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"Hyperoperator" family of operators is defined recursively like so: $$ a\langle{0}\rangle{b} = a+b\\ a\langle{1}\rangle{b} = ab = a+a+a+a... \; \text{(w/ b a's)}\\ a\langle{2}\rangle{b} = a^{b}= aaaa... \; \text{(w/ b a's)}\\ ... $$ We can also extend the hyperoperators to "-1st operator" using the method described in this old post: $$ a\langle{-1}\rangle{b} = a@b = \{a\neq{b}:\text{max}(a,b)+1,a=b:a+2\}\\ $$ This operator (called zeration) satisfies the property that $a+b = a\langle-1\rangle{a\langle-1\rangle}a...\text{(w/ b a's)}\\$.
My question is: Can we extend hyperoperators below zeration indefinitely while preserving the property I described above?

Nirvana
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  • OP, I suggest you change zeration to $\langle 0 \rangle$ to match both the post you link, and the Answer you currently have. Then your question is whether there exists $\langle -1\rangle$ Etc doing what you want – Calvin Khor Mar 02 '21 at 11:09
  • @CalvinKhor I can change it, but I can't edit the older comments that still use the -1 notation, because of SE's annoying 5-minute cap. So it will be confusing. – Nirvana Mar 03 '21 at 02:23
  • You raise a good point...i still think changing it is better since it has to be confusing either way, but it’s ofc up to you. Maybe add a remark about older notation in the main body – Calvin Khor Mar 03 '21 at 02:27

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Your "zeration" definition is unusual. It is usualally defined as increment by one operation:

$$a\langle0\rangle b=b+1$$ Its repeating $b\langle0\rangle b$ for $c$ times is addition $c+b$. Increment is not commutative and does not depend on the first argument.

Moreover, Since $\Gamma(1)=\Gamma(2)=1$, addition and multiplication are both commutative, but since $\Gamma(0)$ goes to infinity, this means that zeration does not depend on the first argument at all. And this is true for any hyperoperation of negative integer order as well. Because the Gamma function has poles there.

Anixx
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  • @Nirvana n incremented n times is n+n – Anixx Mar 02 '21 at 06:47
  • @Nirvana the author of the blog you cited tried to make increment commutative by ad-hoc applying the "max" operation, which is unnatural. For some unknown reason he also wants $a\langle0\rangle a$ to be defined 2, also unnecessary and unnatural idea. – Anixx Mar 02 '21 at 06:52
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    Sorry, I didn't finish the comment. aa=a^2, and a+a=a2. But if we define $a\langle{-1}\rangle{b} = b+1$, then $a\langle{-1}\rangle{a} = a+1 \neq a+2 $ – Nirvana Mar 02 '21 at 06:52
  • @Nirvana you were asking about zeration $\langle0\rangle$, not about $\langle-1\rangle$. Yes, $a+a=2\cdot a$, $a\cdot a=a^2$ etc by induction. But this is not the case for $0$th hyperoperation and for negative orders. – Anixx Mar 02 '21 at 06:55
  • @Nirvana artificilally one can define $a\langle0\rangle a=2$, but I have never seen such definition except in the cited blog, this makes the operation discontinuous, and artificial max makes it non-differentiable as well. – Anixx Mar 02 '21 at 06:58
  • Just to be clear, I am starting with addition as $a\langle{0}\rangle{b}$, multiplication as $a\langle{1}\rangle{b}$, exponent as $a\langle{2}\rangle{b}$, tetration as $a\langle{3}\rangle{b}$, etc, so "0" operator is just addition. You are correct about the zeration definition, but the post I linked to also has an alternate continous definition: $a\langle{-1}\rangle{b}={b>a+1:b+1,else:a+2}$ – Nirvana Mar 02 '21 at 07:02
  • @Nirvana addition is usually considered $a\langle1\rangle b$. Zeration word that you use in the title comes is from "zero", so it is $a\langle0\rangle b$. – Anixx Mar 02 '21 at 07:05
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    I think your Answer makes it quite clear that you are talking about something completely different (ignoring the differing notation) than what OP describes, whether or not it is standard or natural. (Also, "the author" wanted associativity, not commutativity, and Math.SE is not a "blog".) – Calvin Khor Mar 02 '21 at 09:49