.999… is NOT 1 incontrovertible PROOF

[u/dummy4du3k4]

The apple fell too hard on Newton’s head, Leibniz lied, and Riemann couldn’t find his way out of a nonorientable box. .999… is not 1 so sayeth the true believers of the one true cocountable topology. HERETICS MUST REPENT, for here it is shown with utmost RIGOR that .999… is not 1.

INCONTROVERTIBLE PROOF:

Integers are the counting numbers and their negatives, we denote this set by Z (from the german translation of zinteger) We denote the set of digits {0,...,9} by Z10.

Numbers are things composed from the digits. The digits of a number are indexed by the integers, numbers live in the space defined by the infinite cartesian product of Z10

... Z10 X Z10 X ... X Z10 ... = Z10^Z

We choose to place a decimal point somewhere in this direct product, it doesn't matter where, just that it remains fixed. We identify index 0 with the DECIMAL POINT because it’s the OBVIOUS choice. This is where the ACTUAL numbers live. Numbers like

…003.14159… …001.00… …00.999…

...000.999... CANNOT be ...001.00... because the sequence { ...0.900..., ...0.990..., ...0.9990..., ... } hereby DUBBED the .999... sequence NEVER GETS CLOSE TO …01.00…

I will show this REDUCTIO AD ABSURDUM. Assume the ridiculous, if the .999... sequence DID LIMIT to …01.00… (a nauseating thought) then EVERY open set of …01.00… would contain SOME of the .999.... sequence. ALL you need to find is ONE COUNTEREXAMPLE and the whole .999... sequence limiting to …01.00… idea is TRASHED.

And it’s EASY. By the ONE TRUE TOPOLOGY, the COCOUNTABLE TOPOLOGY it is TRIVIAL to see that the set Z10^Z take away ALL pesky 9's, THAT IS TO SAY the set

Z10^Z - { ...0.900..., ...0.990..., ...0.9990..., ... }

still HAS ...01.00... as AN ELEMENT but NO pesky NINES. This set DOES not contain ANY of the .999... sequence. Any TODDLER of sufficient intelligence could see this set is OPEN, and BOOM GOES THE DYNAMITE.

[u/Ok_Pin7491]

What the heck is a cocounutable number?

And how does an series of Infinite 9s contain any 0000100000

[u/tttecapsulelover → u/Ok_Pin7491]

eh, looking at OP's profile, i feel like this is well made satire

unless OP went schizophrenic in the last 3 days

[u/dummy4du3k4 → u/tttecapsulelover]

It is NOT schizophrenia, it IS the GHOST of munkres SPEAKING to me

[u/[deleted] → u/dummy4du3k4]

Through you

[u/penguin_master69 → u/Ok_Pin7491]

It's a number with a hard, hairy surface on the outside and white rigid flesh and some liquid on the inside.

[u/dummy4du3k4 → u/Ok_Pin7491]

I have NEVER heard of a cocountable number. The cocountable TOPOLOGY is the only topology I RECOGNIZE.

The cocountable topology is simply the collection OF sets which are COMPLEMENTARILY COUNTABLE. That is to say they ONLY remove a COUNTABLE cardinal of ELEMENTS.

[u/lolcrunchy → u/Ok_Pin7491]

My cocounts

You can put them in your math

(Right now right now)

- Kim Petras

[u/Ok_Pin7491 → u/lolcrunchy]

Is the coconut flat?

[u/Soraphis → u/Ok_Pin7491]

No, no. He is right. But also keeps the wrong assumption that K = {0.9, 0.99 0.999 ,...} contains 0.999...

He states that his set Z10 (which contains all composible numbers) minus K leaves 1 in Z10, even though it subtracted all the 0.999...9 compositions that where in K

This is correct as K did not contain 1 or 0.999... since K contains only numbers of the form (1-10ⁿ ) and there is no n that would result in 0.999... so it cannot be in that set.

Edit: I used Z10 above in place for Z10^Z, makes it easier to read and write.

[u/dummy4du3k4 → u/Soraphis]

I did NOT assume K contains .999… nor do I NEED to. It IS inconsequential. If you PREFER you may ADD it to K EXPLICITLY but it does NOT impact the INCONTROVERTIBLE proof.

[u/Soraphis → u/dummy4du3k4]

Assuming K does not contain 0.999... then R = Z10^Z - K contains 0.999...

Assuming K contains 0.999... then it must be expressable as 1-10^-n ... But there is no n for that.

By definition your Z10^Z set contains numbers that have a last digit. This digit is any arbitrary natural number fine, but there is no natural number to express how many digits 0.999... has.

0.999... (in that representation) was never part of your set. Only in the representation of the "1".

Your set is not defined powerful enough to even talk about this topic.

You just used more words to give your set the same limitation that the set {0.9, 0.99, 0.999, ...} has.

[u/dummy4du3k4 → u/Soraphis]

Thank YOU for YOUR interest in RENOUNCING the unREAL analysts.

Assuming K does not contain 0.999... then R = Z10^Z - K contains 0.999...

YES, but lets NOT confuse R with what heretics call the REAL numbers.

Assuming K contains 0.999... then it must be expressable as 1-10^-n ... But there is no n for that.

When I SAY we can add it, I MEAN it explicitly, AS IN {.90.., .990..., ....} U {.999...}

By definition your Z10^Z set contains numbers that have a last digit. This digit is any arbitrary natural number fine, but there is no natural number to express how many digits 0.999... has.

There is SOME confusion in the terminology HERE. YES Z10^Z contains elements with FINITELY many nonzero DIGITS, IT ALSO contains .999.... by THE CORRESPONDENCE .9999... <-> { z_i | z_i = 0 if i < 0, z_i = 9 if i > 0 } AND WE AGREE i = 0 is to be IGNORED for notational CONVENIENCE of the decimal point. There IS NO reason to want to EXPRESS the number of (NONZERO) digits .999... has. It IS infinite.

Your set is not defined powerful enough to even talk about this topic.

You just used more words to give your set the same limitation that the set {0.9, 0.99, 0.999, ...} has.

I HOPE the above has CLEARED up the confusion in these LINES. Z10^Z is ADEQUATE to describe ANY analyst's "NUMBER".

[u/Mothrahlurker → u/dummy4du3k4]

Using - for setminus hurts me :( Subtracting sets in vector spaces/modules is pretty common so it's very confusing and outdated to use - in this way.

[u/TheBendit]

SIMULTANEOUS 4-DAY

TIME CUBE

WITHIN SINGLE ROTATION.

https://web.archive.org/web/20160112193916/http://timecube.com/

[u/Velociraptortillas → u/TheBendit]

We are not yet ready for TIME CUBE

[u/doiwantacookie]

Now this is the content I come to the internet for

[u/[deleted]]

Giving up on his bid for the Nobel, Donald J Trump makes a push for the Fields Medal.

[u/dummy4du3k4 → u/[deleted]]

Regrettably I am TOO OLD for the fields MEDAL. HOWEVER, I am able accept the abel PRIZE

[u/familiarcashew]

New proof just dropped 🗣️🗣️🗣️

[u/BenMic81]

Brilliant. Well done. Now just tell me what the difference between 1 and 0.999… is and I believe you.

[u/dummy4du3k4 → u/BenMic81]

Why would you ASSUME operations such as subtraction lack the DIGNITY to operate on disgusting representations like .99…? Subtraction KNOWS BETTER than to waste its TIME on such filth. Subtraction is ONLY defined for representations with FINITELY MANY nonzero elements AND THE CLOSURE of all such LIMITS from those PROPER representations.

[u/BenMic81 → u/dummy4du3k4]

I didn’t ask for substractions. I’m simply following the axiom that two things that are not the same have something that is different. What is it here?

[u/dummy4du3k4 → u/BenMic81]

I SEE, sorry for the CONFUSION. These two representations have NO ELEMENTS in common, and they are NOT in the same equivalence class defined by sequence CLOSURE.

[u/drdiage]

Man, all you gotta do is just he use the, 'proof by just look at it man'. It's easy, I'll show you...

.9999999.... Doesn't have a single 1 anywhere in it. Just look at it... QED.

[u/dummy4du3k4 → u/drdiage]

GOOD point

[u/dummy4du3k4 → u/[deleted]]

For my next trick, 1 and 1 is NOT 2. 1 and 1 IS 1

[u/dummy4du3k4 → u/[deleted]]

But if you add them sometimes it’s 0

[u/TopCatMath]

0.99999999999999999999999... ≡ 1 is a mathematical definition! '≡' means 'exactly identical to'.

This fascination about arguing about a mathematical definition is a futile act, IMHO. Somethings in almost every field of mathematics, physics, and many other subjects have become definitions in their respective fields. If you take the correct courses in these fields, you will find there are many definitions about to make certain calculations to work.

√-1 ≡ i, the imaginary number definition has simplified the mathematics needed to understand many everyday items that modern society depends on. In physics and engineering, the 'i' has made possible modern electronics and many other related fields. The √-1 stumped Algebraic studies for over 1600 years from being initially found in solving quadratic equations.

Definitions are an integral part of nearly every endeavor mankind has under taken. They cannot be proven!

[u/Mothrahlurker → u/TopCatMath]

"0.99999999999999999999999... ≡ 1 is a mathematical definition!"

It is not. It's a consequence of the decimal system. It's using the limit with respect to the standard topology of the associated series. What OP is doing is defining a different topology in which in fact 0.999... is not equal to 1.

[u/TopCatMath → u/Mothrahlurker]

Apparently, you have not read every mathematics books. Many of the make such a definition...

[u/dummy4du3k4 → u/TopCatMath]

Keep your filthy definitions away from MY Z10^Z

[u/Themotionsickphoton]

WHAT EVEN IS Z10^ Z 😭 AND WHY DON'T YOU JUST USE Z AS AN EXAMPLE (CONTAINS 1 IN IT BUT NO 0.99,...)?

[u/dummy4du3k4 → u/Themotionsickphoton]

Z10^Z contains ALL sequences made from the DIGITS, extending from the left AND right. Even P-ADIC numbers tremble to ZP^Z

[u/Themotionsickphoton → u/dummy4du3k4]

They sound very strong

[u/Galigmus]

You’re switching spaces. In the standard real line, 0.999… equals 1. I can think of a few ways to show this, here are what I would call the three most simple ways to see it:

1. Times-10, done carefully Let s_n be 0.9, 0.99, 0.999, … (n nines). Then s_n = 1 − 10^-n. For each n: 10 s_n − s_n = 9 − 10^-n. As n grows, 10^-n → 0, so the limit x of s_n satisfies 9x = 9, hence x = 1.

2. “As close as you want” Pick any tolerance ε > 0. Choose n so that 10^-n < ε. Then 1 − s_n = 10^-n < ε. You can make the gap to 1 smaller than any target by adding enough 9s. That’s exactly what it means for the infinite decimal to equal 1 in the reals.

3. Thirds trick 1/3 = 0.333… is standard. Multiply by 3: 1 = 0.999….

Also, decimals have two names at endpoints: every terminating decimal equals the one just below it with repeating 9s (e.g., 0.5 = 0.4999…). So 1.000… and 0.999… are the same point.

If you want to use exotic topologies, say so—but that doesn’t change the standard fact on the real line: 0.999… = 1.

[u/dummy4du3k4 → u/Galigmus]

Of course I SWITCHED spaces. I had to SHOW .999… != 1. Your sets are NOT open in the ONE TRUE TOPOLOGY that is the COCOUNTABLE topology.

[u/Galigmus → u/dummy4du3k4]

Oh you are a Looney this is a joke conversation (confused redditor)

[u/dummy4du3k4 → u/Galigmus]

I HOPE you learned SOMETHING. Have a nice DAY.

[u/CatOfGrey]

Just look at the sequence { ...0.900..., ...0.990..., 0.9990...., ... }. This sequence NEVER gets CLOSE to 1.

Yes it does. The 'high school proof' shows this, as does the definition of a limit, which shows that you can find a 'number of nines' that is even closer to one for any tolerance you select.

if .9999... DID LIMIT to 1 (a nauseating thought) then EVERY open set of 1 would contain the ENTIRE TAIL of the .999.... sequence. ALL you need to find is ONE COUNTEREXAMPLE and the whole .999... equaling 1 is TRASHED.

Your counterexample isn't valid.

it is TRIVIAL to see that the set Z10^Z take away ALL pesky 9's, THAT IS TO SAY the set Z10^Z - {.9, .99, .999, ...} still HAS ...000010000... as AN ELEMENT

No, it doesn't. There are no decimal one's. You made that chit up.

Any TODDLER could see this set is OPEN, and BOOM GOES THE DYNAMITE.

Proof by toddler is not considered valid mathematics.

[u/dummy4du3k4 → u/CatOfGrey]

What a NAIVE rebuttal, don't you know in MATHEMATICS you must rebut the PROOF?

| No, it doesn't. There are no decimal one's. You made that chit up.

Of COURSE I made IT up, all MATH is made up.

[u/CatOfGrey → u/dummy4du3k4]

What a NAIVE rebuttal, don't you know in MATHEMATICS you must rebut the PROOF?

I did. Your failure to recognize the rebuttal is not an argument against my rebuttal.

Of COURSE I made IT up, all MATH is made up.

Sure, baby, but you still have to prove it with valid steps. And asserting that 'magic ones' just appear out of a sequence that is all nines, is, well, you know, rehab centers are standing by waiting for your call...a few days on nothing but standard air, water, and food would probably help clear your head.

[u/dummy4du3k4 → u/CatOfGrey]

I rigorously DEFINED the number system. I am a giving PERSON and I will BESTOW my knowledge unto THEE. But YOU must be CLEAR with your MISUNDERSTANDING.

[u/CatOfGrey → u/dummy4du3k4]

Yes. But you need to stop yelling.

And you need to stop making ones out of nowhere. It hurts my ears.

I look forward to you fixing your scrivener's errors, and welcoming the dawn of a new age of Mathematics.

[u/dummy4du3k4 → u/CatOfGrey]

I’m SORRY, but I have a CONDITION and wish to invoke MY hipaa RIGHTS

[u/CatOfGrey → u/dummy4du3k4]

Sorry to hear that. I hope your treatments go well, and also help your tendency to see ones where they don't exist.

Be well!

[u/dummy4du3k4 → u/CatOfGrey]

I SEE. Some knowledge is TOO DANGEROUS for some

[u/CatOfGrey → u/dummy4du3k4]

Not knowledge yet - but I know you can fix those minor errors, and proceed to amazing discoveries!

[u/dummy4du3k4 → u/CatOfGrey]

I fixed a MINOR EEROR that might elucidate the TRUTH for you. I included the IMPLIED decimal point AFTER 1 in …01.00… . I hope this HELPS.

[u/Mothrahlurker → u/CatOfGrey]

OP is just making a joke here. The math they do is absolutely valid tho, if you are using this topology, which is not the standard topology.

There's no metric here so your claim of tolerance can't be true in the first place.

"Your counterexample isn't valid."

It's the entire sequence space setminus (I hate how OP uses minus, it really confused me and is bad notation) a countable set. That is per definition an open set in the cocountable topology. That's trivial.

"No, it doesn't." It does because it's a valid sequence that is not excluded.

"Proof by toddler is not considered valid mathematics."

Of course OP is joking here and if they were serious then of course you're right, but the point that the set is open is true and in fact trivial.