Why 0.9 (repeating) is Not a 'Real' Number

In this video, we demonstrate quite conclusively that 0.9 (repeating) is not a real number.

[zbopzbop]

I may have just misunderstood your vid, but my first take on it is that it confuses numbers with our common representation of numbers using decimal digits. That is, "1" is not a number, it is how we represent the abstract concept of the number 1 in our written language. As your vid shows, you can (if you understand limits) also represent the number 1 as a decimal of infinitely repeating nines. (0.9999...) But that's just a quirk of the decimal system we commonly use to represent numbers. Having different representations for the same number also happens with rational numbers. For example, you can write 1 as 1 or 1/1 or 2/2 or 3/3 and so forth. Yes, we refer to all of these representations as simply "a number", but that's just a linguistic short-cut we take for convenience. If you instead always refer to it as, perhaps, "a written representation of the abstract concept of a particular number" it would get very tedious very quickly, no?

[PV]

You didn't prove it doesn't exist. You just proved it's equal to 1 and hence a real number.

[hsianglinchang]

You’re a very brave man, who go against most people’s believe, that 0.9 repeating is equals to 1.

I think most people missed the fact that even 9 over 9 and 10 over 10 are both equal to 1 but they are not the same 1, one is base 9 and the other is base 10 system. The example you used 1 over 9 showed in base-10 system is 0.1 repeating but in base-9 system is 0.1 ( no repeating); 9 over 9 as you said should be 0.9 repeating in base-10 system but since it is base-9 system so it should be 0.9, but again, 0.9 in base-9 system is 1.

[Chris-]

It's very simple, the sum of a series >>IS<< the limit of the sequence of its partial sums (if the limit exists). That's a definitional fact. If you don't agree with that, then you are in a world of your own where all mathematics is a fantasy.

You also don't seem to be aware of the difference between a number (a pure abstract concept), and a numeral (a literal symbol) that represents a number.

0.999... = 1 means that those two symbols >>represent<< the same number. Specifically:
the sum of 0.999... := lim n->oo 0.999...9 (n 9s) = lim n->oo 1 - 1/10^n = 1
the latter following from the epsilon-N definition of limit (for a sequence).

There is no deity given meaning to any mathematics as you seem to think is required. 0.999... is what we want it to be. The choice that is made is easily the most sensible one that can be made.

In answer to your disingenuous question at 3:36 you asked for a pair on integers whose ratio is 0.999... Just pick any non-zero integer z, then z/z = 0.999.... It also easily follows from the following semi-general argument that 0.999... = 9/9.
Let 0.(N) represent the n digit string N being repeated. Then
10^n * 0.(N)) = N.(N)
=> (10^n - 1) * 0.(N) = N
=> 0.(N) = N / (10^n - 1)
Clearly that means that 0.(N) is rational.

If N = 9 then n = 1 and 0.(9) = 9/(10^1 - 1) = 9/9 = 1.
If N = 428571, then 0.(428571) = = 3/7

What beats me is that after proving that 0.(9) = 0.999... = 1, you just reject the math and completely illogically assert that means that 0.999... is not a real number.

[johnlabonte-chul]

Middleschool math here. You missed an important fact in the video about the algorithm x=".99...". It is circular and designed to give the exact value it started with. Many claim this is the "proof" that ".99..." is 1.
Making a minor change of adding x instead of subtracting, then dividing by 11 instead of 9.
10x="9.99..."
11x="10.99..."
x=".99..." as it should.
What happened.? Can infinite numbers be used in arithmetic?
Adding ".11..." to ".11..." to get ".22..." seems safe. There is a one to one correspondence not affected by infinity. (Math says you can't add to infinity, °°+x=°°)
You can "count" to ".99..." without a problem.
Try adding ".99..." and ".11...". Is the answer 1.00...?
Let's see. .9+.1=1.0, .09+.01=.10, .009=.010.
So far I've got 1.1100...
I give, can't get rid of the 0 at the end. (Yes, 10 divided by 9 results in "1.11..." but it is incomplete. )
1-".99..."=".00...", where did that 1 go. Again 1.0-.9=.1, .00-.09=-.09, .000-.009=-.009,
So far .1-.099=.001
1/3 step by step. 1.0/3.0= wait for it! WAIT FOR IT!!



UH, No. I can't wait for it, the 3s just won't end. It will always be incomplete and ever changing, unless you say it is finite by ignoring the 3s past some mathical point. (Which is exactly what math is saying, infinity is that point, where ever that is. Where is infinity exactly? How many digits of pi are necessary for the Webb telescope to determine the width of the furthest star or piece of dust in the universe? Do we know? The continuum of Real Numbers keeps us in the game.)

The valuable concept of infinity is dangerous, incomplete, inconsistent and imprecise.

[johnlabonte-chul]

I agree with you in general. ".99..." is not 1. I don't believe it is a good representation of 1. I believe it is not a real number. It is better to say it is a "real number space" composed of an infinite amount of numbers that are real numbers with a finite number of the digits, 9 that followi the decimal point in decimal place notation less than 1.
(That was a mouthful)

Also why mess up decimal notation by defining a integer with digits other than 0 after the decimal.

In the late 1500's, infinite numbers in decimal notation became accepted as rational numbers instead of fractions.
You pointed out that infinite digits could be obtained by dividing by 9 with a counter example being 9 itself.
You could say those, such as ".33...", are number spaces less than the fraction they represent.

On a sum of an infinite series of >0 terms, how can it be unique and precise number without someone declaring it to be by formula. (Not by basics)

The Archimedean property says we can't define a real number adjacent to 1 as there are always numbers between any 2 real numbers we can conceive of. It is difficult to show that there is no number in base 10 that is closer to 1 than the number space of ".99...". It is always at least 1/10^(n+1) from 1. But in number bases higher than 10, like base 60, 1/60^(n+1) is closer. (Unless you believe 1/n^n^°° equals 0. Then 1/10 and 1/60 represent the same number. [Yes I could have used 1/n^°° but 1/n^n^°° better shows my point, which brings me to...] )

Infinity is incomplete, inconsistent and imprecise.

[EDIT Real numbers are unique and precise. ]

[brandonszpot]

I understand your point but it seems more like this is an issue of symbology rather than mathematics. I’d suggest be careful with this kind of argumentation, because the notion of 0.9… is already very difficult for many students to get their heads around.

[cupiodissolvi]

0.99... = 1 In this equality, the only number is 1.
There doesn't exist any number that would be composed of a whole part equal to 0 and of an inifinite decimal extend with repeating 9. This number doesn't exist because it is the number 1 (whole part = 1 and infinite decimal extension .0000...)

[johnlabonte-chul]

So what is ".99...". I liked your comment that 1 and 0 are your favorite numbers. 1 is the basic unit, the starting and controlling counting number. 0 is more exotic. It is a defined non-value number, the additive identity. In the decimal place notation system it has relative value in holding a place value in the notation when that place value is not present. O is a unique and precise non-value. All real numbers have a unique and precise value.
Between 0 and 1 all the real numbers are previewed and then repeated from 1 through the infinite real numbers.

So what is ".99..."? [An extended real number in the extended real number system.] In decimal place notation. ".99..." is a infinite decimal. A infinite decimal is a notation that represents a sequence of numbers less than, and have a real number limit of, that decimal place notation can't accurately represent such as certain fractions and transcendental numbers.

For centuries, rational numbers have been defined to be the ratio of 2 integers. I see no reason to change it. It seams that non-standard (modern?) wants to change this definition to add repeating decimals but do not change the definition in their texts. " 99..." does not meet the definition of a rational number. [Is it a lone example of a repeating decimal not being rational?] Further, math wants to define ".99..." as a integer while for centuries indicating that in decimal place notation there are no non-zero digits to the right of the decimal point in a integer notation.

I believe that ".99..." represents a set of real numbers less than 1. An open set between the real number 0.99..9, a indescribably large finite number of 9, and those larger real numbers less than 1 that cant be represented in the decimal place system.

The decimal place system has been in use for centuries and modified to be more rigorous. Why mess it up for infinity which is incomplete, inconsistent and imprecise.

[Chris-]

LOL. So you proved that 0.999... =1 and then just arbitrarily claimed that 0.999... is not a real number. You didn't make any attempt to justify your claim. You just gave another proof that 0.999... = 1.

You even said something like if 0.999... = 1 then why do we need 0.999... in the first place. So does that mean that 2/5 + 3/5 = 1 somehow implies that 2/5 + 3/5 is also a fantasy/pseudo number?

[johnlabonte-chul]

Wrong again, Karen. The definition of rational numbers does not state that using the formula to find the integer fraction of a repeating decimal. The definition of a rational number asks for an integer ratio or two integer when divided result in the repeating decimal. 9 divided by 9 results in 1, not ".99...". Learn the basics.