I would say that .333… does not equal 1/3 but it approaches it and is the only thing that can be done in the decimal system to approximate 1/3. It’s an artifact of the numbering system we use. If we had a base 3 system, the answer would be exactly 1.

The only problem I see with that proof is that you’re shifting all the 9’s up one decimal place. Multiplying anything by 10 always causes a zero to be put at the “end”. That seems to suggest the subtraction requires an “end” to the nines because there is a zero somewhere out there at the actual “end”, meaning you end up somewhere out there not with a zero, but a nine. If you do the subtraction, that leaves you with the result of 9.000…9, which you are then dividing by 9 you would end up with 1.000…1

I’m letting stand .999… = 1 for the sake of discussion. Somewhere out there it does not click over to 1.0 if there were a calculator with infinite digits, yet we let it equal 1.

I need this explained to me though and then I’ll come back with my question afterward.

Let’s change the subject to the infinitesimal which I would write as “.000…1”

First, Is that equal to zero? The ‘1’ is somewhere out there but of course you never reach it.

If it’s equal to zero, my next question is:

Given the highest number you can name, what percentage of that number is infinity?

Would it be “.000…1” as I wrote above to express the smallest infinitesimal?

Now, if that’s equal to zero, that makes all named numbers equal to zero as compared with infinity.

Last question:

It was proved that .999… equals one.

This seems to lend to the idea that “.000…1 (which is just as “short” of zero as .999… is of 1.0), is allowed to assume the number it’s trying to reach (I believe that’s what is called its ‘limit’).

I personally don’t think .000…1 equals zero but I’m not aware of a proof for that.

I use that to describe the ultimate infinitesimal because .000…0 would for sure be zero.

Thanks!

Yeah sure! Here you go:
https://youtu.be/w-I6XTVZXww