Sythe Users: What Do They Know? Do They Know Things?? Lets Find Out! #7

I used to think I was intelligent until I read your posts. Holy cow.

 

Don't confuse intelligence with malice.

 

I haven't.

 

Answer is 9+10 .

Or that 1/9 = 0 .11111111111...
and 2/9 = 0.2222222222222222...
so 9/9 = 0.9999999999..
But 9/9 = 1 ?
So 0.999999... 1

^ only works if there are infinite amount of 9's after 0 tho

Probably been said before but sorry I didn't want to read everything above, maybe that answer has been given already tho

 

21?

...

 

Don't confuse ostentatiousness for rudeness.

 

I'm reviving this because I want to.

Doing it with 1 = 0.9999... is not an issue because the partial sums converge. The same cannot be said of -1 = 999... (the partial sums diverge therefore the sum has no meaning).

0.999... = 1 exactly. For some intuition -- try to come up with a number between 0.999... and 1. (You won't be able to, while you can for any distinct numbers).

There is no infinitesimal amount between 1 and 0.999... They are the exact same number, just a different representation. Just like 2/4 = 1/2.

That being said, infinity is not in R, you need to take (R U {\inf}) to be able to use it. If you insist on including infinity and infinitesimals though, there is already a system which does just that: the hyperreals. (Conveniently denoted *R rather than R*).

But the way this has been developed doesn't make much sense.
For one, R* isn't well ordered. Consider the set R > 0. This will include epsilon. But also epsilon/2. And epsilon/4, and so on, so it will have no minimum element. Also, according to the definition, epsilon < |x|, but does that say anything about 2 * epsilon? 1000 * epsilon? So to be more rigorous we'd have to include an "\forall a \in R^+" and then multiply the epsilon by this a in the definition. If we do this, though, we are essentially preventing any type of interaction between epsilon-elements and regular elements. So we might as well make it clearer and define it as R^2 - define comparison of (x, y) with (w, v) as comparison by the first elements, then by the second if the first elements are equal. These are equivalent formulations.

However, the issue is that the argument you are making at the end is analogous to (10, 0) - (10, 1) = (0, -1). If we think of this in terms of complex numbers (real, imaginary). This doesn't say anything about the equivalence of the "real" parts, which is what we were trying to show all along.

The issue arises in your LHS = RHS side. There are no epsilon terms on the right hand side so a = 1, b = 0.


As for your second argument, there are a bunch of issues, but the easiest to address is that there is still no "smallest number you can expression that isn't zero", as the set isn't well ordered.