"For readers who just skimme..."

https://arbital.com/p/448

by Eric Rogstad Jun 9 2016


Let's say that a coin costs 1¢\. Because $~$n$~$ coins can be used to encode any one of $~$2^n$~$ different possibilities, you could get the job done with three trillion coins, which would let you encode the number \(in binary\) for the low low cost of thirty billion dollars\. \(At the time of writing this, three trillion bits of data will actually run you about \$60, but leave that aside for now\. Pretend you can't use modern transistors, and you have to use comparatively large objects such as coins to store data\.\)

For readers who just skimmed over $~$2^{3,000,000,000,000}$~$, and didn't parse it as "two to the three trillion" (e.g. Eric on his first reading), it may be helpful to more explicitly call out that this "three trillion" is the same number that was in the exponent earlier.

(I tried previewing a couple of page edits, like replacing $~$2^{3,000,000,000,000}$~$ with $~$2^\text{3 trillion}$~$, but wasn't able to find anything that I liked.)


Comments

Nate Soares

I think we don't need that much handholding. You've got to let people figure out some of the connections themselves :-p