0.99999~ = 1 ?

No, mathematically they're precisely equivalent - they are, in fact, different ways of writing the same thing. What's more every terminating decimal has two ways of being written, i.e.0.5214 = 0.521399999~
7.2 = 7.199999~And so on.

Nope, they're completely the same, philosophically and otherwise. Nope. There is no uncertainty, 0.9999~ is 1. In fact all decimal representations are infinite sequences, it's just some of them finish in an infinite number of 0s. This is not, as it may sound, a cute aphorism but actually fundamental to the construction of the real numbers. Real numbers are limits of inifnite sequences.

Yeah, but finitism is silly.

The difference isn't infinitesimally small, it's 0.0.9999999~ is the limit of 0 + 9/10 + 9/100 + 9/1000 + 9/10000 + ....The limit. And the limit of that series is 1, not something infinitesimally different from one.

Yes, but the point is that 0.99999~ is a limit. There is no 0.9999~ that isn't a limit. All real numbers are limits, that's how they're constructed. Oh, sure, it feels intuitively wrong, no doubt. But as in higher level physics, intuition is a shoddy guide to higher maths.

Aye, it was a comment, not an argument

0.999999~ is the asymptote, not the value approaching it.To clarify the nomenclature: a curve is said to be asymptotic to a line if it approaches it without ever reaching it (i.e. 1/x is assymptotic to 0), that line itself is the assymptote.0.9 0.99 0.999 0.9999 0.99999 etc. is a sequence, that sequence approaches but never reaches 1, in a way that resembles an assymptotic curve. 0.999~ is not a member of that sequence, it is its limit. That is, it's the value that 0.9 0.99 0.999 0.9999 0.99999 etc. approaches but never reaches.Edited by Mr Jack, : Additional

It's not necessary to write them as 0.9999~ for that proof to work, concluding with an infinite series of 0s also works.