by Eli Tyre Sep 18 2017

The other thing that this interpretation tells us is that the digit worth exactly half an $~$n$~$\-digit is the $~$\\sqrt{n}$~$\-digit, both because \(1\) if you want to use an $~$n$~$\-digit to store two $~$x$~$\-messages for some $~$x$~$, you need $~$x \\cdot x$~$ to be at most $~$n$~$; and \(2\) if you have a 50% chance of getting to use an $~$n$~$\-digit and a 50% chance of nothing, then that's equivalent on average to a certainty of a $~$\\sqrt{n}$~$\-digit\.