True but addition becomes a lot less efficient in this representation :)
To make both addition and multiplication O(n), you can store numbers as their residues modulo a bunch of different primes and appeal to the Chinese Remainder Theorem. However, then size comparison becomes difficult.
alt Hacker News
To make both addition and multiplication O(n), you can store numbers as their residues modulo a bunch of different primes and appeal to the Chinese Remainder Theorem. However, then size comparison becomes difficult.