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I'm inferring that any hyperoperation you could apply here using two for every value would result in four. Why is this?

Isaac Philo
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  • Yes $2+2=2\times 2=2^2$. Is that what you are asking? Not sure that it makes much sense to ask "why" arithmetic operations yield the result they do. – lulu Jan 07 '22 at 16:22
  • In fact, in Conway chains and Bowers arrays, a start with $2-2$ always gives the value $4$, apparently a degenrated case. In the case of Knut's up-arrows , it is easy to show by induction that we have for every $n \ge 1$ : $$2 \uparrow^n 2=4$$ – Peter Jan 07 '22 at 16:23
  • That $2+2=2\cdot 2$ , but $3+3\ne 3\cdot 3$ has the reason that we have two summands. This is similar with the other operations. – Peter Jan 07 '22 at 16:25
  • It's because $2+2=17$ leads to contradictions. – Gerry Myerson Jan 07 '22 at 16:38

1 Answers1

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Suppose $n$ is an integer and $n+n=n\cdot n$.

Then $2n=n^2$ so $n^2-2n=0$. Factoring gives $n(n-2)=0$.

So either $n=0$ or $n=2$.

But if we also have that $n+n=n\cdot n=n^n$ then we have $n^2=n^n$ which gives $n=1$ or $n=2$, so $n=2$ is the common solution to the two equations.

John Wayland Bales
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  • I didn't realize that a perfectly algebraic solution existed for this. Thank you! I will attempt to employ this strategy before asking such questions in the future. – Isaac Philo Jan 08 '22 at 18:28