# 1=0.999... infinities and box of chocolates..Phliosophy of Math...

I think my Kindergarten teacher's words were something like, "Three take away three equals zero."

It's all in the context one uses '0', that's why it has caused so much aggro and 'undefined' stuff when used without really understanding what it means in the context under study.

For example, it is a 'balanced state' when two opposing forces are in equilibrium but are still active.

Or, for another example as in your context above, one can call the initial state a 'neutral 0' state where no apples were present. Then you introduced 3 apples and that was an UNbalanced state (+3). Then you took away 3 apples and returned the state to 'neutral 0'.

Note well the REALITY therein: AT NO STAGE (except philosophically in your imagination) was there EVER any NEGATIVE PHYSICAL/MATHEMATICAL/LOGICAL apples (-3) state existing BEFORE you put the +3 apples on the table! And then the -3 state was only the REVERSAL ACTION (removal) of the original +3 apples that existed on the table you put them on.

You get the idea. And the maths/logics involving 'zero' are infested with cross-contextual and purely unreal/philosophical notions/constructs which CONFUSE the whole maths/number system/operations and causing 'undetermined' undefined etc 'outputs which are of NO USE TO ANYBODY and actually are damaging and dooming mathematics as currently axiomatically 'constructed/conducted' to perpetual incompleteness. For these and all the other reasons I have pointed out over the years and recently.

So go carefully, irrespective on which 'side' you think you are on. The maths will confound you on BOTH 'sides' unless it's cleared up from scratch according to reality CONTEXT as I am doing. Good luck and take care, MD, everyone.

PS: MD, it's 11:07 PM here, so G'night!

I don't think Hansda understood this argument, because there wasn't an question or disagreement with a particular part of the argument, but a denial of the conclusion. That's no basis for a meeting of the minds.
Having established that no number less than 1 could be an upper bound of the family $$S_n$$ and having established that the family $$S_n$$ is strictly increasing with each successive value larger than all those previous, I came to the conclusion that the only possible value of $$\lim_{n \to \infty} S_n$$ is 1.
Identically, if I were to establish that no number greater than 0 (no positive number) could be a lower bound of the family $$10^{-n}$$ and also establish that the family $$10^{-n}$$ is strictly decreasing, then I would come to the conclusion that the only possible value of $$\lim_{n \to \infty} 10^{-n}$$ is 0.

Just as I considered $$\lim_{n \to \infty} 10^{-n}$$ briefly in the first post, so did I consider the lower bound of the family $$T_n$$ and the value of $$\lim_{n \to \infty} T_n$$ in the second post.

Correct! In fact I said this myself before.
I never said there was a value of n that made $$T_n$$ equal to zero. I already said there is no value of $$S_n$$ that equals 1 -- but that's not surprising because there also is no member of $$S_n$$ that equals $$S$$ and the value of $$S$$ is the point of this whole thread. All members of the family $$T_n$$ are positive. But if you name any positive number, no matter how small, then most of the family $$T_n$$ are going to smaller than that positive number. So no positive number can be a lower bound on the family of $$T_n$$. Thus 0 is the greatest possible lower bound on the family of numbers $$T_n$$ and no matter how small any particular $$T_n$$ might be all further values ( $$T_{n+1}, T_{n+2}, T_{n+3}, \dots$$ ) must be smaller still. Thus 0 is also the value that the sequence $$T_n$$ approaches for arbitrarily large values of n. And the way you write this is $$\lim_{n \to \infty} T_n = 0$$.
I believe all that you have done was to demonstrate that you didn't understand my last two posts on the subject of limits. If you understood the argument then you would either be able to tell me what the specific flaw is or have a specific question about some part of the argument you didn't understand. Instead you deny the consequence of the argument without specific objections. Instead of objecting to parts of my argument you simply rephrase points I had already made.

Let me summarize my understanding of your posts on this issue.

Essentially you are trying to tell that in the case: $$n \to \infty$$;

$$S_n = S$$ and $$T_n = T$$ where $$S = 1 = 0.999...$$ and $$T \ne 0$$.

Am i right?

IF i am not right, let me know:

as $$n \to \infty$$ ;

$$S_n = ?$$ and $$T_n = ?$$.

OR, is it that:

as $$n \to \infty$$;

$$S_n = S = 1 = 0.999...$$ and $$T_n = T = 0$$.

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Unless your ‘unreal maths’ PROOFS covers ALL possible instances/formulations of the ‘problem/concept’, then it is NO PROOF of anything except ‘special unreal cases’ which are self-referenctial and trivial therefore, because they only ‘satisfy’ the unreal boundary conditions YOU put by trying to EXCLUDE those REAL PHYSICAL cases where your ‘unreal’ maths ‘proof’ OBVIOUSLY BREAKS DOWN.

So, in reality, the ‘proof/exclusion’ shoe is on the other foot than the one you tried to convince MD and others here it was on. Yes?
Okay, so your issue is that the mathematical concept of infinity does not exist in reality? That it is not applicable in reality?
Otherwise, what issue do you have with a mathematical proof of a mathematical concept?

I thought this thread was with regard the mathematical principle that 0.999... = 1 ?
Clearly you are arguing for the "triviality" of the matter due to it not being physically possible to divide something an infinite number of times.
Mathematical proofs relate to mathematical principles.
To argue that they are trivial or invalid ("no proof of anything") due to inapplicability in the real world is not an issue with the maths or the principle, but an issue with the use you are putting the principle to.

Mathematical proofs relate to mathematical principles.

Good. The very next time I see math in a physics question, I am going to blow a head gasket. Don't you dare use math in physics if it's only proof of math. Now, the question is, when are you going to invent a system to calculate physical realities that relate to the real world? You know, I had 10 jelly beans and I ate .999.... of them, so I didn't have any left.

Total, complete, utter, NON-SENSE!

I repeat, if your math is self reliant, keep it out of physics, because it's BS in the real world, and physics is about real physical phenomena in the real world, not some imaginary mathematical illusion that has its own rules different from the real world.

Don't you dare try to use math in physics again, else you are a hypocrite!

Can the universe get any more ridiculous? Oh yes.

One little point about physics and mathematics. In physics, you observe something and explain it; the explanation is a theory that predicts future observations. You can't prove that a theory is true, but you can test that its predictions are 'correct'.

In mathematics, there are no observations. If you want to prove something you can only do so logically. Mathematical proofs don't require observations to support them, because something that is logically true is always true.

And since you use numbers to count real physical objects, this proves that numbers can always be used to count such objects. The numbers are not physical but are logically "real".
When you count camels, how do you keep mathematics out of the physics? That's a patently ridiculous notion.

Oh yeah, you can just count numbers, so counting doesn't require that you are counting anything physical (like camels). How can we possibly sort this "big problem"? What can we do?

/sarcasm

Good. The very next time I see math in a physics question, I am going to blow a head gasket.
I do so hope that I am fault here for not realising this is a complete wind-up?

Don't you dare use math in physics if it's only proof of math.
It is not only a proof of math - it can be a proof of physics if the mathematical principle is applicable.
For example, if I want to divide an atom into an infinite number of pieces, mathematically I can model it.
Physically I can't carry it out.
The maths is sound, but the applicability is not there due to the limitations that physical reality forces upon us.
Now, the question is, when are you going to invent a system to calculate physical realities that relate to the real world? You know, I had 10 jelly beans and I ate .999.... of them, so I didn't have any left.
Actually, you'd have 9 left.
But be that as it may, what you are looking for is an accurate model of physical realities to which the maths is then applied.
Maths in and of itself is abstract.
Total, complete, utter, NON-SENSE!
That is how I am beginning to find your posts - but then I'm probably unaware that this is just a wind-up, right?
I repeat, if your math is self reliant, keep it out of physics, because it's BS in the real world, and physics is about real physical phenomena in the real world, not some imaginary mathematical illusion that has its own rules different from the real world.
The physical world adheres to maths on the whole, but you can create abstract notions in maths that do not exist in the physical world.
Infinities are thought to be one, as I understand it.
Don't you dare try to use math in physics again, else you are a hypocrite!
Why would I be a hypocrite for using maths in physics?
Oh, I see.
This is a complete wind-up, right?
I'm on Candid Camera or some such, right?

Let me summarize my understanding of your posts on this issue.
You can't because you don't understand the notation you are manipulating and so you convey only misunderstandings.

$$S_n$$ refers to an entire family of numbers. $$S_n$$ doesn't have any specific value until n is specified.
For every counting number n, all of the following are true $$S_n = 1 - 10^{-n}, \; S_n \lt S_{n+1}, \; S_n \lt 1, \; S_n \lt S$$
Thus for every number X which is less than 1, it follows that most of the family $$S_n$$ are larger than X. We can be specific and write $$m(X) = \frac{- \ln ( 1 - X ) }{\ln 10}$$ and prove that for every n greater than m(X), $$X \lt S_n \lt 1$$ is true.
Proof:
$$\begin{eqnarray} m(X) & \lt & n & \lt & n+1 \\ \frac{- \ln ( 1 - X ) }{\ln 10} & \lt & n & \lt & n+1 \\ - \log_{10} ( 1 - X ) & \lt & n & \lt & n+1 \\ \log_{10} ( 1 - X ) & \gt & -n & \gt & -(n+1) \\ 1 - X & \gt & 10^{-n} & \gt & 10^{-(n+1)} & \gt & 0 \\ X - 1 & \lt & -\left(10^{-n}\right) & \lt & -\left(10^{-(n+1)}\right) & \lt & 0 \\ X & \lt & 1- 10^{-n} & \lt & 1 - 10^{-(n+1)} & \lt & 1\\ X & \lt & S_n & \lt & S_{n+1} & \lt & 1 \end{eqnarray}$$​
The number 1 is special in the function m() in this proof because if it were any larger number then the proof would be trivial and if it were slightly less than 1, the proof would fail. Thus 1 is the least possible upper bound to the entire family $$S_n$$.
Thus no matter how close any particular member of the family $$S_n$$ is to 1, all of the higher-numbered members of the family (most of them) are closer to 1. That's the meaning of $$\lim_{n \to \infty} S_n =1$$ for a family indexed by counting numbers.
http://en.wikipedia.org/wiki/Limit_of_a_sequence
For the same reasons, $$\lim_{n \to \infty} S_n = S$$ and since equals to one thing must be equal to each other. $$S = 1$$.

Likewise for the family $$T_n$$:
$$T_n = 9 \times 10^{-n}, \; T_n \gt T_{n+1}, \; T_n \gt 0, \; T_n \gt \lim_{k \to \infty} T_k$$, For any X greater than 0, for all values of n greater than $$-\log_10 \frac{X}{9}$$, $$0 \lt T_n \lt X$$ and $$\lim_{n \to \infty} T_n = 0$$​

Essentially you are trying to tell that in the case: $$n \to \infty$$;
That's not a case, that's a whole family of cases that says "as n takes on ever increasing successive counting numbers". The full expression for $$\lim_{n \to \infty} S_n = S$$ reads as "The unique limiting value approached by the family $$S_n$$ as n takes on ever increasing successive counting numbers is 1."

There is no case where there is a value of n (which is limited to counting numbers) such that $$S_n = 1$$, but because all such n are finite, neither is there a value of n such that $$S_n = S$$.

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I believe we're looking at another example of limit points of sets, the limit point of the family $$S_n$$ is 1, and there is no n such that $$S_n = 1$$, so 1 is not in the union of the family of sets.

Which accords with this from WIkipedia:

In mathematics, a limit point of a set S in a topological space X is a point x (which is in X, but not necessarily in S) that can be "approximated" by points of S in the sense that every neighbourhood of x with respect to the topology on X also contains a point of S other than x itself.
And it says $$S_n$$ is not closed (since it doesn't contain 1):
Note that x does not have to be an element of S. This concept profitably generalizes the notion of a limit and is the underpinning of concepts such as closed set and topological closure. Indeed, a set is closed if and only if it contains all of its limit points, and the topological closure operation can be thought of as an operation that enriches a set by adding its limit points.

All limits of sequences are necessarily limit points of the set of all values of the sequence.
But the reverse is not true. Consider $$A_n = \frac{(-1)^n n}{2 n + 1}$$ which goes like: $$- \frac{1}{3}, \frac{2}{5}, -\frac{3}{7}, \frac{4}{9}, \dots$$.
This family $$A_n$$ has two limit points of $$-\frac{1}{2}$$ and $$+\frac{1}{2}$$ but the series has no limit, because a limit is necessarily a unique value.

@Bds,
My question:
Do you think that in mathematics, 0.999... = 1 means that the 0.999... sequence terminates at 1
no I keep looking at 0.999... and thinking something is missing to make it equal 1
This just beggars the question:

What is the number 1?
From what I understand the number 1 is the sum of it's infinite constituents.
so therefore the sum of 1 = 0.999.... [and all 0.999... means]
(where the value 1 becomes the upper boundary of 0.999... perhaps )

*edit: I think this was cleared up later in the thread [still reading]

What is the number 1?
Depends on which number system one is describing.

So 1 in the ordinals is the "next" number after zero, while 1 in the cardinals is the property of being capable of being paired up with any bag that holds 1 item. For finite ordinals and finite cardinals these definitions are equivalent to the point that the natural numbers and counting numbers are kind of abstract. Positive Rational numbers are more abstract in that the rational number are effectively ratios of natural numbers in lowest terms, so that 1 is the ratio of natural numbers 1 and 1. Positive reals are defined as Dedekind cuts in the ordered world of rationals, so correspond closest with the geometric idea of cutting a segment of a certain length from a line. Specifically you can cut the line not just at the rational places but also at places like $$\sqrt{2}$$ that fall between the rationals on either side. Zero and negative numbers are re-invented by the invention of an equivalence class relating pairs of positive quantities, (a,b). Here $$(a_1, b_1) = (a_2, b_2) \; \textrm{iff} \; a_1 + b_2 = a_2 + b_1$$, and zero re-invented this way naturally has all the properties of zero in the natural numbers. Complex numbers are created by pairing up numbers again and having special addition and subtraction rules.

So 1 has many definitions but a few distinct properties.
Examples: $$1 \times x = x \times 1 = x / 1 = x$$ for all x.

@rpenner,
Would you disagree with the following then? [in the context of this thread]
From what I understand the number 1 is the sum of it's infinite constituents.
so therefore the sum of 1 = 0.999.... [and all 0.999... means]
(where the value 1 becomes the upper boundary of 0.999... perhaps )
btw, thanks for your post.. most interesting!

No -- 1 cannot be the defined sum of it's infinite constituents because there are an infinite number of disjoint infinite series that sum to 1. How would you pick between the alternatives?

$$1 = \sum_{k\geq 1} \frac{1}{2^n} = \sum_{k\geq 1} \frac{2}{3^n} = \sum_{k\geq 1} \frac{3}{4^n} = \sum_{k\geq 1} \frac{4}{5^n} = \dots = \sum_{k\geq 1} \frac{9}{10^n} = \dots$$

No -- 1 cannot be the defined sum of it's infinite constituents because there are an infinite number of disjoint infinite series that sum to 1. How would you pick between the alternatives?

You would have 4 quarters in your hand, and you would say, hmmm, how much does 1/4+1/4+1/4+1/4 equal? Then you'd whip out your calculator, and you'd say, oh, heck, I don't need the calculator for this, it's simple, 1/4+1/4+1/4+1/4=.999...=1 (rolls eyes)

0.25
0.25
0.25
0.25

0.999...

4 .2's is .9
4 .05's is .09
and then to make up the difference, you slip in an extra spoon full of .00999... just to seal the deal.

BTW rpenner, if zero is the first number, then 5 is the 6th number, right?

Now I'm getting confused. If 0.999...=1, then does it even mean anything to say, "first number?" Shouldn't we call it the 0.999...st number or the zeroest number?? Since the numbers are offset by 1, then 2 shall be the number! and 2 it shall be. Not 3, not 1. 4 is totally out of the question. 2 is the number!

You would have 4 quarters in your hand, and you would say, hmmm, how much does 1/4+1/4+1/4+1/4 equal? Then you'd whip out your calculator, and you'd say, oh, heck, I don't need the calculator for this, it's simple, 1/4+1/4+1/4+1/4=.999...=1 (rolls eyes)
hang on that's:
(0.999...)/4 + (0.999...)/4 + (0.999...)/4 + (0.999...)/4 + = 1(0.999...)
remember 1 always = 0.999...

hang on that's:
(0.999...)/4 + (0.999...)/4 + (0.999...)/4 + (0.999...)/4 + = 1(0.999...)
remember 1 always = 0.999...

Oh, so it should be

0.999.../4 + 0.999.../4 + 0.999.../4 + 0.999.../4 =1

Read, "This is how to add 4 quarters (.25 cent pieces) together in math."

Obviously, everyone knows that 4 quarters equals 1 dollar.

All for .999... and .999... for all.

Oh, so it should be

0.999.../4 + 0.999.../4 + 0.999.../4 + 0.999.../4 =1

Read, "This is how to add 4 quarters (.25 cent pieces) together in math."

Obviously, everyone knows that 4 quarters equals 1 dollar.
well there you have it, the number one is simply a shorthand equivalent of 0.999... a mere convenience really...
Saves a lot of time on the key pad..

well there you have it, the number one is simply a shorthand equivalent of 0.999... a mere convenience really...
Saves a lot of time on the key pad..

Plus, that little piece that used to be swept under the rug just turned into infinity divided by 4, which is approximately equal to 1/4. Never mind the man behind that curtain.

Here's something about sequences, series and limits from someone whose native language is Greek:
Nick Karagiaouroglou said:
Let's bring in our minds some very elementary things, that will help us understand what a condensation point is. First we need to remember what a point set is.

For the time being it suffices to accept that such a set is given, when we can prove for any point of some space, if this point is member of the set or not. Notice that "space" and point" both don't need to be the space in which we live and its points.

We could take as space for example all integers, and prove if some given integer (the point) is member of some given set or not. Such a set, for which we can prove what are its members, is bounded, when a number exists that is bigger than the distance between any two points of the set. Often we measure the distance of such points by means of their coordinates.

It suffices to consider such point sets on a straight line, on which we bring the coordinates of points. We can use more than one coordinates if necessary, that is when our set has more than one dimensions. For some one-dimensional set, if we denote the coordinates of two points with x[sub] i[/sub] and x[sub] k[/sub] , then the distance of the two points is given by | x[sub] i[/sub] − x[sub] k[/sub] |.
I think it's a lot more understandable than the contributions from some of the posters here.

An attempt to elaborate: as rpenner pointed out, a limit is a limit point, but a limit point is not necessarily a limit. For instance in rpenner's example of the space $$S_n$$, n is greater than or equal to 1, so n is always greater than 0. However $$S_0$$ is not defined.

That example is only one way to 'decompose' the unit interval into an infinite sum of real terms, and there is no smallest distance from zero in the reals, so zero has a neighbourhood containing an infinite set of points, and zero is a condensation point of the open interval (0,1).

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