2
$\begingroup$

See here :

http://googology.wikia.com/wiki/Arrow_notation

for the definition of the up-arrow function.

Can $10\uparrow^n m<2\uparrow^n (m+2)$ be formally proven for all $m\ge 1$ and $n\ge 3$ ?

With Saibians theorem we get $$10\uparrow^n m<(2\uparrow^n 3)\uparrow ^n m<2\uparrow^n (m+3)$$ Also the claim is trivially true for $m=1$, but I did not manage to complete the induction step.

$\endgroup$
  • $\begingroup$ Not to sound rude or anything, but does my answer suffice? $\endgroup$ – Simply Beautiful Art Aug 4 '18 at 21:44
  • $\begingroup$ Ah, sorry. I've edited to include that now :-) $\endgroup$ – Simply Beautiful Art Aug 6 '18 at 14:59
  • $\begingroup$ I will have to go through the details, but apparently, you have proven the claim :) $\endgroup$ – Peter Aug 6 '18 at 15:02
1
$\begingroup$

For $n=3$, we have:

For $m=1$, we have $10\uparrow^31=10$ and $2\uparrow^33=65536$.

Assume $10\uparrow^3m<2\uparrow^3(m+2)-3$ holds for some $m\ge1$. Then we have

\begin{align}10\uparrow^3(m+1)&=10\uparrow^210\uparrow^3m\\&<10\uparrow^2(2\uparrow^3(m+2)-3)\\&<(2\uparrow^23)\uparrow^2(2\uparrow^3(m+2)-3)-3\tag0\\&<2\uparrow^22\uparrow^3(m+2)-3\\&=2\uparrow^3(m+3)-3\end{align}

Hence $10\uparrow^3m<2\uparrow^3(m+2)-3<2\uparrow^3(m+2)$.

$(0)$ is due to the fact that up-arrows are strictly increasing in all arguments.


Assume it holds for some $n\ge3$ and all $m\ge1$. Then we get

\begin{align}10\uparrow^{n+1}m&=(10\uparrow^n)^{m-1}10\\&<(2\uparrow^n[2+)^{m-1}10]\tag1\\&<(2\uparrow^n)^{m-1}13\\&<(2\uparrow^n)^{m-1}2\uparrow^33\\&<(2\uparrow^n)^{m-1}2\uparrow^n2\uparrow^n2\\&=(2\uparrow^n)^{m+1}2\\&=2\uparrow^{n+1}(m+2)\end{align}


$(1)$ is the result of

\begin{align}2+2\uparrow^nm&<3+2\uparrow^nm\\&<2\uparrow^{n-1}2\uparrow^nm\\&=2\uparrow^n(m+1)\end{align}

$\endgroup$
1
$\begingroup$

Can $10\uparrow^n m<2\uparrow^n (m+2)$ be formally proven for all $m\ge 1$ and $n\ge 2$ ?

No, because $10\uparrow^2 2 > 2\uparrow^2 (2+2)$.

$\endgroup$
  • $\begingroup$ Sorry, I will edit my post. $\endgroup$ – Peter Jan 15 '18 at 18:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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