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

You are right, but what fraction is .999999.... ?
I and others, have PROVEN .9999... = 1/1
but essence of it is:
... I first illustrate, several of the infinite numbers of examples, of true statements concerning terminating or infinitely Repeating Decimals. (And then a procedure for finding the Rational Fraction that equals to ANY given RD):

1/3 =0.333333.... and 1/1 = 0.99999.... are rational fraction numbers with a "repeat length" of 1 in their equivalent decimal versions.

12/99 = 0.12121212... and 19/99 = 0.1919191919... and 34/99 = 0.343434... are rational numbers with a "repeat length" of 2 in their equivalent decimal versions.

In general, any integer less than 99 divided by (and not a factor of) 99 will produce a decimal repeating with length 2. Some of the factors will too. For example 3/99 = 03/99 = 0.03030303... does; but not 11 or 33. I.e. 11/99 =0.111... and 33/99 = 0.3333333333... In ALL cases with 99 as the denominator, the decimal repeats two numbers in blocks of two) is still true, as well as in "blocks" of one for some cases as bold type helps you see.

Likewise any integer less than 999 divided by 999 will be a decimal fraction with repeat length not more than 3 and always will repeat in blocks of 3, but for somecases, like 333 /999 = 1/3 the least long repeat block is less than 3. Check with your calculator if you like. Etc. For example, 678 /999 = 0.678,678,678, .... and that is slightly larger than 678 /1000, which equals 0.678 and should given you a hint of the proof to come.

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

The proof I and others have given that 1 = 9/9 = 0.99999.... is just particular case of the fact ALL rational fractions like a/b or a/9 (both a & b being integers and a < b) are equal to an infinitely repeating decimal (if they are not a finite decimal when b is a factor or product of factors of the base).

For example, the general proof of this goes like:
Rational Decimal, RD = 0.abcdefg abcdefg abcdefg .... Where each letter is one from the set (0,1,2...8,9) and the spaces are just to make it easier to see the repeat length in this case is 7.
Now for this repeat length 7 case, moving the decimal point 7 spaces to the right is not a multiply operation, but a notional change with the same effect on meaning as multiplying RD by 10,000,000. I. e. 10,000,000 RD = a,bcd,efg . abcdefg abcdefg ... Is a 2nd equation with comas for easy reading the integer part.

Now, after noting (10,000,000 - 1) = 9,999,999 and subtracting the first equation from the second, we have:
a,bcd,efg = 9,999,999 x RD. Note 9,999,999 certainly is not zero so we can divide by it to get: The Rational Fraction, RF = RD = a,bcd,efg / 9,999,999 I. e. the rational fraction of two integers exactly equal to the infinitely long repeating (with repeat length =7 in this case) decimal, RD.

Now lets become less general and consider just one of the repeat length = 7 cases. I. e. have a=b=c=d=e=f=g = 9 and recall RD was DEFINED as 0.abcdefg...so is now in this less general RD = 0.9,999,999,... and from green part of line above, The Rational Fraction which equals RD is 9,999,999 / 9,999,999, which reduces to the fraction 1/1 which is unity as the numerator is identical with the non-zero denominator. I.e. the least numerator rational fraction equal to 0.999,999... is 1/1.

By exactly the same procedure the RD = 0.123,123,123,.... is a case with repeat length of 3, can be shown be equal to the RF = 41/333 (as your calculator will show, as best as it can, if used to divide 41 by 333.
To prove this one moves the decimal point of the RD 3 places to the right get: RD' which is 1000 times larger than RD. I.e. RD' = 123.123123123... as the 2nd equation and then subtract the first from it to get after division by (1000 -1): RD = 123 /999, which reduces to 41/333 as the RF = 0.123123123...

The objectors to 1 =0.99999 need not only to give extraordinary proof for their objection but also need to explain why ONLY 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.
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 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.

1/7 is an exception to both of your generalities. It repeats itself every 6 digits - 0.(142857) I'm fairly certain the fraction equivalent of 0.(12345)
Is another one. I forget what the fraction equivalent of that one is but rpwnner was throwing it around earlier.

1/7 is an exception to both of your generalities. It repeats itself every 6 digits - 0.(142857) I'm fairly certain the fraction equivalent of 0.(12345)
Is another one. I forget what the fraction equivalent of that one is but rpwnner was throwing it around earlier.

$$\frac{1}{7} = \frac{142857}{999999} = \frac{142857}{10^6 - 1}= 0.\bar{142857}, \; \frac{4115}{33333} = \frac{12345}{99999} = \frac{12345}{10^5-1} = 0.\bar{12345}$$

I fail to see how either of these are exceptions to Billy T's description of repeating rationals.

A better case might be $$0.58\bar{3} = \frac{5.8\bar{3} \, - 0.58\bar{3}}{9} = \frac{5.83 - 0.58}{9} = \frac{5.25}{9} = \frac{525}{900} = \frac{7}{12}$$ but even here the repeat length is 1 and so the intermediate denominator is of the the form $$(10^n - 1)10^m$$ and both n and m can be read off the decimal fraction as the number of digits in the repeating section and the number of digits in the non-repeating section of the fraction, respectively. And since $$(10^n - 1)10^m$$ divides evenly into $$(10^{p \cdot n} - 1)10^{m+q}$$ for any positive integer p and any non-negative integer q, you don't even have to be especially precise when inspecting the repeating decimal.

$$\frac{1}{7} = \frac{142857}{999999} = \frac{142857}{10^6 - 1}= 0.\bar{142857}, \; \frac{4115}{33333} = \frac{12345}{99999} = \frac{12345}{10^5-1} = 0.\bar{12345}$$

I fail to see how either of these are exceptions to Billy T's description of repeating rationals.

A better case might be $$0.58\bar{3} = \frac{5.8\bar{3} \, - 0.58\bar{3}}{9} = \frac{5.83 - 0.58}{9} = \frac{5.25}{9} = \frac{525}{900} = \frac{7}{12}$$ but even here the repeat length is 1 and so the intermediate denominator is of the the form $$(10^n - 1)10^m$$ and both n and m can be read off the decimal fraction as the number of digits in the repeating section and the number of digits in the non-repeating section of the fraction, respectively. And since $$(10^n - 1)10^m$$ divides evenly into $$(10^{p \cdot n} - 1)10^{m+q}$$ for any positive integer p and any non-negative integer q, you don't even have to be especially precise when inspecting the repeating decimal.

Because apparently I either misread what Billy T said, or interpreted it too literally or too concretely.

Mea culpa.

$$\frac{1}{7} = \frac{142857}{999999} = \frac{142857}{10^6 - 1}= 0.\bar{142857}, \; \frac{4115}{33333} = \frac{12345}{99999} = \frac{12345}{10^5-1} = 0.\bar{12345}$$

I fail to see how either of these are exceptions to Billy T's description of repeating rationals.

A better case might be $$0.58\bar{3} = \frac{5.8\bar{3} \, - 0.58\bar{3}}{9} = \frac{5.83 - 0.58}{9} = \frac{5.25}{9} = \frac{525}{900} = \frac{7}{12}$$ but even here the repeat length is 1 and so the intermediate denominator is of the the form $$(10^n - 1)10^m$$ and both n and m can be read off the decimal fraction as the number of digits in the repeating section and the number of digits in the non-repeating section of the fraction, respectively. And since $$(10^n - 1)10^m$$ divides evenly into $$(10^{p \cdot n} - 1)10^{m+q}$$ for any positive integer p and any non-negative integer q, you don't even have to be especially precise when inspecting the repeating decimal.

Because apparently I either misread what Billy T said, or interpreted it too literally or too concretely.

Mea culpa.

For non-idiots but people not fully understanding the “base, place & decimal point” notation’s meanings (statement numbered for ease of reference if not clear of not believed.):

(1) A “line segment” is a math defined term, which is 1 dimensional (1D etc here after) with two ends. They can be called “a” & “b” or “0” & “1” for more convenience as then one can say “length' of the line segment is 1.

(2) If a second equal to the 0 to 1 line segment is joined to the end called 1, the total length is twice the length of the first alone. This can conceptually be done with out end as there is no “largest number,” N. (N+1 is larger.)
To cope with this fact and yet retain the idea that any well defined number can correspond to one and only one point on a “line,” we define a line to be 1D and like a line segment, but with no ends. Zero has a special roll to play. By common convention, all the numbers of / on the line to the left of 0 are “negative” and to the right are positive, but I don't need to say more about negative numbers, so won't.

Subtraction is defined as the length between the two number points on the number line (where 1 is one unit from 0 is the scale).

(3) All number system notations use a “base” I'll refer to as “B.” The most common in public use has ten different value symbols: 0, 1, 2, 3, 4, 5, 6, 7, 8, & 9 and can be used to describe the length of the longer line segments produced by adding, end to end, more of the 0 to 1 line segments, but then two ends of line segments cease to exit. The pair of line segment of point (2) joined “end to end” have ends called, 0 & 2 and has length two, etc. as “2” = 1+1; 3 = 2 +1; etc. but when we get to 9 + 1 we lack a symbol to call it (except in higher than base ten systems, such as the B =16 system used in many computers. Then 9+1 = a; and a+1 = b etc. up to d + 1 = e, when this “hexadecimal” bases system also runs out of additional symbols. In general the number of symbols, #, is same as the base. # = B.

(4) Man used his fingers to count so and that is why B = ten is so commonly used. (Mental math would be easier if humans had twelve fingers, as division by any factor of the base is relatively easy and twelve has many more factors than ten does. Even dividing a base twelve number by 8 is easy as divide by 2 and then by 4, both factors of the base.) But I digress. Fact that transistors have two well defined states (conducting or not) has made B=2 a very commonly used number system for electronic machine made calculation.

(5) All base systems have the same meaning for 0 & 1 but that is all the symbols the “binary system has, “places with values” in the notational system were invented. I'll count up to base ten 8 in binary to illustrate:
0; 1; 10; 11; 100; 101; 110; 111; 1000. The value of each place is a power of base ten 's 2. i.e. the green ones or even the green 0 last binary number I listed (the one with value 8 in the B =ten system) all are in a “notation place” one to the left of the first place and so have value, if not occupied by 0, of base ten's 2, which is 2^1. likewise the red 1 in the second place to the left of the first place has value of base ten's 2 squared or 2^2 = 4 of base ten. Etc. I. e. a 1 in the fifth place to left of the first place, has value of base ten's 2 raised to the fifth power. All base notation systems work this same way. The value of places, if not occupied by 0, is B^n where n is an integer telling how many places to the left of the first place the space is, if occupied. For example in the hexadecimal base system 1c is the base ten number 16 + 12 = 28 but I am already using the B = ten system when writing this “28.” i.e. the 8 is in what I have been calling the “first place” and the 2 of the 28 is in the first place to the left of the “first place.” So a one, one place to the left of the first place, in any base system, has value = B^1 and the next 1, one to left of it, has value = B^2, as is two to the left of the "first place." etc.

(6) We often mark with a decimal point where the “first place is. For example the hexadecimal number 1b is more clearly written as 1b. And the 1 as stated early is worth B^1 = 16^1 = 16 in the base ten system. The decimal point is essential if we want to write numbers less than 1 as they are written to the right of the decimal point. In the hexadecimal system base ten fraction 1/16 is written 0.1 or if not worried the reader will fail to see the decimal point, just as .1 and a value sixteen times smaller is written as 0.01 etc.

(7) Now with this general background, lets speak mainly of the base ten system:
Any of the ten symbol can be in the first place to the right or left of the decimal point. If for example we have 12. That means twelve. 1.2 means 1 + 1/5 and 0.012 means 1/10 + 2/100 and this is BUILT INTO THE MEANING OF THE NOTATION SYSTEM.

It is true that if I multiply 1.2 by 10, I get 12, but exactly that same effect can be achieved by notational change ONLY. Just move the decimal point found in 1.2 to the other side of the 2. I. e. write 12.
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, which I have not even defined any general algorithm for, but BUILT INTO THE MEANING OF THE base, place & point NOTATION SYSTEM.(BTW, I have not defined an algorithm for division either but only indicated it when writing fractions like a/b.)

For example as 16x16 = 256 then if I wanted to make the hexadecimal number 1b. larger by 256 times I just write it as 1b00. That hexadecimal number 1b. Had value in base 10 of 27 and if in base 10 I want the number which is 256 times larger than 27, I must multiply (or use my calculator) to get 6912, but I can increase the hexadecimal number 1b by 16^4 times easily. It is 1b0000 but to do that to base 10's 27 I will use my calculator to get: 1,769,472. This only hints at the fact large values are shorter to write in larger base number systems. Consider hexadecimal number ed. In decimal notation that is 16x15 + 13 = 253. I. e. take three places, not just two to write and even more to write in a binary system.

As I noted early, base 12 would have been not only much better for “mental calculations” but a good compromise between “writing length” and “complexity” Only two more symbols need to be defined. I think the Devil not God designed our hands.
Challenge to those who don't think 1/1 = 1 = 0.99999 exactly, tell what is wrong with the general procedure I gave in post 1481 for finding the rational fraction equal to any repeating decimal.

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Subtraction is defined as the length between the two number points on the number line (where 1 is one unit from 0 is the scale).

I had to stop reading here, since what you wrote is not true. By definition, length is nonnegative. The distance from 2 to 5 is 3, and the distance from 5 to 2 is 3. So I still don't know what subtraction is.

It's actually harder than it looks to define subtraction, given the counting numbers. Here's one web page I found on the subject. This page describes how to construct the integers given the naturals. It's not difficult but you need to use some technical tricks.

http://web.mat.bham.ac.uk/R.W.Kaye/seqser/constrintegers

I had to stop reading here, since what you wrote is not true. By definition, length is nonnegative. The distance from 2 to 5 is 3, and the distance from 5 to 2 is 3. So I still don't know what subtraction is....

That is a shame, if you had continued perhaps you would have learned something. Note I said that for discussion of proof that 1 =0.999... I did not need to consider negative numbers. Perhaps I should have also said in my definition of subtraction that one subtracts the smaller from the bigger number, but you even note that distance or length of separation on the number line is ALWAYS positive so even what I did not explicitly state is both implied and was recognized by you.

... you even note that distance or length of separation on the number line is ALWAYS positive ...

I said length is defined to be nonnegative. Work on reading comprehension.

That is a shame, if you had continued perhaps you would have learned something.

LOL I stopped reading there too!

I'd like to apologise for going all Sons of Anarchy in this thread, but someone has to keep the bros in line.
On a more serious note, subtraction isn't commutative in the integers, but since the integers are a ring with inverses, isn't subtraction just the addition of a negative integer?

Can't you start with the trivial ring and like, extend it? Or is there a problem since 0 + 0 = 0 - 0?

rr6 said:
I have a MS calculator on my computer so please give us the step step instructions that i punch in, to make infinite 0.999....become finite 1.0.
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?

Come on bro, if you want to earn your patch you need to step up.

I said length is defined to be nonnegative. Work on reading comprehension. ...
Is verbal nit-pick the best you can do- i.e. have no valid objection to the proof given that 1/1 =0.9999....

You also misread /failed to understand that I ONLY built a logical foundation for the math needed for the proof. I did not say anything about limiting processes, multiplication, division, or even addition and subtraction, except when applied to points on the number line with some number of integers space separtions.

I. e. I started with definition of "line segment" quickly specialized to one between 0 and 1, to establish a scale. Then joined two of these unit length line segments to make line segment of length twice as long and introduced the name "2" both as the name of the not zero new end of the longer new line segment and as the sum of two integer. I said 1 + 1 = 2 and then that 2 + 1 = 3. etc. up to 8 + 1 = 9

I did thus define addition of integers, and noted there was no largest one N as N+1 was larger. I also noted we have no symbol for what 9 +1 is equal too.* These two facts forced the introduction of a line (the "number line") with no ends to that 1D item list and the introduction of the place and decimal systems (with B as base) so I could tell that 9 + 1 = 10 in the system in common use by people, and begin to discuss the meaning of the "Base, place, and decimal point" notation systems.

* The hexadecimal (base =16) notation system does have a symbol for 9+1. It is "a"

Note my proof, now well founded on logical extension of the defined "line segment" (with length 1), does NOT use ANY of the normal GENERAL math operation (+ , - , x or /). It needs no algorithms for these operation, except adding and subtracting lengths on the number line. I do write the notation commonly used for fractions, but even there only with integers for a & b in a/b but never describe how the indicated division would be done as the proof has no need of that.

You first objected that my subtraction ideas could produce negative numbers with example 3-5 failing to take notice of my EXPLICT statement than I was not concerned with negative numbers in the proof that 1/1 =0.9999...

Further more, even when in the proof I do use my limited well defined "subtraction" it is only using powers of 10 (like 1000, when the repeat length of the RD is 3) and one step back of magnitude unity towards 0 on the number line. That is the proof does need "integer subtraction" or reduction by 1. For example fact that 1000 - 1 = 999 is a location on the number line one unit closer to 0 than 1000 is. It does also use the algebraic fact that (x+a) - (y+a) = x-y. I.e. two points on the number line that are both greater than x or y by the same amount (a in this case), have length difference or separation of x-y where x > y as I never use or need negative numbers in the proof.

You have a serious reading or comprehension problem, as I never made use in the proof of ANY of the things you are critical of, or making nit-picking verbal objections about**. I don't need to reply to your irrelevant "red herrings" but for the benefit of others have briefly summarized here the foundation logic of the proof that is given in more detail in post 1486, for the first statement (1) about line segment definition and properties thru 6 more numbered items to (7).
Your objections are thus far, irrelevant red herrings. If you have an objection please use the reference numbers to tell what in the proof you are objecting to.

** Yes, a - b, both separated by integer steps, as that is all I defined subtraction for, does have a length of zero if a=b but the proof never subtracts two equal point on the number line.

PS: Not only do I avoid use of all of general math's algorithms for operations including any limiting process, but I have no need to try to defined "infinity" in my proof! The closest to that I come is to use a line with no ends, called the number line, and do all movement on it by unity steps for some starting point on the number line. (My additions and subtractions are not general algorithms, but deal only with integer steps and the subtractions I use in the proof ALWAYS have positive results like 99, or 999 or 9999 etc. but not 999...)

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Origi'sn Proof Explained By R6

I do not believe that 90% of the population is too stupid understand the extremely simple proof I showed. If you wish to claim that your understanding of basic math is this horrendous, well that is your right but don't drag 90% of the population in with you.
Here is a load of proofs, see if there is one that you can understand.
Notice this from the site:

So I tried this on my MS calculator and I got the same without the dots of infinity expression. So right off the bat were seeing a finite #1 being divided into infinite value. ninth( 9th ) of the one( 1 ) is being expressed as an infinite 0.111....Duhh, once the calculator starts giving us infinite values then all bets are off, i.e. yeah infinite set appears to be equal to a finite set. Or so the it appears but again, is this just mathematical illusion?

So it appears that the MS calculator cannot divide the #1 into 9 finite parts or is it that the calculator cannot divide #1 into 9 equal and finite parts?

Ok so now I understand why QQ has been going off on some circle divided into parts to better understand what this means a calculator has to invoke infinite values to divide a one( 1 ) into 9 finite parts.

Lets spell this out for easy understanding, and without subjective or inverctive type terminology.

Does this mean that one( 1 ) is ONLY divided into finite parts by some numbers and not others? Aparrently, at least according to the MS calculator and if so what is the cosmic implications of these two sets of numbers. I can see why QQ started using a circle to get his head around this weirdity because I believe a circle can be divided into a finite set of 9 equal parts.

Ok, now I see that any number divided by 9 is going to have an infinite value on the MS calculator, so infinity is introduced early on in all of this with some numbers and combinations and not others.

1/10 =finite 0.1

1 / 9 = infinite 0.111...

1/8 = finite 0.125

1/7 = infinite 0.142857...

1/6 = infinite(?) 0.666667...
...this one rounds off on the calculator. Why?....
...I belive #6 offers a clue since it is the only one the rounds off the seemingly infinite value....

1/5 = finite 0.2

1/4 = finite 0.25

1/3 = infinite 0.333...

1/2 = finite 0.5

1 / 1 = finite 1.0....
...the zero here is representative of the infinite nothingness( non-occupied space )..
..one is reprsentative of the finite occupied space Universe...

Ok, well that is using a calculator and I've already seen that differrent calculators operate differrently and with some weirdities.

To me it appears that calculators cannot handle finite values with some numbers and number combinations. H,mm that is weird.

Now I can understand how some believe infinite value can equal a finite value.

Again, I believe the #6 rounding off at 7 may offer us a clue to this weirdity, that begins with #3.

No thx needed for my more simple-- Earthly common people ---big dummies guide introduction to grasping the finite = infinite value conundrum(?), paradox(?), weirdity. Obviously this involves some numbers going infinite on us--- going mental ---and others going finite--- going non-mental ---on us.

Perhaps that dude on saturday night live could give us a simple explaination for this 'going mental', infinite value weirdity to us.

r6

rr6,

1 / 9 = infinite 0.111...

You probably shouldn't use the term "infinite" when you mean "infinitely repeating".

1/9 is not an infinite number. It's finite, but it has an infinite number of 1s after the decimal point in base 10 notation.

1/6 = infinite(?) 0.666667...
...this one rounds off on the calculator. Why?....
...I belive #6 offers a clue since it is the only one the rounds off the seemingly infinite value....

Your calculator will always round up if the last digit it retains is in the range 5 to 9. If the last digit it retains is 0 to 4, it rounds down.

To me it appears that calculators cannot handle finite values with some numbers and number combinations. H,mm that is weird.

I'm not sure what you mean. It could be that you're seeing some rounding errors when you do several calculations in a row. That's because the calculator doesn't store numbers with infinite precision.

ThX R.J. for Attempts to Clarify Infinite vs Finite Equalities or NOT Crux

James .."R6 You probably shouldn't use the term "infinite" when you mean "infinitely repeating"."

Infinite value = infinite value whether "repeating" or not, I would think J.R.? Are you sure you not just being nit-picky uneccessarily? If not also an insignificant side tangent?

1/9 is not an infinite number. It's finite, but it has an infinite number of 1s after the decimal point in base 10 notation.

Huh? RJ, are you saying that, 0.111....is not an infinite number?

I'm dealing with a MS calculator, and I have no idea what base it is in.

Your calculator will always round up if the last digit it retains is in the range 5 to 9. If the last digit it retains is 0 to 4, it rounds down.

And by "last digit" you mean that last digit the calculator is able to show us ergo, not cosmic clue to this infinite = finite or infinite conversion to finite crux of connuundrum(?), that I if no one else is having?

So thx, and I would next go to #3 and #7 as offering us a reference clue to the seemingly crux of this connundrum(?) infinite vs finite. I have some ideas at bottom of page that may help resolve this crux/conundrum.

I'm not sure what you mean. It could be that you're seeing some rounding errors when you do several calculations in a row. That's because the calculator doesn't store numbers with infinite precision.

Huh? Did you see they 1/1-10 I posted? Are you saying their incorrect i.e. that I used the calculator incorrectly and that is why I got some or all of those values incorrect? I dont think the list i gave is incorrect and you do not appear to addressing my given list directly as stated with any specifics of that list being incorrect.

Crux = .."A puzzling or apparently insoluble problem"...

First I restate the simple introduction to the finite vs infinite connundrum(?) via a MS calculator.

1/10 = finite 0.1

1 / 9 = infinite 0.111...

1/8 = finite 0.125

1/7 = infinite 0.142857...
..here I would point out a significance differrence regarding #7 and #3( see bottom of page )...

1/6 = infinite(?) 0.666667...
...thx RJ, so the calculator rounds higher after #5 ok, not cosmic...

1/5 = finite 0.2

1/4 = finite 0.25

1/3 = infinite 0.333...

1/2 = finite 0.5

Going back to geometry--- line segments + angle ---we find that it is only with the heptagon that we begin to get an irrrational(?) infinite value for the 7 internal angles, if not the also the external angles.

These are the kinds of cosmically numerical pattern differrentiations that I tend to look for with my less educated-- ergo more simple ---set of abilities.

A tri(3)angle = 3, finite, 60 degree parts or internal angles.

1/3 = 0.333....

So here is simple comparison that may be more complex than you 1D line segment explanations, but 2D polygons with their associated angles, may actually be more visually accessable/grasp-able to larger segment of humanity.

Nona(9)gon has nine, finite, 140 degree internal angles and nine, 40 degree radial angles.

This does not solve or renormalize the infinite vs finite crux but may does assist in seeing from another angle or perspective of viewpoint the crux of finite vs infinite that involves numbers, in a relatively simple way.

One thing I notice here, is that 40 above may be related to the finite resultant of 1/4 = 0.25. i.e. it kind of make sense that the nonagon can be associated with the 1/4 = 0.25 because of the 40. I dunno just looking for clues.

#7 and heptagon definitely resonant with each other. Dont you think so?

Thx again for civility of your attempts to converse, clarify and correct.

r6

You cannot put 0.999... into a calculator unless your calculator has a button that says "this pattern goes on forever." Human beings do have such a tool to work with such numbers, it's called algebra.

Let us partition a number like a=12.34567896789678967896789... into a sum b=12.345 + c=0.0006789678967896789... into a sum like a=12.345 + d=0.0006789 + 0.0000000678967896789...
The repeating part has period n=4.
So a = b + c = b + d + c/10ⁿ and so c = d + c/10ⁿ and so c ( 1 - 1/10ⁿ) = d and so c = 10ⁿ d / (10ⁿ - 1) and so a = b + 10ⁿ d / (10ⁿ - 1)
and so 12.34567896789678967896789... = 12.345 + 6.789 / 9999 = 12345/1000 + 6789 / 9999000 = 12339444/9999000 = 1028287/833250 exactly.

So 12.34567896789678967896789... and 1028287/833250 are two names for the same number.

When I wrote c = d + c/10ⁿ, I was saying c was a pattern that repeated every n digits and that d was just the first repeat. But this is the information you always need to know about a repeating decimal whether you express it as $$12.34567896789678967896789..., \; 12.345\bar{6789}, \; 12.345(6789),$$ etc.

So algebra explains why any repeating decimal must be equal to a rational number, and the division algorithm explains why every rational number has at least one repeating decimal representation (nothing wrong with repeating zeros). And having understood that, 0.999... = 9/9 = 1

... #7 and heptagon definitely resonant with each other. Dont you think so?

Thx again for civility of your attempts to converse, clarify and correct. r6
1/7 = 0.142857.... and any integer multiple of it will except Nx7 (Not with Modulo 7 = 0) also have in its decimal version a repeat length of 6 too. For example if full 360 degree circle is divided into 7 equal parts the apex angle of each will be (in degrees) 51.42857 + 1/ 7,000,000, if you want an exact value with out repeating decimals, that seem to bother you.

You can explicitly write out two repeat patterns and then add a very tiny fraction, 1/ 7E12 I think etc. How accurate do you need the result? Why not happy with:
1/7 = 0.142857... the infinitely repeating blocks of same six numbers.?

I'm dealing with a MS calculator, and I have no idea what base it is in.
r6

I just have to say it... YIKES.

#9, #7 > Infinite--Heptagon > Infinite--NOT the other Polygons-- #6, #3

Billy T.."You can explicitly write out two repeat patterns and then add a very tiny fraction, 1/ 7E12 I think etc."

Huh? B.T. I appreciate your takeing time to address my comment specifically as stated, however, I'm not sure what your point here above is or why you went off that way. Maybe it has something to do with your next comment below.

How accurate do you need the result? Why not happy with:
1/7 = 0.142857... the infinitely repeating blocks of same six numbers.?

I do not recall my asking more accuracy with any of the 9-10 subdivisions of one that I listed.

I snipped out your beginning because your telling me anything I didn't already know and actually had already stated to some degree with involvement of polygons to perhaps find association to those numbers--- 9, 7, 6, 3 and 8, 5, 4 and 2 ---.

As to my not being happy with 1/7 resultant or any number divided by 7 resultant also makes no sense to me cause I never stated anything about being unhappy with 7's infinite repeating decimal.

Again appreciate you attempts to offer something of signiricant potential that is related to the crux/connundrum(?) of the infinite vs finite issue.

Maybe I'm just missing something you find significant in all that. I dunno.

r6

Back To Origins Basic/Simple Proof( ? ) ? Still Dontg Get It

I still don't understand how Origins simple set of formula and little to no explanatory guide is a proof that finite 1.0 = infinite 0.999...or for J.R's sake I will state it here again as infinite 0.999...( repeating decimal ). Seems uneccesary to me, as it is inferred/implied( whatever ) to represent irrational, infinite deciemal number on right side of the decimal place.

We already established that 1/9 = infinite 0.111...Thx Origin and I verified such using my MS calc.

9 * 0.111... = Oh he appears to me to have gone back and deleted some of his comment(s) and added some comment.

So as best as I recall, he was suggesting that, 9 * 0.111... = infinite 0.999...and I verified that on my MS cal.

Next he repeated that suggested that finite 1.0 = infinite 0.999...and that is just repeating the opposite of what I claim i.e. if there is a proof in his given comments as stated, he has refuse to spell it out in nice step-by-step explanatory guding way. Why?

Because he had not really given a proof, or at least one I nor 99% of humanity will every understand.

Infinite can never be equal to finite and ergo an infinite value can never be equal to a finite value.

My best guess is, that, the "mapping" scenario on Wiki that, someone sent me too, is the best way to attempt to have an infinite value = a finite value via mathematical illusion i.e. we make A = B. Ha ha! Ok, so we say that 1.0 = 0.999...and wala, I did a "mapping" conversion and I didn't even have to use words like subjerctive and injecrctive etc......

1/10 = finite 0.1

1 / 9 = infinite 0.111...

1/8 = finite 0.125

1/7 = infinite 0.142857...
..here I would point out a significance differrence regarding #7 and #3( see bottom of page )...

1/6 = infinite(?) 0.666667...

1/5 = finite 0.2

1/4 = finite 0.25

1/3 = infinite 0.333...

1/2 = finite 0.5

Going back to geometry--- line segments + angle ---we find that it is only with the heptagon that we begin to get an irrrational(?) infinite value for the 7 internal angles, if not the also the external angles.

These are the kinds of cosmically numerical pattern differrentiations that I tend to look for with my less educated-- ergo more simple ---set of abilities.

A tri(3)angle = 3, finite, 60 degree parts or internal angles.

1/3 = 0.333....

So here is simple comparison that may be more complex than just a 1D line segment explanation, but 2D polygons with their associated angles, may actually be more visually accessable/grasp-able to larger segment of humanity.

Nona(9)gon has nine, finite, 140 degree internal angles and nine, 40 degree radial angles.

This does not solve or renormalize the infinite vs finite crux but may does assist in seeing from another angle or perspective of viewpoint the crux of finite vs infinite that involves numbers, in a relatively simple way.

It may be significan that the nona(9)gon, which has rational internal angles, is sort of related to 1/4 = 0.25 because there also is a 4, and rational resultant. I dunno if geometry can be used to help clarify or resolve the connundrum?) crux of this infinite vs finite issue.

I beg of someone to please use give us a simple explanatory guide via Origins posting of a simple/basic, two or three formula. I think it is empty hot air, but at least it got me to see that are bets are off, as long the calculator gives us infinite value for some numbers and not others.

If there is simple explanation for that, please, someone explain it. No? Yeah, I didn't think so. Probably take another half page of complex wiki links to do that one also..ha ha!

r6