[tomtomtom1]

if we say that infinitesimals exist. that 1/3 != 0.33.. and 1 != 0.9999... and the probability of possible events is never 0.

what are the properties that we would lose?

[mFixman]

If we say that infinitesimals exist, it still happens that 1 = 0.999…. It just happens that 0.999 ≠ 1 - 𝛚.

0.999… = 1 is a property of the way we write some rational numbers, not of the number system itself.

[tomtomtom1]

wouldn't 0.999.. be equal to 1 - 10^w since it's only a countably finite series of nines.

[mFixman]

0.999...9 with a countable finite amount of nines is clearly less than 1.

0.999... with infinite nines is equal to 1.

[tomtomtom1]

sorry I typoed I meant, countably infinite. by w I meant the ordinal.

[marcosdumay]

I have never seen a number system with infinitesimals where the addition wasn't updated to ignore smaller classes if they come with larger ones.

That is, for any number system I've seen, 1 = 1 + dx, and infinity = infinity + 100.

[MaxBarraclough]

The obvious one seems terrible enough, that division is no longer the inverse of multiplication: (1/3)*3 != 1

[_v7gu]

Nonstandard analysis exists (with infinitesimal and infinite numbers) , but 1/3 and 9/9 is the same there. The problem is that the numbers 0.333... and 0.999... don't really exist.

[LudwigNagasena]

Completeness is one of the most important properties of real numbers. Basically, you will have to completely throw away real analysis.

[2OEH8eoCRo0]

1/3 does equal 0.3.. though.