Is 0.9 recurring equal to 1?

[Zarel]

Like wave and particle?

I said "not sure" and "seem". I didn't dismiss the possibility entirely.

[Tonepoet]

Don't put words in my mouth, I never said anything about 0.999..., my ramblings were an answer to a totally different question.

Does he have to? It's a very easy conclusion to infer, since it is inherent to the nature of the given answers answers given that 0 is the only infinitely small value. Otherwise the orientation of .9~ changes because the difference of 0 wouldn't be the infinity a sequence of .9~ approaches but rather ε.

The only way we could avoid this is if the value isn't sequentially oriented but even then we have the problem of actually placing .9~ in the sequence, as we have very few ways to determine if anything comes after it to actually place it in the sequence, since it comes before itself and it comes after itself giving it infinitesimally cardinal value. We just can't because infinity itself is a constant. The concept of placing .9~ seems contradictory to the nature of infinity just as much as whatever 1-ε is, or 1+ε for that matter. It has multiplicity of self-referring values that never stops.

Therefore I'd suggest:

If ε, then .9~ < 1.

I'd also suggest that:

Numbers are representations known as quantifiers.
Quantifiers quantify.
Therefore anything with a number is quantified.

Numbers can be used to represent points.
If one number represents a point, all numbers must represent a point in relation.
Points are infinitesimally small by definition.

Therefore infinitesimal smallness and quantification are not mutually contradictory.

It also shows a very basic fundamental application of concept ε may already be used in standard mathematics, albeit perhaps not directly in the numbering system. [*coughs* Although I'd suggest a name change for the decimal point in that case.]

Also just today I was reading a very interesting little article about the developmental history of math, the concept of a number zero and how that changed the conventional landscape of math. It was very interesting, however I don't want to link that here due to the potential touchiness of some of the subject matter it contains. It gave a very brief description of the Neumann Hierarchy, which I thought seemed particularly relevant, especially since we're now talking about alternative numbering systems. Would anybody care to go into greater detail as to what that is?

[Cracky6711]

so 0.999 + 0.001 = 1 right? and 0.99999999 + 0.00000001 = 1 and 0.99999999999999999 + 0.00000000000000001=1 and i could go on infiniteley until 0.00000000000000000000000000000001 has so many zeros before the 1 it might as well be 0. but however many 9s you put after 0.999999 there is still going to be an answer which is along the lines of 0.000001, think about it. Think of it as you have infinite number of 9s, but then you put an extra 9 on, so you have to put another 0 on, and this becomes infinitesimally small.

The problem we have is that infinity in itself is an imaginary number, just as is every single recurring number. think about it, infinity is impossible in itself. Also, it is impossible to type on a calculator 0.0r1, because wherever you put the 1, you can always put an extra 0 before it.

Its impossible for infinite 0.9 to ever equal 1, because there will always be a difference of 0.0r with a 1 on the end.
In my opinion, 0.9r is equal to 0.0r with a 1 at the end.

[Vranca]

You just necroed this thred,the last post was 09 Jul 2010...

In my opinion, 0.9r is equal to 0.0r with a 1 at the end.

But if it is requiring it does not end.
form wiki

[Cracky6711]

You just necroed this thred,the last post was 09 Jul 2010...

In my opinion, 0.9r is equal to 0.0r with a 1 at the end.

But if it is requiring it does not end.
form wiki

ok i misfrased that...basically what i mean is that 1 - 0.9r = an infinitesimal value (the smallest possible positive value above 0)

[Joram]

Its impossible for infinite 0.9 to ever equal 1, because there will always be a difference of 0.0r with a 1 on the end.

Only if the number of 9's is finite. But since 0.999... goes on forever, there will never be a 1 to add.

Think about it. No matter how far back you put the 1, there will be 9's beyond it. Therefore, the supposed number that added to 0.999... equals 1 does not exist; not even as an infinitesimal value.

[Iris]

You just necroed this thred,the last post was 09 Jul 2010...

Emphasis mine.

Locked.