O.9999999... = 1?

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Expand view Topic review: O.9999999... = 1?

Re: O.9999999... = 1?

by Spartaculous » Tue Apr 02, 2024 9:30 pm

I referenced the fact that the harmonic series

H = 1/1 + 1/2 + 1/3 + 1/4 + 1/5 + ...

diverges in my previous post. There are several proofs of this, but here is the original proof by the 14th-century French philosopher (and Bishop of Lisieux) Nicole Oresme.

First, we group the terms in the following manner:

H = (1/1) + (1/2) + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + 1/10 + 1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1/16) + (1/17 + ...

After the initial two terms, the numbers of terms in each group are successive powers of 2: 2 terms, 4 terms, 8 terms, etc.. (The next grouping will have 16 terms: from 1/17 to 1/32.)

Now, consider the following series J:

J = (1/1) + (1/2) + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + (1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16) + (1/32 + ...

J comes from replacing each term with the smallest term in its group. So, for the group (1/5 + 1/6 + 1/7 + 1/8), 1/5, 1/6, and 1/7 are each replaced with 1/8. Note that we are always replacing larger terms with smaller terms: 1/5 > 1/8, 1/6 > 1/8, 1/7 > 1/8. This shows that H > J.

But now, count how many terms are in each group. (For example, there are 4 copies of 1/8.) You obtain

J = (1/1) + (1/2) + 2*(1/4) + 4*(1/8) + 8*(1/16) + 16*(1/32) + ...

But each of those products 2*(1/4), 4*(1/8), etc. equals 1/2. This pattern will keep on going, so

J = 1/1 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + ...

Thus, we can see that J diverges to infinity. Since we constructed J in a way so that H > J, H must also diverge to infinity. QED!

[Note: to make this more precise, I shouldn't write H > J, because neither sum converges, but it should get across the gist of the argument.]

Re: O.9999999... = 1?

by sweetandcool » Mon Apr 01, 2024 7:07 pm

You are a hero Spartaculous. It remains to be seen how comprehensible your explanation is to non-math people, but you did a great job keeping it simple while also explaining it well.

In regards to your last paragraph, I had meant to comment previously a thought about the proof Will posted. The main problem with that proof is that it isn't illuminating at all. After reading it, I would still have no idea what is going on. I could see that the proof is "correct", but it really doesn't teach me anything about where the value of -1/12 is coming from.

Whereas, your brief statement on it cuts right to the heart of the matter.

Re: O.9999999... = 1?

by Pengwinja » Mon Apr 01, 2024 7:05 pm

So… 0.9999999… = QWERTY?

Re: O.9999999... = 1?

by Spartaculous » Mon Apr 01, 2024 6:16 pm

kingofthepirates wrote:
Sat Mar 30, 2024 9:07 pm
-1/12 is really wacky. I believe the process to get it is called Ramanujan summation. I'm learning calc rn, so I don't really know how/why it works (my teacher has expressed distaste towards the subject when a friend of mine brought it up), though the logic of the proof seems consistent.
JustAGuyNamedWill wrote:
Sat Mar 30, 2024 9:23 pm
It just feels counterintuitive, even though the math (presumably) checks out.

Like, it should be impossible to add a positive number to another number and get less than what you started with, which is one.
CaptainFritz28 wrote:
Sun Mar 31, 2024 9:15 pm
I still refuse to believe that the sum of all natural numbers is -1/12, however. That just proves, to me, how limited our understanding of infinity is, because it would imply that -1/12 = infinity.
So, yeah, I am not a huge fan of people seeing "1 + 2 + 3 + 4 + 5 + ... = -1 / 12" before they have the mathematical background to put that statement in context, because it really is misleading, and can lead students learning calculus astray.. There are many other cool and surprising math results I would much rather talk about, but here we are.

In calculus (say, calculus 2 in a usual American education), the most important thing about series (infinite summations) is whether they converge (to a particular value) or diverge (which simply means don't converge). A series converges or diverges based on whether or not its corresponding sequence of partial sums converges or diverges. (To be extremely formal here, I should be including an epsilon-N definition of convergence of a sequence.) For example, the series

1/2 + 1/4 + 1/8 + 1/16 + ...

converges to 1, because the sequence of partial sums

1/2, 3/4, 7/8, 15/16, ...

converges to 1. As a non-example, the series

1 + 1 + 1 + 1 + 1 + ...

diverges, because the sequence of partial sums

1, 2, 3, 4, 5, ...

diverges to infinity. (Note: it is not allowed to say that a sequence "converges to infinity".) As a perhaps more interesting non-example, the so-called harmonic series

1/1 + 1/2 + 1/3 + 1/4 + 1/5 + ...

diverges, because the sequence of partial sums

1, 3/2, 11/6, 25/12, 137/60, ...

diverges to infinity (although this is certainly not obvious). So clearly, the series

1 + 2 + 3 + 4 + 5 + ...

diverges, because the underlying sequence of partial sums

1, 3, 6, 10, 15, ...

diverges to infinity.

So now that we have gotten that out of the way, what does it mean when some people say that

1 + 2 + 3 + 4 + 5 + ... = -1 / 12?

The answer is that some mathematicians have asked the question, "Yes, of course, this infinite series diverges. But if we had to assign a real value to it, what value would we assign to it?".

For example, consider the series

1 - 1 + 1 - 1 + 1 - 1 + ... .

It diverges, because the underlying sequence of partial sums

1, 0, 1, 0, 1, 0, 1, 0, ...

diverges. But consider for a moment the geometric series

x ^ 0 + x ^ 1 + x ^ 2 + x ^ 3 + x ^ 4 + ...

If |x| >= 1, then the series diverges, and if |x| < 1, then this series converges to 1 / ( 1 - x ):

x ^ 0 + x ^ 1 + x ^ 2 + x ^ 3 + x ^ 4 + ... = 1 / ( 1 - x ).

(As an example of that, if you set x = 1/2, then you obtain

1 + 1/2 + 1/4 + 1/8 + 1/16 + ... = 1 / ( 1 - 1/2 ) = 1 / ( 1 / 2 ) = 2,

which is almost exactly the same as the first result in this post (just add 1 to both sides).)

But what if you substitute in x = -1 into x ^ 0 + x ^ 1 + x ^ 2 + ... = 1 / ( 1 - x )? You obtain

1 - 1 + 1 - 1 + 1 - 1 + ... = 1 / ( 1 - (-1) ) = 1/2.

We just assigned a value to the divergent series 1 - 1 + 1 - 1 + 1 - 1 + ... . Now, that doesn't mean that it converges - it still very much diverges! And, of course, the answer feels weird: how can we add and subtract a bunch of integers and end up with something that is not an integer? But it is a value that we can assign to this divergent series. In a similar way to how we handled the geometric series, we can show that the formula

1 - 2 x + 3 x ^ 2 - 4 x ^ 3 + ... = 1 / ( 1 + x ) ^ 2

is valid for |x| < 1 (if you know calculus, just differentiate both sides of the geometric series formula above, and replace x with -x). Of course, once again, if |x| >= 1, the series diverges. But if you choose to ignore that, then substituting in x = 1 into the formula gives

1 - 2 + 3 - 4 + ... = 1 / ( 1 + 1 ) ^ 2 = 1 / 4,

which was previously mentioned in this thread. Again, it doesn't make sense - the series diverges! - but it is a way of assigning a value to this divergent series.

So then, the argument that 1 + 2 + 3 + 4 + 5 + ... = -1 / 12 is of a similar flavor. Of course the series 1 + 2 + 3 + 4 + 5 + ... is a divergent series. Of course it doesn't converge to a particular value. Of course, if you add up all of the natural numbers, the sum goes off to infinity. That's what I would hope that students would know and understand. But there exists a way to assign a value to this divergent series, and that value is -1 / 12.

There do exist applications in math of the statement "1 + 2 + 3 + 4 + 5 + ... = -1 / 12", but they are probably all at the graduate level or beyond. My preferred way to "prove" this statement isn't through the proof (mathematicians might prefer the term "heuristic") that Will posted, but rather by observing that, if you construct the analytic continuation of the Riemann zeta function zeta(s) to cover values of s with imaginary part <= 1, then zeta(-1) = -1 / 12, and zeta(-1) corresponds (in some sense) to "1 + 2 + 3 + 4 + 5 + ... ", but that is well outside the scope of this post.

Whew.

Re: O.9999999... = 1?

by Spartaculous » Mon Apr 01, 2024 3:08 am

Diplomacy&Warfare wrote:
Sun Mar 31, 2024 12:32 pm

I have no objection to you flexing your superior mathematical skills. However, can you at least take the time to read my post? I, myself, stated that the smallest positive number was not real. (1/infinity) is not a real number, in the same way that infinity, i, and (i/2) are not real numbers.

Additionally, you also clearly have not read the question I was answering:
Can someone explain that smallest positive number?
Nothing about this question indicated the number had to be a real number.
...
My claim, if you feel like actually debunking it, is:
(1/infinity) > 0
My apologies for not properly responding to your previous post. I think I was mostly hung up on kotp calling 1.[infinite number of zeroes]1 a real number. Let me try to offer a broader view.

So modern mathematics is based on the notion of sets. For example, you could have the set of natural numbers:

N = { 1, 2, 3, 4, ... }

Or, you could have the set Q of all rational numbers (all numbers that can be written as p/q, where p is an integer and q is a natural number), or the set R of real numbers.

When we are dealing with these types of questions ("What is the smallest positive number?"), it is important to be clear about what set of numbers we are working with.

For example, N has the property that there is a smallest positive element of N. (This is 1.)

In my previous post, I showed how R does *not* have this property: there is no smallest positive real number.

If you claim to have a number that is not a real number, then it is important to be able to say what set of numbers it belongs to, and, just as importantly, to be able to describe the properties of this set of numbers.

(Here are some examples of properties. N is closed under addition: add any two natural numbers and you will get a natural number. However, it is not closed under subtraction: 5-7=-2, which is not in N. However, R is closed under subtraction.)

So, when you said "I, myself, stated that the smallest positive number was not real.", you need to be clear about what is the set of numbers that this belongs to.

Now, if you want to do computations involving infinity (such as claiming 1/infinity = 0), strictly speaking, you are not working with the real numbers anymore, as infinity is *not* a real number. You would need to be clear that you are working with the extended reals (https://en.wikipedia.org/wiki/Extended_real_number_line). But the extended reals still have the property that there is no smallest real number.

By the way, my view of the "number" 1.[infinite number of zeroes]1 is that the only way I can see to define it is as the limit of the following sequence of real numbers:
1.1
1.01
1.001
1.0001
...

This sequence converges, and its limit is zero. (Not something slightly greater than zero, but literally zero.)

Maybe I will get around in a day or two to offer comments on 1+2+3+4+... = -1/12: broadly speaking, how to assign real values to certain divergent series. I have a lot of sympathy for this point of view:
kingofthepirates wrote:
Sat Mar 30, 2024 9:07 pm
(my teacher has expressed distaste towards the subject when a friend of mine brought it up)

Re: O.9999999... = 1?

by CaptainFritz28 » Sun Mar 31, 2024 9:15 pm

Perhaps the best way to describe 0.999... is in this format:
lim(x -> 1-) f(x) = x = 1. What tripped me up initially was approaching it from an arithmetic point of view, but once I realized that it's really just a weird way of phrasing the limit of 1, it came together for me.

I still refuse to believe that the sum of all natural numbers is -1/12, however. That just proves, to me, how limited our understanding of infinity is, because it would imply that -1/12 = infinity.

Re: O.9999999... = 1?

by DiplomacyandWarfare » Sun Mar 31, 2024 8:59 pm

*facepalm*
Well, now I feel like an idiot. I forgot that where x is a positive, finite number, x/infinity = 0. That removes my entire point. On the other hand, this does at least mean that 1/infinity = a real number. I will now cease arguing about mathematics until I remember at least half of what I'm talking about.

Re: O.9999999... = 1?

by sweetandcool » Sun Mar 31, 2024 4:01 pm

Dip, I read what you wrote.

I'm not sure how to reply to you because you have contradict yourself in a few places and you say things that unfortunately just don't make sense.

Okay, so in your first paragraph here you claim that 1/infinity is not a real number. You give examples of i and I/2 not being real numbers. So are you claiming 1/infinity is complex number, and not a real number? Regardless, as I said before, 1/indinity is 0. So it is a real number, and also a complex number.

Paragraph two, I did answer the question you were answering. Spartaculous and I just showed you why there is no "smallest positive number". We answered the question.

Paragraph three I don't understand.

Paragraph four, is confusing because by claiming that there is a smallest positive number, particularly in the context of the discussion, implies that you don't believe that 0.9999..... =1.

Your conclusion is not incorrect, because I read everything. I'm sorry if your feelings are hurt. Just as I assume you read what I write, I would like you to believe that I read what you write

Your final claim is incorrect. 1/infinity=0. This is something you would learn in beginning calculus, though the concept could easily be taught in Algebra.

Re: O.9999999... = 1?

by DiplomacyandWarfare » Sun Mar 31, 2024 12:32 pm

sweetandcool wrote:
Sat Mar 30, 2024 6:44 pm
[snip]
I have a Master's in Mathematics. Your proof just shows that 0=0.

If you let x=1/(infinity) then x=0
(well really this should be stated as a limit, but I digress)

So then according to your proof:

0-0=y
0=0
therefore,
0-0=0
and 0=0

------------------------------------------------------------------------------------------

Spartaculous has covered why there is no smallest positive real number.

But I would also like to mention that in Mathematics once something is proven (yes, in very high level Math sometimes mistakes in proofs occur, but then someone will notice and it will eventually be disproven), then it is ironclad true. It's not a matter of opinion. Either something is true or it is false.

It is 100% true that 0.9999.....=1.0.

And this is not even a wacky result.

Another math result that truly is incomprehensible, but true (but I won't blame anyone for disputing): The infinite sum, 1 + 2 + 3 + 4+ ...... = -1/12.

That is, the sum of all Natural numbers is equivalent to -1/12.
I have no objection to you flexing your superior mathematical skills. However, can you at least take the time to read my post? I, myself, stated that the smallest positive number was not real. (1/infinity) is not a real number, in the same way that infinity, i, and (i/2) are not real numbers.

Additionally, you also clearly have not read the question I was answering:
Can someone explain that smallest positive number?
Nothing about this question indicated the number had to be a real number.

Yes, my proof ends up being something like x=x, and if x = 0 then it's just 0=0. However, the point of that particular proof was pointing out how useless a particular line of reasoning was.

Finally, you made the claim that 0.999999999... = 1 while implying I had disagreed. Which I had not. I am happy to defer that problem to you.

In conclusion, you failed to read the claim or reasoning you were disagreeing with, failed to read the question I was providing an answer to, failed to read what I was proving with my proofs, and "debunked" me by answering an unrelated question.

My claim, if you feel like actually debunking it, is:
(1/infinity) > 0

Re: O.9999999... = 1?

by sweetandcool » Sun Mar 31, 2024 12:25 am

JustAGuyNamedWill wrote:
Sat Mar 30, 2024 9:23 pm
It just feels counterintuitive, even though the math (presumably) checks out.

Like, it should be impossible to add a positive number to another number and get less than what you started with, which is one.
Agreed, it makes sense to just use common sense. A positive integer plus a positive integer is always going to be a positive integer, so the result doesn't really make sense except for in the proper context.

The answer is infinity, but infinity is cool and likes to break the rules, so the answer is really -1/12 in the context of the function they were using to yield the result.

Re: O.9999999... = 1?

by sweetandcool » Sun Mar 31, 2024 12:22 am

kingofthepirates wrote:
Sat Mar 30, 2024 9:07 pm
-1/12 is really wacky. I believe the process to get it is called Ramanujan summation. I'm learning calc rn, so I don't really know how/why it works (my teacher has expressed distaste towards the subject when a friend of mine brought it up), though the logic of the proof seems consistent.
Probably because the common sense answer is that it equals infinity. Off the top of my head I don't actually know much about the -1/12 result. I think it has some relation to zeta functions. It really is less of a matter of literally taking the sum of the Natural numbers and getting -1/12 and rather a matter of doing some funky things with some zeta function and the function spitting out -1/12 as an answer.

That is, it's essentially assigning the value of -1/12 to this divergent infinite sum, and it's a result that makes sense in the context of the function.

However, it doesn't really make sense from a layman's perspective.

Your teacher probably dislikes it because they either don't understand the math that led to that result (likely) or they hate that the result is being taken out of context (also likely).

Re: O.9999999... = 1?

by JustAGuyNamedWill » Sat Mar 30, 2024 9:56 pm

Even funnier is the numbers are the most stable concept we’ve come across. Yet we still don’t understand them

Re: O.9999999... = 1?

by kingofthepirates » Sat Mar 30, 2024 9:30 pm

yeah. numbers to wacky things. like you can uniquely map every number between 0 and 1 to the entire number line. crazy things lol.

Re: O.9999999... = 1?

by JustAGuyNamedWill » Sat Mar 30, 2024 9:23 pm

It just feels counterintuitive, even though the math (presumably) checks out.

Like, it should be impossible to add a positive number to another number and get less than what you started with, which is one.

Re: O.9999999... = 1?

by kingofthepirates » Sat Mar 30, 2024 9:07 pm

-1/12 is really wacky. I believe the process to get it is called Ramanujan summation. I'm learning calc rn, so I don't really know how/why it works (my teacher has expressed distaste towards the subject when a friend of mine brought it up), though the logic of the proof seems consistent.

Re: O.9999999... = 1?

by JustAGuyNamedWill » Sat Mar 30, 2024 8:47 pm

I think?? I understand? At least the concept.

Still is a weird answer tho lmao

Re: O.9999999... = 1?

by sweetandcool » Sat Mar 30, 2024 8:30 pm

So really that proof is a bit of a cheat, because now you need to prove that Y=1/4, which I assume is an advanced proof. Or it cheats by utilizing the result of a different advanced proof, etcetera.

Re: O.9999999... = 1?

by sweetandcool » Sat Mar 30, 2024 8:19 pm

JustAGuyNamedWill wrote:
Sat Mar 30, 2024 7:30 pm
IMG_0679.jpeg

I think i found it

Ill let the smarter people figure this out lmao
Oh, okay, I guess there is an elementary proof. Very nice!

I'll explain it for you Will.

Line 1: Let S be the sum of all Natural numbers.

Line 2: Let Y be the alternating sum of all Natural numbers. We are going to use this to calculate S.

Line 3: Consider S - Y.

Line 4: Distribute the negative (or just consider S + (-Y)) and rearrange the terms of the sum S - Y. In doing so, a pattern becomes clear.

Line 5: S -Y = 4 + 8 + 12 + 16 + .....

Line 6: Since all terms of S - Y are divisible by 4, we can factor it out. So
S-Y=4(1+2+3+4+....)

Line 7: But wait! That's just 4 times the sum of all natural numbers! So really
S-Y=4S.

Line 8: Basic algebra yields S=-Y/3. Since it is known that Y=1/4 (an entirely separate proof), we get S=-(1/4)(1/3). We evaluate and arrive at our conclusion:

S= -1/12

Re: O.9999999... = 1?

by JustAGuyNamedWill » Sat Mar 30, 2024 7:30 pm

IMG_0679.jpeg
I think i found it

Ill let the smarter people figure this out lmao

Re: O.9999999... = 1?

by sweetandcool » Sat Mar 30, 2024 7:17 pm

JustAGuyNamedWill wrote:
Sat Mar 30, 2024 7:13 pm
sweetandcool wrote:
Sat Mar 30, 2024 6:44 pm

Another math result that truly is incomprehensible, but true (but I won't blame anyone for disputing): The infinite sum, 1 + 2 + 3 + 4+ ...... = -1/12.

That is, the sum of all Natural numbers is equivalent to -1/12.
? Explain to us simpletons
Heh, I don't understand the proof either, really.

I think the lesson to be learned here is that you ever encounter anything related to infinity, just mark your mental map with "Here be dragons" and just accept you might be facing results that don't make sense.

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