In binary the one on the left is meaningless, and therefore the two cannot be compared. In any base in which they can be compared, the one on the left is smaller.

[–] itslilith@lemmy.blahaj.zone 28 points 10 months ago* (last edited 10 months ago) (3 children)

Base ⅒

[–] Eagle0600@yiffit.net 13 points 10 months ago

Alright, you've got me there.

[–] Sonotsugipaa@lemmy.dbzer0.com 2 points 10 months ago (1 children)

Wouldn't that require the number of available digits to be 1/10?

[–] itslilith@lemmy.blahaj.zone 6 points 10 months ago

Fractional bases are weird, and I think there's even competing standards. What I was thinking is that you can write any number in base n like this:

\sum_{k= -∞}^{∞} a_k * n^k

where a_k are what we would call the digits of a number. To make this work (exists and is unique) for a given positive integer base, you need exactly n different symbols.

For a base 1/n, turns out you also need n different symbols, using this definition. It's fairly easy to show that using 1/n just mirrors the number around the decimal point (e.g. 13.7 becomes 7.31)

I am not very well versed in bases tho (unbased, even), so all of this could be wrong.

[–] Klear@sh.itjust.works 1 points 10 months ago

Based.