessimistic as to the solvability of such functional equations,
i.e, that we may devise a generally valid procedure to procure the answer.
I think we'll end up with a "guess&try"-method.

3. Jan 4, 2005

### matt grime

You appear to be hiding a lot in the ellipsis "and so on", how many operations are you going to allow? An isolated example won't tell you anything, as you noticed.

4. Jan 4, 2005

### Alkatran

X can be any positive integer. I guess it all comes down to: Can we take our standard operators (not even including the inverses, like / or sqrt() or log()) and represent them in this way? Shouldn't there be some way to do this? After all the relationship is simple enough: A + is a series of increments, a * is a series of +, etc.

I guess what it would eventually come to is "What if x isn't an integer?"
ex: 3 o(x) 4 = 5, solve for x. (yikes!)

edit: ... Come to think of it, 3 OR 4 = 5
011 OR
100 =
111

5. Jan 4, 2005

### arildno

I don't see why you can't just let the operators be seen as functions from one set of numbers to another set of numbers?

6. Jan 4, 2005

### Alkatran

Alright, so let's say o(3, 4, x) = 5

Where x is the operator. Hmmm... Could use some 4d graphing tools or something. lol

7. Jan 4, 2005

### matt grime

There are an infinite number of binary operators on N such that 3*4=5. I mean, define x*y to be sqrt(x^2+y^2) when this is an integer and 0 otherwise, or xy-7, or x^2-y, or... do I need to continue?
You definition of what constitutes basic is entirely arbitrary, and there is no reason why anything of the nature you desire "must" be true.

8. Jan 4, 2005

### jcsd

Are you saying something along the lines of let: o(1) = +, let y o(n) y = y o(n+1) 2 and y o(n) y o(n)....o(n) y = y o(n+1) p where (p is the number of 'y's' on the LHS)? Cos it's not obvious to me exactly what o(x) is menat to represnet when x isn't equal to 1,2 or 3.

9. Jan 4, 2005

### StatusX

If you intend o(n) to be defined as jcsd described, which I'm betting you do, then this problem is made complicated by the fact that for n above 2, o(n) is neither commutative nor associative. An alternate way of defining o(n) which still has o(1) -> + and o(2) -> * and maintains commutativity and associativity for all n is:

o(1) -> +

exp(ln(x) o(n) ln(y)) = x o(n+1) y

or

ln(exp(x) o(n) exp(y)) = x o(n-1) y

This allows you to extend it to negative integers as well, so it would probably be better to call addition o(0). The downside is that the operation above multiplication is not exponentiation but instead is x^ln(y).

Last edited: Jan 4, 2005
10. Jan 5, 2005

### Alkatran

Simply put:
y o(n) y = y o(n+1) 2 when n is a positive integer

Then comes the problem of what -, /, sqrt(), log() are...
Come to think of it: + and * only have one inverse while ^ has two... ^^ has more than two!

3^^4 = 3^3^3^3 = ((3^3)^3)^3 or 3^(3^3)^3 or ...
And you should be able to get 3 and 4 from all of those answers using some function!

IE: F[((3^3)^3)^3, 3) = 4

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