See here :

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.

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

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


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


$(1)$ is the result of


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

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

  • Sorry, I will edit my post. – Peter Jan 15 at 18:17

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.