MATH: inverse of a function with two variables
Some foreword, for the fun of it, I'm redefining basic arithemetic operations such that:
+'(a,b) = a + b + ((a+b)/2)
*'(a,b) = a +'a +' ... +' a, b times
^'(a,b) = a*'a *' ... *' a, b times
and so on
However, I don't think it would be possible to define /' such that it is the inverse of *'. What it breaks down to is the inverse of a function with two variables, and I don't know how I would begin to do that, let alone if it would be possible. I've done some quick googling, but it seems my google foo has failed me.
Thus, I wonder, is it possible to define a function /'(a,b) such that it is the inverse operation of
sum(a+'a, b)?
+'(a,b) = a + b + ((a+b)/2)
*'(a,b) = a +'a +' ... +' a, b times
^'(a,b) = a*'a *' ... *' a, b times
and so on
However, I don't think it would be possible to define /' such that it is the inverse of *'. What it breaks down to is the inverse of a function with two variables, and I don't know how I would begin to do that, let alone if it would be possible. I've done some quick googling, but it seems my google foo has failed me.
Thus, I wonder, is it possible to define a function /'(a,b) such that it is the inverse operation of
sum(a+'a, b)?
-′(+′(a, b), b) = a
-′((3/2)(a+b), b) = a
So -′(x,y) = (2/3)x-y
That's the inverse of +′
Given that *′(a,b) = ∑′a from 1→b (with ∑′ being a sum using +′)
I'd think we could reverse *′ to /′ just by repeatedly applying -′ instead of +′
-′((3/2)(a+b), b) = a
So -′(x,y) = (2/3)x-y
That's the inverse of +′
Given that *′(a,b) = ∑′a from 1→b (with ∑′ being a sum using +′)
I'd think we could reverse *′ to /′ just by repeatedly applying -′ instead of +′
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