[split] 0.999... equals 1

[IATIA]

Any number raised to the power of infinity is undefined. 0<x<1 raised to the power of infinity will converge and the limit is zero, but the number is not zero. Infinity + 1 or +2 or + any number still equals infinity. Any formula based on infinity deals with undefined terms, only the limit is valid, but the result is just that, a limit. The proof for the geometric series is a limit based on this undefined number, so the geometric series formula is technically invalid. And again, it is acceptable at this time until we come up with a better way to handle infinities.

Infinity+1=infinity subtract infinity from both sides 1=0 This works with any equation in one form or another, ergo, undefined.

[LastPoet]

You don't get it do you? Let me put it in an informal way, its the numbers Decimal Representation that is infinite, not the number itself. The number is by definition finite. You are making such a mess in your head about this, mixing completely unrelated stuff.

Adrian's proof its a series, or a sum, not a sequence. 1 is not infinite, it has two decimal representations 1.000.... and 0.999..... both infinite, but the Real numbers themselves are the same and finite.

Tell me, you know what a dense order is? Its a property of the Real Set and states: for all a,b that belong to the set, where a < b, there exists a number c so that a < c < b. Make a = 0.999... and b = 1, now, if you're right about 0.999... =/= 1 surely you have the conditions met a < b, now, find me a number c so that 0.999... < c < 1.

[Tiberius]

Any number raised to the power of infinity is undefined.

True.

0<x<1 raised to the power of infinity will converge and the limit is zero, but the number is not zero.

False. As previously stated, due to the completeness property of all real numbers, each "limit" is itself a real number. The limit of the geometric sequence 0.9 + 0.92 + 0.93 + ... is 1, and hence the sequence (representing the number 0.999...) is also equal to 1.

See: http://en.wikipedia.org/wiki/Real_number#Completeness

Infinity + 1 or +2 or + any number still equals infinity.

No it doesn't. Infinity is not a number that can be used in arithmetic calculations.

Any formula based on infinity deals with undefined terms, only the limit is valid, but the result is just that, a limit.

Unless that limit is dealing with a geometric sequence determined from a real number, in which case the limit is also a real number.

The proof for the geometric series is a limit based on this undefined number, so the geometric series formula is technically invalid. And again, it is acceptable at this time until we come up with a better way to handle infinities.

No. You can hold that the geometric series formula is "invalid" if you want, but mathematicians do not. They hold it as perfectly valid, and there are various proofs of its validity. It is not just something we "accept" until we come up with a better way to handle things. Mathematics is not science; it doesn't deal with empirical methods, but rather ones of logic. How mathematics works is how our logic determines it to work, and how we've designed it to work. The quirks we get out of mathematics aren't going to change.

Infinity+1=infinity subtract infinity from both sides 1=0 This works with any equation in one form or another, ergo, undefined.

Again, you cannot use infinity in arithmetic calculations; it is undefined and so your calculation above is wrong.

[IATIA]

As I stated earlier, because the 'space' between 1.000... and 0.999... is indivisible, they are accepted to be the same number.

However, the proofs used to establish it's value all relate to limits in one form or another. I have yet to see a proof that a limit is a real value. A limit is only accepted as such.



the f(x) becomes discontinuous (undefined) at x=1

f(0.999...)=1.999...

[Tiberius]

Again, read up on the completeness property of real numbers. I've posted the link twice now I believe.

[IATIA]

I have and it still does not alter the fact 0.999... is not discontinuous and 1.000... is discontinuous at f(x).

Obviously this going to go nowhere, so I acquiesce, go ahead and have the last word.

[Tiberius]

Technically, both are not discontinuous. 1 is the simplified form of 1.000..., and there is no difference at all between an infinite number of zeroes and an infinite number of 9s (in terms of the way they are treated).

Both 1.000... and 0.999... are real (and rational) numbers, and they must abide by the rules of rational / real numbers. That includes being complete, which means that 0.999... is not simply a limit, but at actual value, equal to the value of 1.000...

[LastPoet]

I have and it still does not alter the fact 0.999... is not discontinuous and 1.000... is discontinuous at f(x).

Obviously this going to go nowhere, so I acquiesce, go ahead and have the last word.

Don't get me wrong IATIA, but you sound like a 'math creationist', we have bombarded you with proof that 0.999...=1, direct proofs, both algebraic and analitic, I even handed to you in a plate the proof by Reductio in my last post and explained why this is, but still you have offered no proof sustaining your premise, and mixing stuff that doesn't have anything to do with the problem at hand, because it does not 'feel' right to you. Its a very common misundertanding the one you're having about this.

Either way, I'd advise you to get some books on analisis and algebra, university level and read them at the same time you do the math in a notepad.

I may not be a mathematical master, or have a degree on it, but this is a wrater easy problem in math. In my Analisis book, this problem appears right in the first chapter, and believe me, it will only get harder from there on.

[edk]

IATIA, if you're still in this thread, can you come up with any way of obtaining the number 0.999...?

[Modular Moog V]

IATIA, if you're still in this thread, can you come up with any way of obtaining the number 0.999...?

I can tell you it won't work in current computer modeling. It needs constructionist models. Infinity will make it crash