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I here define the hierarchy of basic mathematical operators and their respective "inverse" operation (see hyperoperation).

$$ \begin{array}{c|c|c|} & \text{Operator} & \text{"Inverse"} \\ \hline \text{Incrementation} & a+1 & a-1 \\ \hline \text{Addition} & a+b & a-b \\ \hline \text{Multiplication} & ab & \frac{a}{b} \\ \hline \text{Exponentiation} & a^b & \sqrt[b]{a} \\ \hline \text{Tetration} & ^ba & \sqrt[b]{a}_s \\ \hline \text{} \vdots & \vdots & \vdots \\ \hline \end{array} $$

Now, it is clear that each "level" is simply the previous one except the process is done several times. Adding two integers is like incrementing one integer many times. Multiplication is addition of the same number many times.

My question is: does anything precede incrementation?

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How could anything precede the successor operation? By definition, there are no natural numbers strictly between $n$ and $n + 1$... (Also, if you're interested in higher levels, you might want to look at how the Ackermann function is constructed.) –  Zhen Lin Jul 13 '12 at 16:31
No, because the successor operation is no longer a binary operation. I really don't see how you plan on breaking down a unary operation as "repeating", though I'm more than willing to be proven wrong. –  Robert Mastragostino Jul 13 '12 at 16:46

2 Answers 2

up vote 3 down vote accepted

No, because the successor operation enters the stage already in the definition of natural numbers.

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We can notice that

$a^{S(n)}=mul_a(a^n)=a\times (a^n)$

$a \times S(n)=add_a(a\times n)=a+(a\times n)$


here the seqence breaks, since successor is not a binary operation. but we can continue finding a function $f$ that is the "$(-1)$ step of the sequence (if the Successor is the step $0$).

$S(S(n))=f(S(n))$ that becomes

$n+2=f(n+1)$ and for $n=m-1$ we get


so $S(n)=f(n)$: "successor precedes" the successsor in the sequence

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