# 1 is 0.9999999999999............

Give it a break SG. Take RP's post before yours, as an example. Do you really expect 99% of humanity to have the foggiest idea what all the symbols and numbers mean much less verify that they are valid, and not just mathematically illusionary, mental masturbation?

Well that's bad logic. 99% of average people probably can't tell you the underlying principle that causes a lightbulb to go on when you flip the room switch. If you asked the average person, how many would say, "Closing the switch completes a circuit between the lightbulb and the power company?"

That doesn't mean electrical engineering is false. It just means that 99% of the people don't understand the technical details of the world around them. And why should they? I like to drive my car but I'd be hard-pressed to explain the workings of a modern computer-controlled fuel injection system.

1) give us a rational, logical, common sense and relatively simple explanatory guide, as to how/why finite 1.0 = infinite 0.999...,

You can't find a number that's between .999... and 1.

"Closing the switch completes a circuit between the lightbulb and the power company?"
A pedant would point out that the physical circuit that closes traces to the EMF induced in the secondary windings of the local transformer, which may very well be owned by the power company.

But then pedants are horrible, horrible people and none survive to be pointed at as examples on the Internet.

You can't find a number that's between .999... and 1.

Isn't that the same as saying x=y because there is no letter between x & y?

that 0.999...=1 because there is no number between them?

Isn't that the same as saying x=y because there is no letter between x & y?

So you reckon that if there is no world between yours and mine they would be equal worlds?

hee hee.. boy are you in trouble...!

We are still waiting for your erudite explanation as to why your earlier proof quoted from wiki falls over on it's face..
but that may take more cerebral effort than you are prepared to put in hey? :sleep:

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Isn't that the same as saying x=y because there is no letter between x & y?

Mr. Q you made me LOL.

Mr. Q you made me LOL.
hee hee, well I was just thinking about the way limits are used when they say that :
0.999... = 1 because the reducing value of magnitude between the digits is less than any real number...
Could be interpreted as to means exactly the same thing that x = y because there is no real letter between them...
Which I find rather amusing for some strange reason....

There is no property of the letters of the alphabet (or the integers) that always either halfway any distinct two of them is a third distinct value or somewhere between any distinct two of them is a distinct third value. Some (simple) codes even equate integers and the alphabet in their natural orders so that c is naturally halfway between a and e, but there is no letter anywhere between x and y, just as there is no integer between 24 and 25.

Rationals and Reals are dense and its is always true that there are distinct rational or real numbers between any distinct pair. But since $$0.\bar{9}$$ and $$1.\bar{0}$$ don't admit a distinct decimal representation between them. All candidates either start with a zero and are less than or equal to $$0.\bar{9}$$ or start with a number other than zero and are greater than or equal to $$1.\bar{0}$$.

No Banana for Rpenner--Yet--Mapping = Mathematically Illusonary, Masturbation-imho

" so why do you think ignorance of the real numbers is making your viewpoint attractive in public discourse?"

Huh? RP, I'm not here at RP to appear attractive. I'm here to seek truth and especially cosmic truths, and a Theory of Everthing that connects them all. I have an ego and surely do want what I have discovered to be recognized and acknowledge. However, so does every other poster here.

Because I believe I'm, correct in my assessment of the crux/connundrum of finite 1.0 NOT equal to infinite 0.999...and you and those who believe a finite value is equal to infinite value, are incorrect. It is that simple irrespective of you continual offer very complex mathematics to myself-- if not some others ---who cannot even begin to understand much less verify.

You have used the word "mapping" in several of your posts and I originally went to a given wiki page that explained this "mapping", and what I could glean from it, and I repeat here again, is take finite or infinite value will call A in column Y and place it in Column Z and then call it D.

I call this mathematically illusionary,mental masturbation, that in no way addresses my comments, as stated i.e. give us a rational, logical, common sense and relatively simple explanatory guide, as to how/why, the finite value 1.0 can be equal to an the infinite value 0.999....?

How many posts have gone under the water, with nobody, NOT one individual, offer such explanation as I challenge you and others to do.

That no one does, and usually gives very complex mathematics and links or offer simple formulas that really do not prove the crux/connundrum of finite value vs infinite value, or just continue their head slapping of r6.

Sorry RP, no banana for you on the specific issue of this topic/thread and as always, i will not hold my breath waiting for to address my challenge to you and others as stated.

r6

Hi Billy T.

...

Likewise 1.23456789 can be made ten times larger by move of decimal point one space to the right to get: 12.3456789 or a ten thousand (which is 10^4) times larger by move of decimal point 4 spaces to the right. I. e. 12345.6789 is ten thousand times larger than 1.23456789 is. This is NOT an operation of multiplication,...

...

Thanks for that confirmation of what I tried to tell certain people earlier on, mate. Namely, that 'moving the decimal point' is not a mathematical operation, but rather a mere formatting of the results of a mathematical operation. In other words, logical mathematical treatment is more fundamentally based than formatting convention 'treatment'. Cheers, Billy T, everyone.

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I don't need induction because I am not proving a novel statement for all n, but rather for one specific n. You already agree with me that the partial sums are of the form $$1 - \frac{1}{10^k}$$ so I don't need to prove that anew with induction.
You aren't even using the correct domain of mathematics. This thread is about the real numbers and analysis.
$$y = 10^x$$ is continuous and one-to-one mapping all real numbers to positive real numbers, therefore $$x = \log_{\tiny 10} y$$ is continuous and maps all positive numbers to the real numbers. This is very basic analysis.

Therefore if $$0 \lt 0.999... \lt 1$$ then $$1 - 0.999...$$ is positive and therefore $$\log_{\tiny 10} \left( 1 - 0.999... \right)$$ is a real number less than 0, therefore $$\left\lfloor \log_{\tiny 10} \left( 1 - 0.999... \right) \right\rfloor$$ is an negative integer and $$1 - \left\lfloor \log_{\tiny 10} \left( 1 - 0.999... \right) \right\rfloor$$ is a positive integer, and therefore a natural number. We call this natural number n.

Now because we assumed $$0.999... \lt 1$$, we know n exists and that $$0.999... \lt 1 - \frac{1}{10^n}$$ but because $$1 - \frac{1}{10^n} \in \left{ 0, \, \frac{1}{10}, \, \frac{1}{10^2}, \, \frac{1}{10^3}, \, \dots \right}$$ then and $$0.999... \; = \; \textrm{sup} \, \left{ 0, \, \frac{1}{10}, \, \frac{1}{10^2}, \, \frac{1}{10^3}, \, \dots \right}$$ it follows that necessarily, $$1 - \frac{1}{10^n} \leq 0.999...$$ and from the transitive axiom of ordering, it follows that $$0.999... \lt 0.999...$$ which contradicts $$0.999... = 0.999...$$ and so we reject the hypothesis $$0.999... \lt 1$$.

At no point did I mean to write $$\color{red} \textrm{sup} \, \left{ 0, \, \frac{1}{10}, \, \frac{1}{10^2}, \, \frac{1}{10^3}, \, \dots \right} \lt 0.999...$$ because that contradicts what I am asserting $$0.999... = \textrm{sup} \, \left{ 0, \, \frac{1}{10}, \, \frac{1}{10^2}, \, \frac{1}{10^3}, \, \dots \right}$$ which in layman's terms is 0.999... is the smallest number at least as big as any number of the form $$1 - \frac{1}{10^k}$$ where k is any natural number. I see I did do it as a typo,

Obviously, $$1 - \frac{1}{10^{n-3}} \lt 0.999... \lt 1 - \frac{1}{10^n} \lt \textrm{sup} \; \left{ 0, \, 1 - \frac{1}{10}, \, 1 - \frac{1}{10^2}, \, 1 - \frac{1}{10^3}, \dots \right} = 0.999...$$ was meant. But it doesn't matter much since both $$1 - \frac{1}{10^n} \lt \textrm{sup} \; \left{ 0, \, 1 - \frac{1}{10}, \, 1 - \frac{1}{10^2}, \, 1 - \frac{1}{10^3}, \dots \right}$$ and $$1 - \frac{1}{10^n} \lt 0.999...$$ are true for any natural number n, and since by hypothesis $$0.999... \lt 1 - \frac{1}{10^n}$$ then $$0.999... \lt 0.999... \lt 0.999...$$ and it follows that $$0.999... \lt \textrm{sup} \; \left{ 0, \, 1 - \frac{1}{10}, \, 1 - \frac{1}{10^2}, \, 1 - \frac{1}{10^3}, \dots \right} \lt 0.999...$$. That one mistaken hypothesis (assuming $$0.999... \lt 1$$) leads to more than one faulty conclusion is called the Principle of Explosion.

Whether or not I was consciously meaning to assert that explicitly, I do not recall. But I didn't call it out in text, and that is a mistake of sorts.

1) I don't need induction because I am not proving a novel statement for all n
What math template are you using that is not based on induction. I mean, I am operating from the mainstream and induction is required to establish any results dealing with finite ordinals. So, give me your justification of some method that proves some statement for all n and does not require induction.

2) You aren't even using the correct domain of mathematics. This thread is about the real numbers and analysis.

The problem with this statement is that proofs are based on infinite summation as you did. You see, you are adding some object for all n. That is based on the natural numbers. Oh, BTW, the result is a real number. The issue is can you perform an infinite summation based on n.

So, what is your justification from the mainstream. I already showed a link that indicates there is no justification for adding an infinite number of items.

Here it is again.

We obviously can't add up an infinite number of terms, but we can add up the first n terms, like this
http://www.math.utah.edu/~carlson/teaching/calculus/series.html

Now, if you provide a link that shows you can add an infinite number of terms:
b) Prove it based on set theory.

3) You then claim your supremum operator is justified because you say without proof it is equal to .999..., whatever that means.
However, you still are comparing on the left side a collection of terms based on n. You did not use induction. What is your template available under logic that allows you to do this? You have been asked this before. Are you ever going to answer?

4) Your assertion above only has a template in induction, which is usually called the pinching theorem. You continue to refuse to justify your proof methodology since you setup the problem in a semi-pinching theorem template and then violated it. You have yet to show your justification.

There is in analysis, because analysis deals with the real numbers, not just the natural, rational, or algebraic numbers.

Here's just one textbook on the subject: http://ramanujan.math.trinity.edu/wtrench/texts/TRENCH_REAL_ANALYSIS.PDF (This was Google's first hit for "analysis textbook" but you are welcome to try any others.)

It begins on page 200 to discuss infinite summation of a sequence (a set indexed by a mapping from the natural numbers onto that set).
Theorem 4.3.8 on page 205 reads:
with proof following on the next page.

But as long as you are looking at this textbook you might as well look at Theorem 1.1.3 and Theorem 1.1.6 on pages 4 and 6 which says if $$0.999... < 1$$ then there must be a real number equal to 1 - 0.999... and there must likewise be a rational number between 0.999... and 1. The fact that no such numbers have ever been named seems to indicate a problem with the hypothesis $$0.999... < 1$$.

I am looking at page 201 of your link and I am glad the mainstream has not contradicted itself on an infinite series.

The ﬁnite sum An is the nth partial sum of

Note how this author is defining the infinite series as I said, a sequence of finite partial sums.

But as long as you are looking at this textbook you might as well look at Theorem 1.1.3 and Theorem 1.1.6 on pages 4 and 6 which says if $$0.999... < 1$$ then there must be a real number equal to 1 - 0.999... and there must likewise be a rational number between 0.999... and 1. The fact that no such numbers have ever been named seems to indicate a problem with the hypothesis $$0.999... < 1$$.

I am not sure you understand what these proofs mean.

1.1.6 says between any 2 reals I can find a rational.
In particular, for this case, between .9999(n) and 1, I can find for example, a rational .9999(n+1).

Since this is true for all n, this indicates the sequence of partial sums .9, .99, .999, .... cannot ever reach 1 because I can always find a rational between the nth partial sum and 1.

In order for .999... to emerge into 1, this theorem would need to be violated.

Thanks for the post!

Hi Trippy, everyone.

chinglu said:
You are right, but what fraction is .999999.... ?

9/9

That is 1.

Where is the .999....?

9/9

While I do appreciate, Trippy, your succinctness (and your humour) of those "9/9" responses to chinglu, I would again point out my earlier generally directed observations regarding starting an argument from a facile 'construction'; or even a UNITARY state and not from a fractional state. Because naturally if one starts from 1, all arguments will lead back to 1. Circuitous. Not a way to 'proofs' at all of the fractional case of 'becoming one'. It is a case of now having to 'prove' the definition of .999...=1, rather than just effectively 'restating' it tacitly by using a UNITARY statement like 9/9, or 8/8, -----1/1 (not to mention what is 0/0 = ??? in that same line of UNITARY (or like/like) 'construction'?).

This was not intended to offend anyone. Only putting again my longstanding 'take' so far as it applies to such kinds of responses/assumptions. This was just a general timely observation/reminder, Trippy, everyone, that as far as I can observe the discourse/arguments on these things, it may not be enough to dispel the confusions/dissension with responses involving such facile 'constructions' which may lead to inevitably circuitous, and sometimes very 'peculiar' (as in 0/0 ?) things?

1.1.6 says between any 2 reals I can find a rational.
CORRECT!
In particular, for this case, between .9999(n) and 1, I can find for example, a rational .9999(n+1).
Where 0.9999(n) is non-standard notation for the rational (and therefore real number) $$\sum_{k=1}^{n} \frac{9}{10^k} = 1 - 10^{-n}$$, I agree. CORRECT.
Since this is true for all n,
CLARIFICATION: All natural numbers, n
this indicates the sequence of partial sums .9, .99, .999, .... cannot ever reach 1 because I can always find a rational between the nth partial sum and 1.
CORRECT. For all finite n, there is always a rational number between $$1 - 10^{-n}$$ and $$1$$. In fact there is a rational number exactly half-way between them: $$1 - 5 \times 10^{-n-1}$$. There is also at least one irrational number between them: $$1 - \sqrt{27} \times 10^{-n-1}$$.
In order for .999... to emerge into 1, this theorem would need to be violated.
INCORRECT. Because
• for all natural numbers n, $$1 - 10^{-n} \lt 0.999...$$
• Also for all natural numbers n, $$1 - 10^{-n} \lt 1$$
• thus if $$S = \left{ 1 - 10^{-n} \; | \; n \in \mathbb{N} \right}$$, then both 0.999... and 1 must be upper bounds of the set S, $$0.999... \; \geq \; \textrm{sup} \, S \; \leq \; 1$$ and
• so looking at numbers less than 0.999... cannot by itself settle the question of whether $$0.999... =^? 1$$.
In other words, you have not demonstrated your claim that in order for 0.999... = 1 that any theorem would need to be violated. Because $$0.999... \not\in S$$.

To prove 1 is the least upper bound of set S, it is sufficient to demonstrate that any number less than 1 has a non-empty subset of S which are greater than that. i.e
• $$\forall x \in \mathbb{R} \, \exists n\in \mathbb{N} \, x \lt 1 \rightarrow x \lt 1 - 10^{-n}$$ but if $$x \lt 1$$ means $$\epsilon = 1 - x \gt 0$$ and thus $$\epsilon \in \mathbb{R}^{\tiny +}$$
• $$\forall \epsilon \in \mathbb{R}^{\tiny +} \, \exists n\in \mathbb{N} \, 1 - \epsilon \lt 1 - 10^{-n}$$ which is equivalent to saying
• $$\forall \epsilon \in \mathbb{R}^{\tiny +} \, \exists n\in \mathbb{N} \, n \gt -\log_{\tiny 10} \epsilon$$ and since $$\log_{\tiny 10}$$ is a continuous mapping between $$\mathbb{R}^{\tiny +}$$ and $$\mathbb{R}$$ it is only necessary to show
• $$\forall y \in \mathbb{R} \, \exists n \in \mathbb{N} \, n \gt y$$ which is a super-generic property of the number-line -- there are always larger numbers.
And by using the base-10 logarithm and the floor function you can compute the smallest n that meets that criteria for any alleged number x smaller than 1. But for 0.999... to both be smaller than 1 and larger than any number smaller than 1 means it is larger than itself, a contradiction. And that is how you prove 1 is the only such least upper bound. But then since we both reject $$0.999... \gt 1$$ then and we both reject $$0.999... \in S$$ then it follows $$\textrm{sup} \, S \leq 0.999... \leq 1 = \textrm{sup} \, S$$ and therefore $$0.999... = 1$$.

CORRECT!
Where 0.9999(n) is non-standard notation for the rational (and therefore real number) $$\sum_{k=1}^{n} \frac{9}{10^k} = 1 - 10^{-n}$$, I agree. CORRECT.
CLARIFICATION: All natural numbers, n
CORRECT. For all finite n, there is always a rational number between $$1 - 10^{-n}$$ and $$1$$. In fact there is a rational number exactly half-way between them: $$1 - 5 \times 10^{-n-1}$$. There is also at least one irrational number between them: $$1 - \sqrt{27} \times 10^{-n-1}$$.
INCORRECT. Because
• for all natural numbers n, $$1 - 10^{-n} \lt 0.999...$$
• Also for all natural numbers n, $$1 - 10^{-n} \lt 1$$
• thus if $$S = \left{ 1 - 10^{-n} \; | \; n \in \mathbb{N} \right}$$, then both 0.999... and 1 must be upper bounds of the set S, $$0.999... \; \geq \; \textrm{sup} \, S \; \leq \; 1$$ and
• so looking at numbers less than 0.999... cannot by itself settle the question of whether $$0.999... =^? 1$$.

To prove 1 is the least upper bound of set S, it is sufficient to demonstrate that any number less than 1 has a non-empty subset of S which are greater than that. i.e
• $$\forall x \in \mathbb{R} \, \exists n\in \mathbb{N} \, x \lt 1 \rightarrow x \lt 1 - 10^{-n}$$ but if $$x \lt 1$$ means $$\epsilon = 1 - x \gt 0$$ and thus $$\epsilon \in \mathbb{R}^{\tiny +}$$
• $$\forall \epsilon \in \mathbb{R}^{\tiny +} \, \exists n\in \mathbb{N} \, 1 - \epsilon \lt 1 - 10^{-n}$$ which is equivalent to saying
• $$\forall \epsilon \in \mathbb{R}^{\tiny +} \, \exists n\in \mathbb{N} \, n \gt -\log_{\tiny 10} \epsilon$$ and since $$\log_{\tiny 10}$$ is a continuous mapping between $$\mathbb{R}^{\tiny +}$$ and $$\mathbb{R}$$ it is only necessary to show
• $$\forall y \in \mathbb{R} \, \exists n \in \mathbb{N} \, n \gt y$$ which is a super-generic property of the number-line -- there are always larger numbers.
And by using the base-10 logarithm and the floor function you can compute the smallest n that meets that criteria for any alleged number x smaller than 1. But for 0.999... to both be smaller than 1 and larger than any number smaller than 1 means it is larger than itself, a contradiction. And that is how you prove 1 is the only such least upper bound. But then since we both reject $$0.999... \gt 1$$ then and we both reject $$0.999... \in S$$ then it follows $$\textrm{sup} \, S \leq 0.999... \leq 1 = \textrm{sup} \, S$$ and therefore $$0.999... = 1$$.

INCORRECT. Because for all natural numbers n, $$1 - 10^{-n} \lt 0.999...$$ [*]Also for all natural numbers n, $$1 - 10^{-n} \lt 1$$

This is amusing.

1) , a telephone is black

2) , the night is black

Therefore, a telephone is the night.
You have not proven anything with this nonsense.

You wrote.

$$S = \left{ 1 - 10^{-n} \; | \; n \in \mathbb{N} \right}$$, then both 0.999... and 1 must be upper bounds of the set S, $$0.999... \; \geq \; \textrm{sup} \, S \; \leq \; 1$$

There is no math template for this reasoning.

Use your simple calculus link or use Rudin to support this template of reasoning. None exists.

So your conclusions have no basis in the mainstream.

Next, you go on about a least upper bound of a set.

That may all be true.

But for .999(n) = 1, you must prove theorem 1.1.6 of your link collapses to false.

Can you do that?

Clearly, you do not understand applicable templates of logic and math.

Otherwise, as I have said over and over, provide a valid math template which proves the infinite sequence of finite partial sums actually becomes its infinite limit. That means a provable infinite sequence of all finite elements has an infinite element. That requires emergence and magic, not math.

Undefined said:
Thanks for that confirmation of what I tried to tell certain people earlier on, mate. Namely, that 'moving the decimal point' is not a mathematical operation, but rather a mere formatting of the results of a mathematical operation.
I guess that's pedantically true.

But since you can write 999.0 = 0.999 x 10[sup]3[/sup], what's wrong with saying the representation on the left is the same as the one on the right with its decimal point shifted three places "to the right". So "x 10[sup]n[/sup]" is equivalent to a shift function f(n) acting on the decimal, with n an integer? This function preserves "the operation", so what's wrong with it?
In other words, logical mathematical treatment is more fundamentally based than formatting convention 'treatment'.
The writing conventions, or rules, also preserve something, can you guess what it is?

Hi arfa brane.

I guess that's pedantically true.

But since you can write 999.0 = 0.999 x 10[sup]3[/sup], what's wrong with saying the representation on the left is the same as the one on the right with its decimal point shifted three places "to the right". So "x 10[sup]n[/sup]" is equivalent to a shift function f(n) acting on the decimal, with n an integer? This function preserves "the operation", so what's wrong with it?
The writing conventions, or rules, also preserve something, can you guess what it is?

Actually, mate, it wasn't from pedantry that I pointed out the difference between them 'fundamentally'. It was to point out that any formatting convention for displaying a 'result' of a mathematical operation comes after that operation is 'effected' in the axiomatic rules/actions/logics.

And whether or not any formatting system is also amenable to being used for 'shortcuts' to a 'result', is a fortuitous use/benefit of 'extrapolating' using that formatting system/convention.

Naturally, any further exploitation of a formatting/display system/convention must also remain 'true' to the axioms (as far as practicable, of course....but it sometimes leads to "undefined" and "undetermined" and "infinity etc outcomes that the axioms are necessarily mute on because the results do not conform to what the axioms 'expected', as it were).

I'm not sure what you are getting at in your last sentence/question. Can you just tell me?....as I have to log out very soon! .

Edit: Sorry, arfa, couldn't wait any longer. I'll catch you up tomorrow if I have time. Thanks!

... But since you can write 999.0 = 0.999 x 10[sup]3[/sup], what's wrong with saying the representation on the left is the same as the one on the right with its decimal point shifted three places "to the right". So "x 10[sup]n[/sup]" is equivalent to a shift function f(n) acting on the decimal, with n an integer? This function preserves "the operation", so what's wrong with it?...
Absolutely Nothing Wrong! In fact it is the heart of my very simple proof that 1/1 =1 =0.999... which does not need any multiplication as a result and is built "from the ground up" starting from a definition for the 0 to 1 length line segment, even then defining the meaning of 2, 3, ... 8 & 9 ! and then noting that (1) there is no largest integer, N, as (N+1) is larger, plus (2) fact there is (in base 10 or less) no symbol for 9+1, so we need to develop the "Base, Place, & decimal point" notation system, which does have the property you describe with decimal point shift having the same effect on values as multiplying (in base 10) by some power of 10^n.

rr6 keeps posting his same confusion and ignorance and asking for simple proof that is given at the link, in only English (No text notation) and FULLY derived from definitions given. He does not even know the meaning of several of the math words he used - assumes they have their common use meanings, etc. He thinks it is a big "crux" as a result of his math ignorance and wonders why his calculator gives terminating decimal for some divisions and infinitely long decimals for others, when exactly why I explained much earlier at the link given below. Etc. - Not worth your time explaining (and useless to do so).

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BT = 1/1 = 1--WOW! How Enlightning Is That?---NOT---1 = 0.999...NOT

rr6 keeps posting his same confusion and ignorance and asking for simple proof that is given at the link, in only English (No text notation) and FULLY derived from definitions given. He does not even know the meaning of several of the math words he used - assumes they have their common use meanings, etc. He thinks it is a big "crux" as a result of his math ignorance and wonders why his calculator gives terminating decimal for some divisions and infinitely long decimals for others, when exactly why I explained much earlier at the link given above. Etc. - not worth your time explaining (and useless to do so).

BT, I've skimmed and/or read your documents previously and have to repeat to you again, give us a big dummies guide and or even better, a rational, logical, common sense and relatively simple--- your long explanation is not relatively simple ---explanatour guide to finite 1.0 = infinite 0.999....

Your given simplest answer is a as bad or worse the Origins i.e. you give the following;

1/1 = 1 = 0.999.... ha, good grief dude!, how/why does any of that show that finite 1.0 = infinite 0.999...and the answer is it does not just as Origins given couple formula--- with little to no explanation in no way address finite 1.0 = infinite 0.999....

This is real simple BT, finite value vs infinite value, and your incorrect if you keep stating that a finite value 1.0 = infinite value 0.999...

We can give you a banana for a long explanation attempt whereas Origin gave none. Oh wait, he revised in later with red print about latex condums in a string, and RPenner keeps posting that complicated and meaningless to me red print.

Drop you ego and face the facts BilyT, you are incorrect and no amount of r6 is ignorant of this or that is ever going place you into the correct catagory.

Come back and talk to me when you have a rational, logical, common sense and relaltively simple explanatory guide to finite 1.0 = infinite 0.999...

As always, I will not be holding my breath for you or others who keep repeating things like 1/1 = 1, as if that is and explanation. NOT.

1 = 0.999...is then how you end. Duhh, so youve repeated the topic thread, how does that make you correct? NOT.
I agree there is no crux/connundrum, because you are incorrect and cannot admit such nor can you or others offer anything that addresses the challenges by myself. You got nada/ZIP/Zero in ways of adddressinng my comments as stated.

r6

... This is real simple BT, finite value vs infinite value, and your incorrect if you keep stating that a finite value 1.0 = infinite value 0.999...
... r6
No 0.999... Is an infinitely long (in base 10) way to express the FINITE value 1, or 0.999... which are only two different names, not values, for one finite and rational same point on the number line.

You ignorance about math's basic concepts is appalling. But what is worse (as all are ignorant about some things) you have no desire to learn! Only to post your confused nonsense.
Why do you think 0.999... is infinite? Is it because you're an idiot, or is it because you don't know how long division works? ...
Nice try at guessing, but correct answer is: BOTH.

Hey, Alpha, I bet you five coin drop clicks, you can't get this idiot to understand that in base 2 the following is true:

1 = 0.11111111111111111111111111111... He thinks the right side is an "infinite value" and one certainly is not.

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