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

I and others, have PROVEN .9999... = 1/1
but essence of it is:The objectors tend to fall into two classes: idiots and those not fully understanding the meaning of the "Bases, places and decimal point" notational system, so I will explain it in another posts, soon.

I would say idiots are those that do not understand when a proof by induction applies.

But, since you are so smart, can you explain why induction should not be applied.

Or, can you explain your conclusions are based on induction?

How can Chinglu's ravings be acceptable but Tach's not? I truly don't understand the standards here.

If you can claim I am raving, write your proof. Support your assertions by proof.

No one hates a crackpot more than someone that can't prove their assertions.

Either way, I am not opposed to Tach either.

Here is one of may very simple proofs, this is from wiki:

$$\frac{1}{9} = 0.111...$$

$$9 \times \frac{1}{9} = 9 \times 0.111...$$

$$1 = 0.999...$$

It isn't really that hard to understand.:shrug:

How do you support your multiplication since infinite multiplication has no basis in Peano arithmetic? It only supports finite multiplication by recursion.

What basis in math do you support your conclusion?

9/9 = .999999....

Please indicate why induction it not needed.

Or please indicate why it is needed.

Thanks

0.999... is not an irrational number. Being equal to 1, it is clearly rational.

So, a number that cannot be described by any fraction, ie .9999.... is a rational number?

How is this true?

No one hates a crackpot more than someone that can't prove their assertions.

I love crackpots. That's why I participate in .999... threads.

Sure it does. All the $$a_k$$ terms are non-negative because they are digits, right?
$$\sum_{k=0}^{m} \frac{a_k}{10^k} \; = \; \textrm{sup} \, \left{ x: \; \exists n \in \mathbb{N} \; n \leq m \wedge x = \sum_{k=0}^{n} \frac{a_k}{10^k} \right}$$, right?
So $$\sum_{k=0}^{\infty} \frac{a_k}{10^k} \; = \; \textrm{sup} \, \left{ x: \; \exists n \in \mathbb{N} \; x = \sum_{k=0}^{n} \frac{a_k}{10^k} \right}$$, because I just knock out the finite upper bound on the index. Right?
And the completeness axiom says $$\textrm{sup} \, A$$ exists when $$A \neq \emptyset$$ and $$A$$ has an upper bound which is satisfied by $$a_0 + 1$$ but equally well by $$a_0 + 2$$.

As I did where? Also, why not? Where am I using the pinching theorem? I'm saying that IF $$0.999... \lt 1$$ then there exists an n where $$1 - \frac{1}{10^{n-3}} \lt 0.999... \lt 1 - \frac{1}{10^n}$$ and this contradicts $$0.999... \; = \; \textrm{sup} \, \left{ 0, 1 - \frac{1}{10}, 1 - \frac{1}{10^2}, 1 - \frac{1}{10^3}, \dots \right}$$ which says that for all n, $$1 - \frac{1}{10^n} \lt 1 - \frac{1}{10^{n+1}} \leq 0.999...$$ -- there is no pinching, just contradiction in assuming $$0.999... \lt 1$$.

Then I go beyond what is necessary and tell you how to calculate n if you insist $$0.999... \lt 1$$, so it is no mere existence proof which some people find unsatisfying, but a constructive proof.

Further it is not the only such contradiction that arises if you assume $$0.999... \lt 1$$.

All sources on modern mathematics agree $$0.999... = 1$$ so it appears you are trying to invoke an authority you yourself ignore.

OK, all modern mathematics applies the pinching theorem which is proven by induction on all sides.

Here is what you did.

This follows from continuity and monotonicity of the exponential because $$\forall k\in \mathbb{Z} \; \frac{1}{10^{k+1}} \lt \frac{1}{10^k}$$ and so if $$0 \lt 0.999... \lt 1$$ then $$n \quad = \quad 1 - \left\lfloor \log_{\tiny 10} \left( 1 - 0.999... \right) \right\rfloor \quad \in \mathbb{N}$$ and therefore $$1 - \frac{1}{10^{n-3}} \lt 0.999... \lt 1 - \frac{1}{10^n} \lt \textrm{sup} \; \left{ 0, \, 1 - \frac{1}{10}, \, 1 - \frac{1}{10^2}, \, 1 - \frac{1}{10^3}, \dots \right} \lt 0.999...$$ which is a contradiction that proves $$0.999... \not\lt 1$$.

Now, let's note obvious error 1.

You claimed continuity is responsible for proofs regarding natural numbers. Continuity requires the real number system. You cannot derive continuity from only natural numbers. So, you are wrong here.

Next, you wrote,

$$1 - \frac{1}{10^{n-3}} \lt 0.999... \lt 1 - \frac{1}{10^n} \lt \textrm{sup} \; \left{ 0, \, 1 - \frac{1}{10}, \, 1 - \frac{1}{10^2}, \, 1 - \frac{1}{10^3}, \dots \right} \lt 0.999...$$

You see, on the left side of your equation you are using n, which therefore requires induction, and then you compare it to the supremum operator on an infinite set on the right side.

There is no valid math template that allows you to do this.

But, since you say you are right show us all your justification for this move.

I love crackpots. That's why I participate in .999... threads.

Is that so you can know you are a crackpot?

Why would you like that?

Rpenner in your post #1456, I asked you for the justification for the infinite summation operator. I explained to you it was finite in nature and you did not respond.

You see, natural number summation is based on finite recursion. There is no such thing as infinite summation based on natural numbers.

The dots mean that infinitely many terms follow. 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

So, you can read this article that the "infinite summation operator" in this case is based on a sequence of finite partial sums as I explained to you. There is no infinite summation.

Prove this wrong.

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 claimed continuity is responsible for proofs regarding natural numbers. Continuity requires the real number system. You cannot derive continuity from only natural numbers. So, you are wrong here.
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,
This follows from continuity and monotonicity of the exponential because $$\forall k\in \mathbb{Z} \; \frac{1}{10^{k+1}} \lt \frac{1}{10^k}$$ and so if $$0 \lt 0.999... \lt 1$$ then $$n \quad = \quad 1 - \left\lfloor \log_{\tiny 10} \left( 1 - 0.999... \right) \right\rfloor \quad \in \mathbb{N}$$ and therefore $$1 - \frac{1}{10^{n-3}} \lt 0.999... \lt 1 - \frac{1}{10^n} \lt \textrm{sup} \; \left{ 0, \, 1 - \frac{1}{10}, \, 1 - \frac{1}{10^2}, \, 1 - \frac{1}{10^3}, \dots \right}$$$$\color{red} \lt$$$$0.999...$$ which is a contradiction that proves $$0.999... \not\lt 1$$.
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.

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There is no infinite summation.
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:
The theory of series $$\sum a_n$$ with terms that are nonnegative for sufficiently large n is simpler than the general theory, since such a series either converges to a finite limit or diverges to $$\infty$$, as the next theorem shows.
Theorem 4.3.8 $$\quad$$ If $$a_n \geq 0$$ for $$n \geq k$$ then $$\sum a_n$$ converges if its partial sums are bounded, or diverges to $$\infty$$ if they are not. These are the only possibilities and, in either case,
$$\sum_{n = k}^{\infty} a_n \; = \; \textrm{sup} \, \left{ A_n | n \geq k \right}$$​
where
$$A_n = a_k + a_{k+1} + \dots + a_n, \quad n \geq k$$.​
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 $$0.999... < 1$$.

Is chinglu really a spambot?

Stay tuned, folks.

Is that so you can know you are a crackpot?

Why would you like that?

Well, I don't think I'm a crackpot. But of course crackpots don't think they're crackpots either, so that doesn't prove anything.

There are two types of people who participate in .999... threads online. One, crackpots. These are people who have an ax to grind about the subject and grind away with gusto. These folks have always been with us. The circle squarers, the angle trisectors, the Cantor deniers. Almost always male; and generally very intelligent. Often in technical occupations such as mechanical engineering, disciplines where they use a lot of math but don't actually study math as mathematicians do.

The other type is people who understand that .999... = 1, that it's an established mathematical fact, and who are only too glad to explain this counterintuitive fact to people encountering it for the first time.

But once it's clear that providing a mathematical explanation won't help; and that the crackpots are going to refuse to accept the conclusion no matter what you say; why then stay in these threads? There is something a little odd about those people who willingly participate in these threads despite knowing that it's a waste of time. I'd call these people the anti-crackpots. I include myself in this group. Anti-crackpots are people who understand the math and not only enjoy explaining it to earnest students; but who also enjoy debating with crackpots.

I'm not sure why I do it. I have a love-hate relationship with .999... threads.

... The other type is people who understand that .999... = 1, that it's an established mathematical fact, and who are only too glad to explain this counterintuitive fact to people encountering it for the first time.

But once it's clear that providing a mathematical explanation won't help; and that the crackpots are going to refuse to accept the conclusion no matter what you say; why then stay in these threads? There is something a little odd about those people who willingly participate in these threads despite knowing that it's a waste of time. I'd call these people the anti-crackpots. ...
I agree. I'm unsubscribing from these two threads, but may start a new one in Physics & Math focused on the least foundations (axioms) needed to find rational fractions equal to decimal expressions, etc. some what as I did at post: http://www.sciforums.com/showthread.php?136842-1-is-0-9999999999999&p=3140116&viewfull=1#post3140116

There I surprised myself learning that a great deal can be well established without any of the four general math operations or the limiting process, or even mention of "infinity" just with a good understanding of the meanings inherent in "bases, place & decimal point" notational systems. I think I will play around with this infinite repeating decimals is what rational fraction question at first in the hexadecimal system to see what I can learn.

BTW, I think I falsely stated that e was the largest hexadecimal symbol but think it is f now.

I'm leaving as not an "anti-crackpot" but did post here as until recently some of the many crackpots seemed capable of learning.

Top > Pathway( Complex and Simple ) > Bottom--Essential Truths

someguy.."But once it's clear that providing a mathematical explanation won't help";

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?

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

2) those who agree with Origins supplied and relatively simple set of formula as a proof of the above, please add rational, logical, common sense and relatively simple explanatory guide to accompany Origins two or three formula.

No one takes on my challenge and that certainly makes me suspicous of their integrity-- intellectual and moral ---. So please, when you or the others want exhibit some intellectual and moral integrity, at minimum, make some attempt to address my given 1 and 2 comments above, as stated.

Please, do not just quote them, and act as if that is addressing them. I cannot count the number of trolls who have done that with me.

and that the crackpots are going to refuse to accept the conclusion no matter what you say

ditto the above, again.

why then stay in these threads?

Because your a troll? That is my guess as you do not appear to offer anything that is helpful to those who ask for rational, logical, common sense and relatively simple explanatory guide concerning the crux/connundrum(?) of this issue.

There is something a little odd about those people who willingly participate in these threads despite knowing that it's a waste of time.

It is waste of time, as long as no one offers the truth, and the truth spelled in ways that 99% of humanity will have a chance of understanding and verifyiing others claims. You have none of this so please move on.

I'd call these people the anti-crackpots. I include myself in this group. Anti-crackpots are people who understand the math and not only enjoy explaining it to earnest students; but who also enjoy debating with crackpots.

Your debating your lack moral integrity as you have not offered any of what I mentioned above, at least not that i'm aware of ergo, your just here to troll others who are searching for truths. You or others, stating this or that is true,without giving something that is understandable to 99% of humanity is good for nothing to 99% of humanity.

We don't need more trolls-- as believe you may tend to be ---we need sincere, intellectually moral individuals, who want to help those who do not understand how a finite value can equal an infinite value. I believe you and others who believe that, finite 1.0 = infinite 0.999... are incorrect and that is why you cannot and have not addressed my givens above as stated, and not just quoted.

You would think Sci-forums would be more involve with people searching for truth and less with trolls like yourself. No offense is intended, as only trying to clearly express how troll behaviors of being dismissive, disparaging and other mental head slappings, when uncalled for and uneccessary is NOT helpful to anyone at Sci-fourm and does litttle to noting to advance humanities understandings of truth if not also higher mathematics.

When in comes to mathematics, Im doing baby steps and some others here appear to be flying in a jet plane. Fine, but I'm not here to learn how to fly a mathematical jet plane and "earn my patch" as arfra-brane suggests. I'm here for cosmic truths and rational, logical, common sense and relatively simple pathways to understanding those cosmic truths and some less cosmic truths.

r6

I'm not sure why I do it. I have a love-hate relationship with .999... threads.

Complex > Pahtway > Simple(?)---Exit Stage Right >>>>

I'm leaving as not an "anti-crackpot" but did post here as until recently some of the many crackpots seemed capable of learning.

Putting aside your "crackpots" comments, I never saw you sincerely and earnestly address my comments as stated--- rational, logical, common sensical and realtively simple explanatory guide --- to help those 99% of humanity who will never grasp your, RPenners alledged proofs, that finite 1.0 = infinite 0.999...

I.e wheres the beef? Apparrently some believe the beef lies within the context of "mapping" ergo subjerctive, injerctive and A in columen X becomes C ind column D. I'm not recalling exactly the complexity involved.

Again, I think it is mathematically illusionary, mental masturbation on the side who claims the a finite = an infinite. My evidence for this, is that they cannot address my given comments and requests as stated. Origin claims a simple proof with two or three formula, with little to no explanation of how his given formula prove a finite = and infinite.

No one else has addressed my challenge to use Origins simple formula and explain it.

What I do get from Origins given set, is that MS calculator, gives some inferred/implied infinite values regarding division by some numbers--- 9, 7, 6, 3 ---and others give finite values--- 10, 8, 5, 4 ----.

It is interesting to me, that,;

10 > 0.1

8 > 0.125

5 > 0.2

4 > 0.25--nona(9)gon > 40 degree and 140 degree angles

2 > 0.5

... I never saw you sincerely and earnestly address my comments as stated---....
What I do get from Origins given set, is that MS calculator, gives some inferred/implied infinite values regarding division by some numbers--- 9, 7, 6, 3 ---and others give finite values--- 10, 8, 5, 4 ----....
That "seeing nothing" is your own FAULT*. I.e. You did not read my proof which gives the reason why division by 10, 8, 5, or 4 gives terminating decimals, and even illustrates the point for "slow learners" in now bold text here:
The procedure given below works for ALL Repeating Decimals (all RDs) to produce the exactly equivalent rational fraction, but some stubborn and /or ignorant people posting here make the extraordinary claim that it does not when the RD = 0.9,999,999,999.... and offer zero proof of their claim when extraordinary proof is required! ....(A large part of this post 301 in the "1=0.999... chocolate thread" has been omitted at this point. For those honestly wanting to learn read all at: http://www.sciforums.com/showthread...ophy-of-Math&p=3136195&viewfull=1#post3136195)...

However, any integer divided by a factor of the number base (1, 2 & 5 for base 10) or any product of these factors (like 4, 16, 2^n, 5 or 5^m, {2^n x 5^m} ) will terminate, not repeat. For example 17 /(1x4x5) = 0.85

Then objectors to 1 =0.99999 need not only to give extraordinary proof for their objection but also need to explain why one of the other infinite number of successes of this procedure fails to produce the rational fraction that is exactly equal to the infinite Repeating Decimal , RD.
For example the 8 that troubles you is 2x2x2, or "a product of the base 10 factor 2."

You decided that you calculator could teach you more than Rpenner could (and that is probably true as you have near zero understanding of his precise "Math Text" notation); but I explain why some divisions you do on your calculator terminate and others do not. There is no "crux" only your ignorance (and unwillingness to learn even when explained to you in simple English) that is causing you the "why some terminate and others don't" problem.

* As I recall you nit-picked my valid statement that the subtraction I DEFINED gave only positive lengths. You said subtraction could give negative lengths and illustrated by 3-5
so you refused to read more and learn. Your nit-picking was irrelevant as I didn't use the more complete general definition of subtraction in my proof, but only my limited definition that had results like 1000 -1 999 or 1, like ALL your comments about 0-1 =9, etc. were. I.e. MY DEFINED SUBTRACTION ALWAYS HAD 9 OR 99 OR 999 etc. as the result, but this could be for example when (1000xRD - 1RD) was found to be 999RD by my defined subtraction where RD was the Repeating Decimal of concern. Please note 1000xRD is value not a multiplication operation, which I never used, except as an indication of my meaning. I.e. to make that 1000RD I shifted the decimal point in the RD three place to the right, using my understanding (which you seem to lack) of the meaning of the "base, place, & decimal point" notation systems.

The main beauty of my very simple proof was that all that it used was defined from starting definition; and I did not use, limiting procedures or any multiplications. My use of division was extremely limited too: only indication (but not doing it) in from a/b or in the case a/a telling the value was 1. My adding was also very limited to unity steps. I.e. after starting with line segment of length 1 (with ends called 0 & 1) I repeatedly added copies to get the line segment 0 to 2 and that also defined the meaning of the symbol "2" etc. up till my sequentially defined integer 8 + 1 made symbol of 9.

Since there is no symbol for 9+1, I had (and did) start to discuss the "base, place, & decimal point system" all in English you could have followed and learned from, but learning was of no interest to you.

Yes I will soon have these threads many crackpot are active in, with no interest in learning, on "ignore" but will look here a day or two more first.

Queiryfull/Questions > Complex > Pathways > Simple > Essential Truths

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 $$0.999... < 1$$.

RP, this appears to me to relate to my MS cal. giving a seeming infinite value--- fills all the spaces of the cal. ---for some division process'es and only uses 1, 2 or 3 spaces for other.

It appears to me, that the reason no such number--- infinite 0.999... < 1 ---is because were dealing with an infinite value to begin with.

Putting aside the mathematical technical labeling of a value as' irrational' or not, infinite values are irrational in their essential nature because they never resolve themselves i.e. infinite values never come to closure with themselves.

Infinity is not finite, tho I still don't understand why 9, 7, 6 and 3 appear to take us off into infinity on the cal. when 10, 8, 5, 4 and 2 do not.

Infinite value can never equal a finite value. If i'm in error then please give rational, logical, common sense explanatory guide for us as to how/why and infinite value = finite value.

No one has done that, to my knowledge, and the reason is-- or so I believe ---is because they cannot offer us something that does not exist.

I cannot understand much less verify any of your given mathematics i.e. it is very similar to any person who knows little about contructions or auto mechanics etcc and they go to some with much more knowledge in that field, and are totally dependent on their being trustworthy individuals.

We know there scamming/conning of people by those with supposedly more knowledge in some fields occurs, to whatever degreee.

'Wheres the realtively simple beef' to my queries in all of these regards? If there only exists very complex mathematics then......we will have infinite queries, disscussion, debate etc.....

In future there may exist less queries, of this nature because some will become more knowledgeable but quieries will exist eternally, as long as there are entities to be quieiryful.

r6

RP, this appears to me to relate to my MS cal. giving a seeming infinite value--- fills all the spaces of the cal. ---for some division process'es and only uses 1, 2 or 3 spaces for other.

It appears to me, that the reason no such number--- infinite 0.999... < 1 ---is because were dealing with an infinite value to begin with.
That's a unhelpful way to look at numbers because it is a bias that means you believe $$\frac{1}{2} + \frac{1}{2} = {0.5}_{\tiny 10} + {0.5}_{\tiny 10} = 1$$ but not $$\frac{1}{3} + \frac{1}{3} + \frac{1}{3} = {0.333...}_{\tiny 10} + {0.333...}_{\tiny 10} + {0.333...}_{\tiny 10} = 1$$ even though I can choose to use base 12 and write $$\frac{1}{3} + \frac{1}{3} + \frac{1}{3} = {0.4}_{\tiny 12} + {0.4}_{\tiny 12} + {0.4}_{\tiny 12} = 1$$ or build a calculator that can do exact math with repeating decimals. Because the decimals repeat that means there is only a finite amount of information encoded in the unending digits, and it is this information that the long division algorithm works with to produce the repeating decimal from the rational number $$\frac{1}{3}$$.

Putting aside the mathematical technical labeling of a value as' irrational' or not, infinite values are irrational in their essential nature because they never resolve themselves i.e. infinite values never come to closure with themselves.
That's a value judgement and not mathematics and not applicable in the case of repeating decimals because once we detect that they repeat, we know them utterly with only finite information. Notation wise $$0.87123123123... = 0.87\bar{123}$$ and from just this we can immediately read that it is equal to some integer divided by 99900. Thus $$0.87\bar{123} = 87036/ 99900 = 7253/8325$$. Now if there wasn't a pattern to the digits, you would be stuck if you needed to know all the digits. But that's why we admit multiple equivalent representations. A certain number might be named various ways but is the same number:
$$\textrm{The Golden Ratio} \; = \; \textrm{The positive root of "} { \small x^2 - x - 1 = 0 } \textrm{"} \; = \; \frac{1 + \sqrt{5}}{2} \; = \; 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \dots}}}} \; \approx \; 1.6180339887+$$ (I don't use "..." but instead use a plus sign to indicate the real number is larger than what is written here but no pattern of repeating digits is implied)
And knowing the number lets us add as many digits to the decimal approximation as we desire.

You are basically embracing ignorance over logical study of the field of numbers. That's really a poor basis for a communication with others because other people see no beauty in your holy ignorance unadulterated with knowledge. Numbers are the invention of mankind, invented to serve our needs. You wouldn't operate an automobile unless you knew the rules of the road and understood the purpose of the automobile, so why do you think ignorance of the real numbers is making your viewpoint attractive in public discourse?

You wouldn't operate an automobile unless you knew the rules of the road and understood the purpose of the automobile, so why do you think ignorance of the real numbers is making your viewpoint attractive in public discourse?
Because ignorance of numbers is socially acceptable and is considered the social norm.