Non-trivial extension of max(n,1)-1 to the reals and its iteration.

[MphLee]

I was very interested in the problem of extending \( max(n,1)-1 \) to the reals (that is actual the "incomplete" predecessor function \( S^{\circ (-1)} \) over the naturals)

I've asked the same question on MSE but it was a bit ignored...I hope because it is trivial!

http://math.stackexchange.com/questions/...its-fracti


The question is about extensions of \( A(n):=max(n,1)-1 \) to the reals with some conditions

A-\( A(x)=max(x,1)-1 \) only if \( x \in \mathbb{N} \)
B-\( A(x) \) is not discontinuous

I just noticed that I've made a lot of errors in my MathSE question, I'll fix it in this post (and later on mathSE)

From successor and inverse successor we can define the subtraction in this way

\( x-0:=x \) and \( x-(y+1):=S^{\circ(-1)}(x-y) \)

with \( A \) , that is a modified predecessor function we could define its iteration using an "esotic subtraction" that is "incomplete" for naturals and is "complete" for reals (like we are cutting all the negative integers)

\( x -^* 0:=x \)
\( x-^* (n+1)=A(x-^* n) \)

In this way we have

\( x-^*1=A(x) \) and
\( x-^*n=A^{\circ n}(x) \)

How we can go in order to extend \( x-^*y \) to real \( y \)?

For example what can we know about \( x-0,5 \) ?

for example \( (x-^*0.5)-^*1=(x-^*1)-^*0.5=x-^*1.5 \)

if we put \( x=0 \)

\( (0-^*0.5)-^*1=(0-^*1)-^*0.5=0-^*1.5 \)

then if \( 0_^*0.5=\alpha \)

\( \alpha-^*1=0-^*0.5=0-^*1.5 \)


if \( \alpha \) is not a natural number

\( \alpha-^*1=\alpha=0-^*1.5 \)

!?? what is going on here?

If \( \alpha \) is natural

\( max(\alpha,1)-1=\alpha=0-^*1.5 \)

in this case we should have that \( \alpha=0 \).
What do you think about this?