According to Wikipedia hyperoperation for positive integers is defined as $$ H_{n}(a,b)=H_{n-1}(a,H_{n}(a,b-1)) $$ with some base conditions. (We take $ n \geqslant 1 $.)

Question:
Recursivly define a sequence of binary operators that inverts hyperoperation for all positive integers.
Meaning we are looking for $I_{n}$ such that $$ \forall n,a,b \in \mathbb{N}_{+}: I_{n}(H_{n}(a,b),a)=b $$

  • This is at most possible for $n \geq 1$, because $H_0$ ignores its first argument. – A.P. Oct 31 '15 at 15:09
  • $I_{0}$ I'd simply define as counting down (by $1$). – MrFrety Nov 5 '15 at 7:02
  • That wouldn't work. The problem is that you are asking for an inverse of $H_n(\cdot, b)$, but $H_0$ is not injective in it's first component. Indeed, $H_0(a, b) = b+1$ for every $a \in \Bbb{Z}_{\geq 0}$, so with your definition we would have $I_0(H_0(a,b), b) = b$ for every $a \in \Bbb{Z}_{\geq 0}$... – A.P. Nov 5 '15 at 11:06
  • I'd rather define $H_{0}(a,b)=a+1$. Mentioned it on the discussion page... Then it would work, wouldn't it? – MrFrety Nov 7 '15 at 20:48
  • No, it wouldn't, because you'd be messing with the base case of the definition of $H_n$. Indeed, consider $H_o'(a,b) = a + 1$. Then $H_1(1,b) = H_0'(1, H_1(1, b-1)) = 2$ for every $b > 0$... – A.P. Nov 7 '15 at 22:46
up vote 0 down vote accepted

Examples:

  • $9/3=1+(9-3)/3=1+(1+(6-3)/3)=1+(1+1)=3$

  • $log_{2}(8)=1+log_{2}(8/2) =1+(1+log_{2}(4/2))=1+(1+1)=3$

  • $log_{3}(27)=1+log_{3}(27/3) =1+(1+log_{3}(9/3))=1+(1+1)=3$

So in general: $$ I_{n}(c,a)=1+I_{n}(I_{n-1}(c,a),a) $$ whereby $I_{n}(a,a)=1$ for $n>1$ and $I_{1}(c,a)=c-a$ for $n=1$.

  • This definition will terminate only when the first argument of $I_n(\cdot, b)$ is of the form $H_n(a,b)$. Can the definition be meaningfully extended to other values? For example, we might say something like $I_n(c,d) = 0$ if $d > c$, to give the "quotient" of the (pseudo)inverse, discarding the "remainder" (I didn't actually test if this is sufficient). – A.P. Oct 31 '15 at 15:15
  • 1
    I have some ideas on how to expand it to other number sets or even construct new number sets from it. But it was too hard, so I restricted myself to positive integers for now. – MrFrety Nov 5 '15 at 6:58
  • This definition doesn't always hold true. It only occurs here due to log rules, mainly $1+\log_b(a)=\log_b(ab)$ – Simply Beautiful Art Jun 8 '17 at 22:23
  • I claim that it holds true for all n purely from the definition of the Hyperoperator. The inverse basically counts the number of repetitions... or it should... I'm always thankful for corrections... – MrFrety Apr 30 at 3:47
  • After some adjustments the variable names are hopefully more consistent, now. – MrFrety Apr 30 at 5:04

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.