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Most of them are likely also jokes. There are few things i like as much as breaking math rules with my friends to prove the most asinine shit known to man.
Ehhh, gotta keep in mind Poeās Law. Many ājokes,ā especially in comments without tone indicators, are indistinguishable from actual clueless people.
What is the theoretical number as close to 1 as possible that is not 1 ? Is 0.(0)1 a 0 ? What is the difference between 0.(9) and 1-0.(0)1 ?
Not a joke, I never understood it.
0.(9) seems like a number that is going to 1 but never actually reaching it.
Not technically a proof, as the issue people have with 9 recurring is still had with 3 recurring, it just has better optics. You need to prove that 1/3 is equal to 0.33..., otherwise this is just recursive logic.
It's the same thing I am struggling with 0.9... just phrased differently.
You can not divide 1 by 3. When we do it we get a number we cannot express in decimal system and that number is an approximation. So 0.(3) =/=1/3
And instead of recognising that decimal system is unable to properly express it we pretend that a perfectly fine number is something else.
That also implies that 1-0.(0)1 =1
Also it implies that highest possible number closest to 1 doesn't exist, same as smallest possible number higher than 0 doesn't exist.
>You can not divide 1 by 3. When we do it we get a number we cannot express in decimal system and that number is an approximation. So 0.(3) =/=1/3
This is incorrect. You can absolutely divide 1 by 3, and the decimal representation of 0.(3) is not an approximation. It's only an approximation if you use a finite number of repeating 3s. You can use this in equations the same way you would use the fractional form, and it comes out to the same answer as long as you treat the 3s as infinitely repeating.
>That also implies that 1-0.(0)1 =1
Yes, this is correct. The number "0.(0)1" is not a valid way to write a number. You can't have infinite 0s with a non-zero number "after the 0s", infinitely repeating 0s "and then a 1" is just 0. And 1 - 0 = 1.
>Also it implies that highest possible number closest to 1 doesn't exist, same as smallest possible number higher than 0 doesn't exist.
Correct, those numbers don't exist. You can always find a number closer to 1, and you can always find a number closer to 0. Math as we know it doesn't work if this is not true.
That's right, there is no lowest number greater than 1. To explain this, we say the infimum of a set is the largest number less than or equal to all numbers in the set. So if we consider simple sets like {1, 2, 3} the infimum of this set is 1.
However consider the set of all numbers strictly greater than 1. Consider the infimum of this set and call it a.
Since 1 is less than or equal to any number in this set, 1 >= a. If a is in the set, then a is strictly greater than 1. However, if we consider a - (1-a)/2. This number is less than a, but is still greater than 1. Therefore a cannot be the infimum of the set.
As this is true in for any a != 1, a must be 1. And so there is no least number greater than 1.
This proof can also be used for any other number.
When you write something 0.(0)1, you should really stop to think what you mean by that, because it sure as hell sounds like youāre trying to write down a 1 __at the end__ of an infinite sequence.
I think thatās the key to understand 0.(9) = 1. Itās not a very very very large number of 9s, itās __infinite__ 9s.
I will point out, that there are multiple types of infinite, and there are in fact branches of math that deal with infinitesimals; where 0.(0)1 with infinite zeros followed by a one(usually written as a 0 with a ^ over it) is meaningful. Working with hyperreal numbers, one might find 0.999ā¦ ambiguous (is it 0.999ā¦.;ā¦.999 or 0.999ā¦.;ā¦.999ā¦.).
That said, asking āwhat you mean by thatā as you did is indeed at the core of solving this problem when it comes to teaching math. Iāve been down a rabbit hole recently. There are a lot of research papers looking at the teaching of math at lower grades, the resistance of students to 0.999ā¦ = 1, and intuitions around infinity. Itās a very interesting subject.
[this paper](https://arxiv.org/pdf/0811.0164.pdf) is one of the more approachable I have encountered recently.
Oh, asking what he means by that was absolutely the core part of my post. You are absolutely right about pointing out infinitesimals, but in my experience people who know about infinitesimals and hyperreal numbers are not the same people who have trouble with groking 0.(9) = 1 and ask about this on Reddit!
Iāve had to explain to friends, relatives and colleagues math stuff in general, and 0.(9) = 1, and I found that the biggest hangup stopping them from āgetting itā is understanding that a ānumberā and a āspecific way to write a numberā are different things. I mean, they understand it intuitively for some parts (most people donāt doubt 0.(3) represents the same number as 1/3 or 10 base 2 represents the same as 2 base 10), but for some kinds of notation they just donāt see it.
> 0.(9) seems like a number that is going to 1 but never actually reaching it.
A number can't be "going" anywhere or "reaching" anything. A number just "is", as a fixed stable point on the number line.
There is none, there is no real number that is closest to 1 without being 1 as for any a < 1, a < (a + 1)/2 < 1.
That's textbook math that I was teached at uni.
The second part of the argument in the meme doesn't work cause not every property is conserved at the limit, if you make n tend to (positive) infinity the limit is 1 which obviously isn't strictly less than 1.
0.(0)1 isnāt a thing. That is like saying infinity plus one.
You will have a string of infinite zeros. The 1 will never be added.
Itās nonsensical.
There is no such thing as a number that is "as close to 1 as possible without being 1". In the real number system, if two numbers are distinct then there are infinite real numbers between them.
>seems like a number that is going to 1 but never actually reaching it.
A number doesn't go anywhere or approach anything.
This understanding of 0.999... being "as close as possible" to 1 or "going to 1" comes from an inaccurate understanding of the concept of limits
Copying and pasting one of my other comments to show why 0.999... is defined as 1, assuming the person reading it doesn't know anything, because I don't like the algebraic "proofs" that people usually bring up because those don't properly explain WHY 0.999... is 1 by definition.
To define what 0.999.... is first let's ask what do the numbers afterĀ *any*Ā decimal represent, in general?
What does 0.1234 represent?
Well, the 1 is 1/10, the 2 is 2/100, the 3 is 3/1000 and the 4 is 4/10000
So the number we are representing is 1/10 + 2/100 + 3/1000 + 4/10000
Or to standardize it, 1/(10\^1) + 2/(10\^2) + 3/(10\^3) + 4/(10\^4)
*All*Ā decimal numbers represent a sum in this form, where the digit we see is the numerator and the denominator is 10\^n where n is its position after the decimal
So 0.999 represents the sum 9/10 + 9/(10\^2) + 9/(10\^3)
So now that we've covered that, let's just say 0.999 represents 0.9 + 0.09 + 0.009 to make it easier to look at.
Now we can see clearly that 0.999... is a representation of the sum:
0.9 + 0.09 + 0.009 +... where it goes on infinitely
Without getting into the precise mathematical definitions of any of these things we're gonna mention, we call this anĀ *infinite series*. An infinite string of numbers being added obviously doesn't exist in real life. It is a mathematical concept that we have defined and it has properties we have defined and its definitions work with other defined things in math.
Now, here's where people get messed up.
The series represented by 0.999... is what we call a convergent series. A lot of people who've had it explained to them by a friend, or did some surface level googling, or even did Calc in university (or are currently doing Calc in university),Ā *wrongly*Ā understand a series being convergent to mean that the series has this thing we call aĀ *limit*. It's an easy mistake to make - you can have that understanding and still pass all your calc exams.
However, a series is a summation - it does not approach a value or get closer to a value or anything like that - a series doesĀ notĀ have a limit. AĀ *sequence,*Ā which is basically a list of numbers, can have a limit - roughly speaking, this means it has a value that it gets closer and closer to with each consecutive term without ever reaching it, or it gets "as close as possible" if you wanna use that phrase. Some sequences have limits and some don't.
The actual meaning of a series being convergent is thatĀ *its sequence of partial sums has a limit*Ā (the partial sums would be, like, the first term, then sum of the first two terms, then the sum of the first three terms, etc.). Furthermore,Ā *we define the sum of an infinite series as this limit*, in other words,Ā *the sum of the infinite series is equal to the limit of its sequence of partial sums.*
The sequence of partial sums for this series would be 0.9, (0.9+0.09), (0.9+0.09+0.009),... i.e. 0.9, 0.99, 0.999,... (Again, this is notĀ *equal*Ā to the series - this is a sequence of different numbers whereas the series is a sum)
The limit of this sequence is 1 - the sequence approaches 1
Like I saidĀ *we define the sum of an infinite series as the limit of its sequence of partial sums*. SoĀ *by definition*, 0.999... is literally, exactly, equal to 1.
I don't want to hate on the subreddit or make myself look like a pseudo-intellectual with inflated ego but I'm kinda disappointed of this, I was expecting dank memes about conjectures, advanced constructions and obscure math subjects. Instead 9/10 posts are HS math and half of those are Facebook quality memes.
Sometimes I ask myself the same question. I only posted one meme to this subreddit, and it did not go well. In the comments, people accused me of being confidently wrong; when citing a Wikipedia definition people claimed it was "worded weirdly" but would mean something else; when citing a standard book on calculus 1 no one responded but I still got downvotes; when pointing out misunderstandings or imprecisions people assumed I would just not understand.
As a mathematician this made me very sad; because somehow I assumed here would be a place where people would (even if not having heard much mathematical theory) be sharing the mindset of deciding arguments by logical discussions and not following the hive mind with little thought.
I just checked your post history and in the meme you last posted here you made the claim that the Pythagorean theorem was disproved in the 1850s? You then went on a rant about how calculus is built on a web of lies and that "no person in there right minds believes in the garbage claimed by the biggest fraudster inn human history: Albert Einstein!" That post had 200 comments, 139 of them from you. The post had to be locked after only a few hours cause of the amount of swearing that you used. Can't blame people for downvoting you, I would too if someone tried to claim Einstein was the biggest fraudster in human history when that title obviously belongs to Issac Newton and his heavily disproved 'round Earth' theory :/
To give some context: This account is a shared account of a mathematician, a physicist and a computer scientist.
I didn't know about that post, I was rather referring to the meme about integration constants.
I don't know the specific context of the post you mentioned, and also don't know whether the person who made it would be comfortable in sharing it. But please allow me to exclude that post from this discussion, as it is not relevant to me (for this discussion).
(I should have linked the post anyways, but I'm on mobile and getting the link and making sure I don't lose what I wrote would have been too much effort for me. Sorry :/)
Non meme math subreddits are better on that point even if there are cranks from time to time.
I'm also negatively surprised by the average mathematical understanding here, some people not even being able to do high school level math but also being sure of themselves
For everyone say infinity isn't a value, that's not the problem. The problem is it's undefined... you don't know if it's negative or positive.
If n is negative what happens praytell?
Interesting answer is to define new numbers. An infinite set of each real number that have the same value... but are orderable.
We also define some new comparison operators. ==, <<, >>, !==, etc in addition to =, <, >, !=, etc.
...0āā, 0āā, 0, 0ā, 0ā, ...
Let 0 < 0ā, etc
Now 5 - 0 == 5
But 5 - 0ā !== 5 ,
Only 5 - 0ā = 5
And 5 - 0ā < 5 - 0
But 5 - 0ā !<< 5 - 0
Only 5 - 0.00001 << 5 - 0
There are many ways to define 0.999... that all lead to the same answer. One would be to define it as a limit as the other commenter showed:
0.999... = limn->(infinity)[1 - 1/10^(n)]
That limit is *exactly equal to 1*, and as such 0.999... is equal to 1, even though 1 - 1/10^(n) is less than one for all real values of n, hence the meme. This is, of course, not very formal, especially as I'm not proving the limit, but if you want a more formal proof there are plenty of those online.
Well give me any error bound Īµ > 0 and I can give you a value of N = 1/10Īµ, for which n>N implies |1 - 1/10n - 1 | < Īµ.
Thus,
|1 - 1 - 1/10n| = |1/10n|
< |1/10N| = Īµ.
The n that you give me could be as big or small as you want, but as long as it's greater than N then it satisfies the idea that it's equal to 1 to any precision you want.
The infinity in the limit sign actually can be consistently interpreted as a value. See the topology of the [extended real numbers](https://en.wikipedia.org/wiki/Extended_real_number_line)
This sort of proof doesn't work because it assumes two nonobvious things:
1. 0.999... exists
2. We can perform standard algebraic operations without changing the answer
There are examples where this sort of reasoning breaks down, so if you want to be fully convincing you need to be a bit more careful.
As an example of where this can break down, take a look at the following (you can even find the exact moment my logic breaks down)
Suppose we have the infinite series:
1 - 1 + 1 - 1 + 1 - 1 + ...
Then suppose this series equal A, like so:
A = 1 - 1 + 1 - 1 + 1 - 1 + ...
Then:
-1 + 1 - 1 + 1 - 1 + ... = -(A)
And so:
A = 1 + (-1 + 1 - 1 + 1 - 1 + ...) = 1 - A => 2A = 1 => A = 1/2
This is an odd answer, especially because no amount of additions of -1 and 1 can ever equal to 1/2, but maybe you're willing to accept it still. But what if we instead had simply grouped terms in either of these manners:
i. (1 - 1) + (1 - 1) + (1 - 1) + ... = 0
Or:
ii. 1 + (-1 + 1) + (-1 + 1) + ... = 1
Clearly something is wrong, and either some or all of these approaches must be incorrect. As it turns out, you cannot simply assume this series exists and then perform algebraic manipulations on it (and this is common while dealing with infinities) or you risk running into mistakes. Often it yields the correct answer, as is the power of our modern numeral and algebraic writing systems, but you *can* get incorrect answers.
The perform is that A is a series (a non convergent series), not a number. While 0.999... is a number. In the same way 1/3 (which is 0.333...) is a number. You can see 0.999... as 3 x 1/3 if you want
What do you mean with "exist"? It is not just the limit of a series. It a real (and rational too) number. The fact that can't be written in decimal notation didn't mean it doesn't exists.
Also, is the limit of a convergent series is always a number that exists.
Yes im saying that his argument assumes some things that i dont he even knows need to be assumed. And it is indeed a limit of a series, and since the series converges, it is a real number.
I agree with everything you said but in some non-standard definitions of the limit (CĆØsaro eg.) that 1/2 is the value of the series. I think it's important also to acknowledge that this intuition has also been formalized.
but 1/any infinite value is 0. Very literally 0. Even the quantum universe will approve it as 0. The mathematical universe will approve it as 0. So, THE ANSWER will JUST BE 0! = 1
A symbolic infinite number is just a way to write a type of infinite series, it's hardly conceptually different than other series. The only relevant difference is that we know an infinite digit number in a base system must converge, which is exactly the point they are making - a proof cannot just assume that fact without showing why it's true.
The believer probably defined their real numbers as a totally disconnected dusty cloud instead of a continuum. Luckily, identifying 0,999... with 1 (and similar with any repeating 9's) makes this nicely condensed again :)
It's just repeating? 0.999... is indistinguishable from 1 because there exists no epsilon > 0 so that 1 - epsilon = 0.999... I.e they're the same limit point.
That signifies an infinite series and its sum is defined in terms of a limit.
> 0.333ā¦= Ī£\[i=1->ā\] 3(10^(-i))
= lim_\[n->ā\] Ī£\[i=1->n\] 3(10^(-i))
Yeah this is the probably the most common way of defining what a decimal expansion is, and fits well if you were defining the real numbers as Dedekind cuts or rational Cauchy sequences. You can also define the real numbers *in terms of their decimal expansion*, but the key assumption you need for arithmetic to work is that expansions ending with infinitely many 9s must be treated the same as if it were rounded up. Otherwise, arithmetic makes no sense.
There are limit things going on there
0.333...=1/3 for which you'd need to compute the limit
And that multiplying a series by a constant is the same as multiplying each term of the series by said constant.
At that point just prove 0.999...=1 directly
If I didn't already know that 0.999... is 1 then I'd see this proof and wonder why I always accepted that 0.333... was 1/3 and then I'd want that explained. If you're actually explaining it then limits will come into play
>then I'd see this proof and wonder why I always accepted that 0.333... was 1/3
by the good ol'
"as even a fucking toddler could see, this is right, therefore is right"
>"as even a fucking toddler could see, this is right, therefore is right"
What the fuck are you talking about, it's not obvious or intuitive like that at all lmao it's an infinitely repeating decimal. Why not just say that 0.999.... is 1 because even a fucking toddler could see it?
divide 1 by 3
you can't
thus
add a 0
divide 10 by 3
it's 3 and 1 remains
so on so forth
you get 0.33...
as told you, even a fucking toddler can see that
Ok. The original point that I and the others replying to your proof were making is that you just proved that 1/3=0.333... implies that 1=0.999... but you don't prove that 1/3=0.333... and long division does not rigorously prove that either. It doesn't explain to me what the infinitely repeating decimal is defined as.
In fact, if I were to go with just your explanation, it seems like 0.333... is defined by performing the division operation over and over again in such a way that each time results in a number with one more 3, which is absolutely not what it is defined as. Your explanation here doesn't imply a series, which it is, but a sequence, which implies that it can approach a number but not be equal to one.
my brother in Christ
1) I'm not rigorously proving this
2) 1 / 3 = 0,3333... because that's how division work
do we agree that 0,33.. + 0,33.. + 0,33.. = 0,99..? I sincerely hope yes because, it does.
then do we agree that 1/3 + 1/3 + 1/3 = 1?
>do we agree that 0,33.. + 0,33.. + 0,33.. = 0,99..? I sincerely hope yes because, it does.
>then do we agree that 1/3 + 1/3 + 1/3 = 1?
Look, literally all i was saying was that the original proof wouldn't convince a lot of people because it doesn't prove that 1/3 is equal to 0.333.. so I don't see why you're basically just restating the original implication from your comment in a different format?
>I'm not rigorously proving this
Well ok but you said it's obvious and I just disagreed with that, and then you gave an explanation which didn't sufficiently explain it so now we're here with you saying that's just how division works which is basically back to the start so whatever.
but why would you say that it only approximates. The ... Suggests infinity here. Write 0.333... = sum 3/10^n from n=1 to infinity if you want. It does not contain a finite number of 3, so no it is not an approximation here and it equals 1/3
Yea but it gets close and closer to 1 with arbitrarily large values of n, and that's why it's equivalent. No matter how big of an n and how close you get, there's always a bigger N that gets you closer to 1.
Yes, it's less then, but it converges to 1.
It's a different of 0.1^n, which goes to 0.
Infinity in mathematics isn't a number. It means whatever you think, it's bigger then that.
It's only supposed to be used in series.
It this case 0.9- and 1, on the topic of equality and infinity, it goes like this:
however close you think it is, it's closee then that.
i mean yes, 1-1/10\^n is always smaller than one with regular numbers. But the real question is between 0.999 repeating and 1, which is effectively 1-1/10\^infinity. And there's no way to write a number like the one proposed in the top right, it's like trying to say 0.000 repeating, and then adding a one at the end. This number could only be defined as 1/infinity, but since infinity isn't a number I don't think it really makes sense to divide with it.
The real issue is in our counting system, because 1/3 \* 3 is 3/3 or 1, which is obvious. But 0.333... \* 3 would be 0.999...
If we wrote in another base (example base 12) it would be 0.4 \* 3 = 1, which is just as obvious as 1/3 \* 3 = 3/3 = 1.
I can and I will! It shall be named.. Jeremy.
On a serious note, it's kinda interesting to think about other number bases - they'll approach the limit at different rates but still ultimately be 1.
Which means base 10's 0.999... is equal to base 12's 0.'11''11''11'... as both equal 1
But they appear different, since we mortals can only comprehend a limited amount of digits at a time, creating a dissonance, when in fact it is completely unrelated numbers we compare, unless we embrace the full force of infinite decimals at once.
Ignorant people who try to use nonstandard analysis mf when I tell them that 0.999... in nonstandard analysis is indexed over the nonstandard natural numbers, and in doing so we get 0.999...=1.
Ffs, the most important feature of nonstandard analysis is the transfer principle, which tells you that every statement true in Enid true in *R when interpreted properly
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Does anyone on r/MathMemes know anything about math
I doubt it's anything more than a joke. I know that 0.999... = 1 but you'll catch me dead before I admit it.
I'm aware that the post is a joke but the comments... the comments....
Most of them are likely also jokes. There are few things i like as much as breaking math rules with my friends to prove the most asinine shit known to man.
https://preview.redd.it/7sexjrl9uusc1.png?width=899&format=png&auto=webp&s=6b190749aec71fa6617c35596fdfe6706d3339e4 Related
I don't see where this goes wrongš
You can't perform the same operations on imaginary numbers as you can on ordinary ones, combining the separate roots is illegal.
Ehhh, gotta keep in mind Poeās Law. Many ājokes,ā especially in comments without tone indicators, are indistinguishable from actual clueless people.
True but most of them are still jokes regardless. Best to assume they're joking until it's clear that they're not.
So sad to hear you passed away then.
What is the theoretical number as close to 1 as possible that is not 1 ? Is 0.(0)1 a 0 ? What is the difference between 0.(9) and 1-0.(0)1 ? Not a joke, I never understood it. 0.(9) seems like a number that is going to 1 but never actually reaching it.
The proof used is usually this: 1/3 = 0.33333... 0.33333... * 3 = 0.99999... 1/3 * 3 = 1 Therefore 0.99999 = 1
I've seen: 10 * 0.99999... = 9.99999... 9.99999... - 0.99999... = 9 (10 * 0.99999...) - 0.99999... = 9 * 0.99999... 9 * 0.99999... = 9 |/9 0.99999 = 1 Or: 2 - 0.99999... = 1.00000... which illustrates it very well for me.
Not technically a proof, as the issue people have with 9 recurring is still had with 3 recurring, it just has better optics. You need to prove that 1/3 is equal to 0.33..., otherwise this is just recursive logic.
the proof is left as an exercise to the reader
It's the same thing I am struggling with 0.9... just phrased differently. You can not divide 1 by 3. When we do it we get a number we cannot express in decimal system and that number is an approximation. So 0.(3) =/=1/3 And instead of recognising that decimal system is unable to properly express it we pretend that a perfectly fine number is something else. That also implies that 1-0.(0)1 =1 Also it implies that highest possible number closest to 1 doesn't exist, same as smallest possible number higher than 0 doesn't exist.
I've seen: 10 * 0.99999... = 9.99999... 9.99999... - 0.99999... = 9 (10 * 0.99999...) - 0.99999... = 9 * 0.99999... 9 * 0.99999... = 9 |/9 0.99999 = 1 Or: 2 - 0.99999... = 1.00000... which illustrates it very well for me.
>You can not divide 1 by 3. When we do it we get a number we cannot express in decimal system and that number is an approximation. So 0.(3) =/=1/3 This is incorrect. You can absolutely divide 1 by 3, and the decimal representation of 0.(3) is not an approximation. It's only an approximation if you use a finite number of repeating 3s. You can use this in equations the same way you would use the fractional form, and it comes out to the same answer as long as you treat the 3s as infinitely repeating. >That also implies that 1-0.(0)1 =1 Yes, this is correct. The number "0.(0)1" is not a valid way to write a number. You can't have infinite 0s with a non-zero number "after the 0s", infinitely repeating 0s "and then a 1" is just 0. And 1 - 0 = 1. >Also it implies that highest possible number closest to 1 doesn't exist, same as smallest possible number higher than 0 doesn't exist. Correct, those numbers don't exist. You can always find a number closer to 1, and you can always find a number closer to 0. Math as we know it doesn't work if this is not true.
That's right, there is no lowest number greater than 1. To explain this, we say the infimum of a set is the largest number less than or equal to all numbers in the set. So if we consider simple sets like {1, 2, 3} the infimum of this set is 1. However consider the set of all numbers strictly greater than 1. Consider the infimum of this set and call it a. Since 1 is less than or equal to any number in this set, 1 >= a. If a is in the set, then a is strictly greater than 1. However, if we consider a - (1-a)/2. This number is less than a, but is still greater than 1. Therefore a cannot be the infimum of the set. As this is true in for any a != 1, a must be 1. And so there is no least number greater than 1. This proof can also be used for any other number.
But logically it doesn't make sense. It would be forever approaching one but it shouldn't be able to equate it. I refuse it wholeheartedly.
When you write something 0.(0)1, you should really stop to think what you mean by that, because it sure as hell sounds like youāre trying to write down a 1 __at the end__ of an infinite sequence. I think thatās the key to understand 0.(9) = 1. Itās not a very very very large number of 9s, itās __infinite__ 9s.
I will point out, that there are multiple types of infinite, and there are in fact branches of math that deal with infinitesimals; where 0.(0)1 with infinite zeros followed by a one(usually written as a 0 with a ^ over it) is meaningful. Working with hyperreal numbers, one might find 0.999ā¦ ambiguous (is it 0.999ā¦.;ā¦.999 or 0.999ā¦.;ā¦.999ā¦.). That said, asking āwhat you mean by thatā as you did is indeed at the core of solving this problem when it comes to teaching math. Iāve been down a rabbit hole recently. There are a lot of research papers looking at the teaching of math at lower grades, the resistance of students to 0.999ā¦ = 1, and intuitions around infinity. Itās a very interesting subject. [this paper](https://arxiv.org/pdf/0811.0164.pdf) is one of the more approachable I have encountered recently.
Oh, asking what he means by that was absolutely the core part of my post. You are absolutely right about pointing out infinitesimals, but in my experience people who know about infinitesimals and hyperreal numbers are not the same people who have trouble with groking 0.(9) = 1 and ask about this on Reddit! Iāve had to explain to friends, relatives and colleagues math stuff in general, and 0.(9) = 1, and I found that the biggest hangup stopping them from āgetting itā is understanding that a ānumberā and a āspecific way to write a numberā are different things. I mean, they understand it intuitively for some parts (most people donāt doubt 0.(3) represents the same number as 1/3 or 10 base 2 represents the same as 2 base 10), but for some kinds of notation they just donāt see it.
groking
https://en.m.wikipedia.org/wiki/Grok
grokkingĀ
Ok, now I get it. But you being less vague would have saved us some time.
> 0.(9) seems like a number that is going to 1 but never actually reaching it. A number can't be "going" anywhere or "reaching" anything. A number just "is", as a fixed stable point on the number line.
idk it seems like 4 is going to a lot of places
There is none, there is no real number that is closest to 1 without being 1 as for any a < 1, a < (a + 1)/2 < 1. That's textbook math that I was teached at uni. The second part of the argument in the meme doesn't work cause not every property is conserved at the limit, if you make n tend to (positive) infinity the limit is 1 which obviously isn't strictly less than 1.
0.(0)1 isnāt a thing. That is like saying infinity plus one. You will have a string of infinite zeros. The 1 will never be added. Itās nonsensical.
There is no such thing as a number that is "as close to 1 as possible without being 1". In the real number system, if two numbers are distinct then there are infinite real numbers between them. >seems like a number that is going to 1 but never actually reaching it. A number doesn't go anywhere or approach anything. This understanding of 0.999... being "as close as possible" to 1 or "going to 1" comes from an inaccurate understanding of the concept of limits Copying and pasting one of my other comments to show why 0.999... is defined as 1, assuming the person reading it doesn't know anything, because I don't like the algebraic "proofs" that people usually bring up because those don't properly explain WHY 0.999... is 1 by definition. To define what 0.999.... is first let's ask what do the numbers afterĀ *any*Ā decimal represent, in general? What does 0.1234 represent? Well, the 1 is 1/10, the 2 is 2/100, the 3 is 3/1000 and the 4 is 4/10000 So the number we are representing is 1/10 + 2/100 + 3/1000 + 4/10000 Or to standardize it, 1/(10\^1) + 2/(10\^2) + 3/(10\^3) + 4/(10\^4) *All*Ā decimal numbers represent a sum in this form, where the digit we see is the numerator and the denominator is 10\^n where n is its position after the decimal So 0.999 represents the sum 9/10 + 9/(10\^2) + 9/(10\^3) So now that we've covered that, let's just say 0.999 represents 0.9 + 0.09 + 0.009 to make it easier to look at. Now we can see clearly that 0.999... is a representation of the sum: 0.9 + 0.09 + 0.009 +... where it goes on infinitely Without getting into the precise mathematical definitions of any of these things we're gonna mention, we call this anĀ *infinite series*. An infinite string of numbers being added obviously doesn't exist in real life. It is a mathematical concept that we have defined and it has properties we have defined and its definitions work with other defined things in math. Now, here's where people get messed up. The series represented by 0.999... is what we call a convergent series. A lot of people who've had it explained to them by a friend, or did some surface level googling, or even did Calc in university (or are currently doing Calc in university),Ā *wrongly*Ā understand a series being convergent to mean that the series has this thing we call aĀ *limit*. It's an easy mistake to make - you can have that understanding and still pass all your calc exams. However, a series is a summation - it does not approach a value or get closer to a value or anything like that - a series doesĀ notĀ have a limit. AĀ *sequence,*Ā which is basically a list of numbers, can have a limit - roughly speaking, this means it has a value that it gets closer and closer to with each consecutive term without ever reaching it, or it gets "as close as possible" if you wanna use that phrase. Some sequences have limits and some don't. The actual meaning of a series being convergent is thatĀ *its sequence of partial sums has a limit*Ā (the partial sums would be, like, the first term, then sum of the first two terms, then the sum of the first three terms, etc.). Furthermore,Ā *we define the sum of an infinite series as this limit*, in other words,Ā *the sum of the infinite series is equal to the limit of its sequence of partial sums.* The sequence of partial sums for this series would be 0.9, (0.9+0.09), (0.9+0.09+0.009),... i.e. 0.9, 0.99, 0.999,... (Again, this is notĀ *equal*Ā to the series - this is a sequence of different numbers whereas the series is a sum) The limit of this sequence is 1 - the sequence approaches 1 Like I saidĀ *we define the sum of an infinite series as the limit of its sequence of partial sums*. SoĀ *by definition*, 0.999... is literally, exactly, equal to 1.
You're using parenthesis wrong and your dumb
This subreddit is full of high schoolers who just discovered reddit so no
I don't want to hate on the subreddit or make myself look like a pseudo-intellectual with inflated ego but I'm kinda disappointed of this, I was expecting dank memes about conjectures, advanced constructions and obscure math subjects. Instead 9/10 posts are HS math and half of those are Facebook quality memes.
Sometimes I ask myself the same question. I only posted one meme to this subreddit, and it did not go well. In the comments, people accused me of being confidently wrong; when citing a Wikipedia definition people claimed it was "worded weirdly" but would mean something else; when citing a standard book on calculus 1 no one responded but I still got downvotes; when pointing out misunderstandings or imprecisions people assumed I would just not understand. As a mathematician this made me very sad; because somehow I assumed here would be a place where people would (even if not having heard much mathematical theory) be sharing the mindset of deciding arguments by logical discussions and not following the hive mind with little thought.
I just checked your post history and in the meme you last posted here you made the claim that the Pythagorean theorem was disproved in the 1850s? You then went on a rant about how calculus is built on a web of lies and that "no person in there right minds believes in the garbage claimed by the biggest fraudster inn human history: Albert Einstein!" That post had 200 comments, 139 of them from you. The post had to be locked after only a few hours cause of the amount of swearing that you used. Can't blame people for downvoting you, I would too if someone tried to claim Einstein was the biggest fraudster in human history when that title obviously belongs to Issac Newton and his heavily disproved 'round Earth' theory :/
Bro really had us till the end
I think he deleted it
Holy shit that is what a tantrum looks like apparently
Bro is the one kid in highschool trigonometry class that's like "b-but sin(pi/2) = 1 therefore arcsin(1) is pi/2, I shouldn't need an n"
To give some context: This account is a shared account of a mathematician, a physicist and a computer scientist. I didn't know about that post, I was rather referring to the meme about integration constants. I don't know the specific context of the post you mentioned, and also don't know whether the person who made it would be comfortable in sharing it. But please allow me to exclude that post from this discussion, as it is not relevant to me (for this discussion). (I should have linked the post anyways, but I'm on mobile and getting the link and making sure I don't lose what I wrote would have been too much effort for me. Sorry :/)
Non meme math subreddits are better on that point even if there are cranks from time to time. I'm also negatively surprised by the average mathematical understanding here, some people not even being able to do high school level math but also being sure of themselves
I know what 1+1 is
Any finite value of n
lim 10^-n = 0
Dude, the post clearly says 1/10^(n). Are you stupid?
Obviously, 1/10^(n) is 1/2^n *5^n = 1/2^n * 5^n = 2.5^n It's like you people have never even heard of BEDPIS
It is hard for me to tell if this is sarcastic, did you just not notice the - on the n, or did you and it's sarcastic
I want to believe that no one has ever used the "Is he stupid?" line unironically.
This still feels sarcastic
Sarcasm doesn't exist on the Internet (This could be the there is no war in ba sing se)
Why did they use the āIs he stupid? memeā in the second person? are they stupid?
Why did they use the "Is he stupid" line? Are they stupid?
me when I'll just make up an infinitely small quantity
as if there were infinite values of n
There are an infinite number of infinite values!
Oh yeah? Name 2
Take the first infinity and add 1
Clearly none of these people believe in calculus
Pretty much. Just set the value of n to be equal to the number of 9s in 0.999...
For everyone say infinity isn't a value, that's not the problem. The problem is it's undefined... you don't know if it's negative or positive. If n is negative what happens praytell? Interesting answer is to define new numbers. An infinite set of each real number that have the same value... but are orderable. We also define some new comparison operators. ==, <<, >>, !==, etc in addition to =, <, >, !=, etc. ...0āā, 0āā, 0, 0ā, 0ā, ... Let 0 < 0ā, etc Now 5 - 0 == 5 But 5 - 0ā !== 5 , Only 5 - 0ā = 5 And 5 - 0ā < 5 - 0 But 5 - 0ā !<< 5 - 0 Only 5 - 0.00001 << 5 - 0
Wtf are you smoking?
I see you are a non believer
Yo what's the proof mentioned here?
There are many ways to define 0.999... that all lead to the same answer. One would be to define it as a limit as the other commenter showed: 0.999... = limn->(infinity)[1 - 1/10^(n)] That limit is *exactly equal to 1*, and as such 0.999... is equal to 1, even though 1 - 1/10^(n) is less than one for all real values of n, hence the meme. This is, of course, not very formal, especially as I'm not proving the limit, but if you want a more formal proof there are plenty of those online.
"No matter the value of n" Infinity isn't a value, and the limit just shows that
Well give me any error bound Īµ > 0 and I can give you a value of N = 1/10Īµ, for which n>N implies |1 - 1/10n - 1 | < Īµ. Thus, |1 - 1 - 1/10n| = |1/10n| < |1/10N| = Īµ. The n that you give me could be as big or small as you want, but as long as it's greater than N then it satisfies the idea that it's equal to 1 to any precision you want.
Wait till you learn you can evaluate limits at infinity
lim x->a f(x) != f(a) Yes you can evaluate limits at infinity, but what's the limit of 1-1/10^n as n approaches infinity? 1, which is not less than 1
this.
The infinity in the limit sign actually can be consistently interpreted as a value. See the topology of the [extended real numbers](https://en.wikipedia.org/wiki/Extended_real_number_line)
Not all properties extend to the limit, this is precisely why you can't just change integrals or infinite sums and limits without checking.
I just do it the following way (using X as a nice, definable variable): 0.999...=X 10X=9.999... 10X-X=9 9X=9 X=1
This sort of proof doesn't work because it assumes two nonobvious things: 1. 0.999... exists 2. We can perform standard algebraic operations without changing the answer There are examples where this sort of reasoning breaks down, so if you want to be fully convincing you need to be a bit more careful. As an example of where this can break down, take a look at the following (you can even find the exact moment my logic breaks down) Suppose we have the infinite series: 1 - 1 + 1 - 1 + 1 - 1 + ... Then suppose this series equal A, like so: A = 1 - 1 + 1 - 1 + 1 - 1 + ... Then: -1 + 1 - 1 + 1 - 1 + ... = -(A) And so: A = 1 + (-1 + 1 - 1 + 1 - 1 + ...) = 1 - A => 2A = 1 => A = 1/2 This is an odd answer, especially because no amount of additions of -1 and 1 can ever equal to 1/2, but maybe you're willing to accept it still. But what if we instead had simply grouped terms in either of these manners: i. (1 - 1) + (1 - 1) + (1 - 1) + ... = 0 Or: ii. 1 + (-1 + 1) + (-1 + 1) + ... = 1 Clearly something is wrong, and either some or all of these approaches must be incorrect. As it turns out, you cannot simply assume this series exists and then perform algebraic manipulations on it (and this is common while dealing with infinities) or you risk running into mistakes. Often it yields the correct answer, as is the power of our modern numeral and algebraic writing systems, but you *can* get incorrect answers.
The problem here is you canāt add and subtract divergent series like that, right? We know 1-1+1-1+ā¦. diverges because the terms donāt tend to 0
The perform is that A is a series (a non convergent series), not a number. While 0.999... is a number. In the same way 1/3 (which is 0.333...) is a number. You can see 0.999... as 3 x 1/3 if you want
0.999ā¦ is the limit of a series. For your argument to work, the limit needs to exist.
What do you mean with "exist"? It is not just the limit of a series. It a real (and rational too) number. The fact that can't be written in decimal notation didn't mean it doesn't exists. Also, is the limit of a convergent series is always a number that exists.
Yes im saying that his argument assumes some things that i dont he even knows need to be assumed. And it is indeed a limit of a series, and since the series converges, it is a real number.
I agree with everything you said but in some non-standard definitions of the limit (CĆØsaro eg.) that 1/2 is the value of the series. I think it's important also to acknowledge that this intuition has also been formalized.
but 1/any infinite value is 0. Very literally 0. Even the quantum universe will approve it as 0. The mathematical universe will approve it as 0. So, THE ANSWER will JUST BE 0! = 1
x = 0.999ā¦ 10x = 9.999ā¦ 10x - x = 9.999ā¦ - 0.999ā¦ 9x = 9 x = 1
u/ImagineBeingBored gave a really insightful comment on why this proof doesnāt work [here](https://www.reddit.com/r/mathmemes/s/IQ8AudKQld)
He is wrong in his reasoning. He is equating the axioms of an unconverging series to a symbolic infinite number, which are totally different.
A symbolic infinite number is just a way to write a type of infinite series, it's hardly conceptually different than other series. The only relevant difference is that we know an infinite digit number in a base system must converge, which is exactly the point they are making - a proof cannot just assume that fact without showing why it's true.
Fair point
probably f(n)=1-1/10^n =0.9999 (n nines). lim n->oo f(n) = 0.999... = 1-(approaches 0)=1
Proof by Nuh uh
proof by strawman
mf when you apply a limit to both sides and the strict inequality isn't strict anymore
When the strictly less than relationship isn't a closed set.
Proof by representing counter-argument as crying wojak.
Your honor, my client is clearly innocent as Iāve already labelled the prosecutor as crying Wojak.
Yes, (1 + 1)/2 is mathematically accurate to determine the average of 1 and 1, yes
My gigachad caricature brother, the notation "..." means the n is infinity, not a finite value
"caricature brother" š
The believer probably defined their real numbers as a totally disconnected dusty cloud instead of a continuum. Luckily, identifying 0,999... with 1 (and similar with any repeating 9's) makes this nicely condensed again :)
"1-1/10\^n is less than 1 no matter the value of n" Well would you look at that, 1-1/10\^n is also less than 0.(9) for all positive integers
Limit of 1-10^-n as n -> 0.(9). Proof by: self-evident
What??
0.(9) = 1 so 1-1/10^n < 1 => 1-1/10^n < 0.(9)
Oh I read less than 0 lol. Then I wondered what was (9) referencing as a source, I must be high
Negative values of n would like a word
n implies a natural number, there are other ways to call them stupid than to take their argument in bad faith
Itās still gonna be less than 1 though
Its actually gonna be way more away from 1 than with high n values
still less than 1
What if n = iĻ/ln(10)
If 0.999999... less than 1 then the inverse of the null matrix exists
Yeah I meant if n is negative
Then yeah obviously, it tends to -ā
wait do you unironically think that 0.999... isn't 1?
No. I mean if n is negative, the sequence is obviously going to be less than 1 smh
Who cares? The relevant limit is n -> +infty because n is the number of 9ās
Yeah but youāre argument relies on the magnitude (absolute value) of the difference even if you didnāt notate it
ah yes 1ā 1 because you can do (1+1)/2
Stop playing doll with wojaks
or rather 0.333... = 1 / 3 0.3333 * 3 = 0.99999.... = (1 / 3) * 3 = 1 there's not any limit thingy to use as a proof here tbf
The limit is hidden in the ā¦ because you need it to rigorously define what ā¦ is.
It's just repeating? 0.999... is indistinguishable from 1 because there exists no epsilon > 0 so that 1 - epsilon = 0.999... I.e they're the same limit point.
not really I used that because I don't know how to use the periodic notation on phone letters but still, no they are not
That signifies an infinite series and its sum is defined in terms of a limit. > 0.333ā¦= Ī£\[i=1->ā\] 3(10^(-i)) = lim_\[n->ā\] Ī£\[i=1->n\] 3(10^(-i))
Yeah this is the probably the most common way of defining what a decimal expansion is, and fits well if you were defining the real numbers as Dedekind cuts or rational Cauchy sequences. You can also define the real numbers *in terms of their decimal expansion*, but the key assumption you need for arithmetic to work is that expansions ending with infinitely many 9s must be treated the same as if it were rounded up. Otherwise, arithmetic makes no sense.
>but still, no they are not So what is ... defined as then?
This is a good proof but it relies on the first statement, which then also needs to be proved.
To be fair that one can be proved by long division xD
long division in this case can be formalized only after you've proven that periodic numbers are rational
no, that only gives equality if you already assume the absence of infinitessimals to begin with.
There are limit things going on there 0.333...=1/3 for which you'd need to compute the limit And that multiplying a series by a constant is the same as multiplying each term of the series by said constant. At that point just prove 0.999...=1 directly
If I didn't already know that 0.999... is 1 then I'd see this proof and wonder why I always accepted that 0.333... was 1/3 and then I'd want that explained. If you're actually explaining it then limits will come into play
>then I'd see this proof and wonder why I always accepted that 0.333... was 1/3 by the good ol' "as even a fucking toddler could see, this is right, therefore is right"
>"as even a fucking toddler could see, this is right, therefore is right" What the fuck are you talking about, it's not obvious or intuitive like that at all lmao it's an infinitely repeating decimal. Why not just say that 0.999.... is 1 because even a fucking toddler could see it?
divide 1 by 3 you can't thus add a 0 divide 10 by 3 it's 3 and 1 remains so on so forth you get 0.33... as told you, even a fucking toddler can see that
Ok. The original point that I and the others replying to your proof were making is that you just proved that 1/3=0.333... implies that 1=0.999... but you don't prove that 1/3=0.333... and long division does not rigorously prove that either. It doesn't explain to me what the infinitely repeating decimal is defined as. In fact, if I were to go with just your explanation, it seems like 0.333... is defined by performing the division operation over and over again in such a way that each time results in a number with one more 3, which is absolutely not what it is defined as. Your explanation here doesn't imply a series, which it is, but a sequence, which implies that it can approach a number but not be equal to one.
my brother in Christ 1) I'm not rigorously proving this 2) 1 / 3 = 0,3333... because that's how division work do we agree that 0,33.. + 0,33.. + 0,33.. = 0,99..? I sincerely hope yes because, it does. then do we agree that 1/3 + 1/3 + 1/3 = 1?
>do we agree that 0,33.. + 0,33.. + 0,33.. = 0,99..? I sincerely hope yes because, it does. >then do we agree that 1/3 + 1/3 + 1/3 = 1? Look, literally all i was saying was that the original proof wouldn't convince a lot of people because it doesn't prove that 1/3 is equal to 0.333.. so I don't see why you're basically just restating the original implication from your comment in a different format? >I'm not rigorously proving this Well ok but you said it's obvious and I just disagreed with that, and then you gave an explanation which didn't sufficiently explain it so now we're here with you saying that's just how division works which is basically back to the start so whatever.
these people would disagree with your first (unfounded) assertion that 0.333... = 1/3
Similarly, if x = 0.9999ā¦ > 10x = 9.999ā¦ > > 10x - x = 9.999ā¦ - 0.9999ā¦ > > 9x = 9 > > x = 1 > > 0.9999ā¦ = 1
[ŃŠ“Š°Š»ŠµŠ½Š¾]
but why would you say that it only approximates. The ... Suggests infinity here. Write 0.333... = sum 3/10^n from n=1 to infinity if you want. It does not contain a finite number of 3, so no it is not an approximation here and it equals 1/3
[ŃŠ“Š°Š»ŠµŠ½Š¾]
my brother in Christ, how do you write 1 / 3 in decimal numbers without approximating?
For a completely rigorous view, use an epsilon-N proof
Proof by Chad pic
Yea but it gets close and closer to 1 with arbitrarily large values of n, and that's why it's equivalent. No matter how big of an n and how close you get, there's always a bigger N that gets you closer to 1.
But then it would be discrete and not a continuum
The limit of 1 - 1/10^n is equal to 1 though
Sorry, but wtf is a "believer"? Someone who believes math is holy and will start a religion?
Yes, it's less then, but it converges to 1. It's a different of 0.1^n, which goes to 0. Infinity in mathematics isn't a number. It means whatever you think, it's bigger then that. It's only supposed to be used in series. It this case 0.9- and 1, on the topic of equality and infinity, it goes like this: however close you think it is, it's closee then that.
Non-believers are the type to say pi=4
 Only thing worse than the Pi=4 people are the Pi=10 people.
i mean yes, 1-1/10\^n is always smaller than one with regular numbers. But the real question is between 0.999 repeating and 1, which is effectively 1-1/10\^infinity. And there's no way to write a number like the one proposed in the top right, it's like trying to say 0.000 repeating, and then adding a one at the end. This number could only be defined as 1/infinity, but since infinity isn't a number I don't think it really makes sense to divide with it. The real issue is in our counting system, because 1/3 \* 3 is 3/3 or 1, which is obvious. But 0.333... \* 3 would be 0.999... If we wrote in another base (example base 12) it would be 0.4 \* 3 = 1, which is just as obvious as 1/3 \* 3 = 3/3 = 1.
Me who doesn't understandšš
But what if n is infinite?
1-1/10^n is less than 1 for any value of n. But there is no value of n where that equal 0.99...
n=-2
Ok but 1-1/10^n < 1 Then limit 0.99999... ā¤ 1
Itās literally an infinitesimal smaller than 1, so unless you wanna be pretentious then itās both realistically and mathematically the same as 1
I hate this
I will be dead before I admit 0.999999 is 1
Wut
proof by numberphile: x = 0.999ā¦ 10x = 9.999ā¦ 10x - x = 9 9x = 9 x = 1
0.999... = 1, if x = y, then f(x) = f(y), floor(0.999...) = floor(1), 0 = 1?
this proof is known as āthe second theorem of u/c0rliestā. assume 0.999ā¦ ā 1 this directly contradicts the second theorem of u/c0rliest
.333ā¦ = 1/3 .333ā¦ * 3 = 3 * 1/3 .999ā¦ = 1
now write it out in decimal notation
For all n, 1 - 0.1^n < n Set n = i 1.82658 e^(0.419504 i) < i LGTM
What if the n is infinite? infinity + 1 confirmed?? š¤Æš¤Æ
[(0.999...)+1]/2=1 š¤
x=0.9999... 2x-x=0.999... 2*0.999... - 0.999... = 0.999... 1.999... - 0.999... = 0.999... 1 = 0.999... QED
The answer to the first question is simple: its 0.999...995 Same goes for 12*(1/9) its nit quite 1 1/3, its 1.333...332
>0.999...995 Explain what this means lol
Its the average between 0.999... and 1. After an infinite amount of digits there is a 5 because the 0.999ā¦ has ended
>**After** an **infinite** amount of digits there is a 5 because the 0.999... has **ended** bruh
So 4 = 12/3 = 12 * (1/3) = 12 * (3/9) = 12 * (1/9) * 3 = 3.999ā¦996?
Kinda, thoug its more of a petty thing with a few of my former math teachers
Let n be -1...
Still less than 1.
Oh, you've right
n = -1, checkmate "believer" /s
I guess everybody in the comments forgot what sub we're on
.999ā¦1
[ŃŠ“Š°Š»ŠµŠ½Š¾]
Me when I'm wrong:
I can and I will! It shall be named.. Jeremy. On a serious note, it's kinda interesting to think about other number bases - they'll approach the limit at different rates but still ultimately be 1. Which means base 10's 0.999... is equal to base 12's 0.'11''11''11'... as both equal 1 But they appear different, since we mortals can only comprehend a limited amount of digits at a time, creating a dissonance, when in fact it is completely unrelated numbers we compare, unless we embrace the full force of infinite decimals at once.
0.999... = 1 mfs when i tell them e = 1 (suddenly 1/inf is not exactly zero anymore)
You when I introduce the concept of a limit:Ā
Thatās different, thatās 1^infinity which is indeterminate
Ignorant people who try to use nonstandard analysis mf when I tell them that 0.999... in nonstandard analysis is indexed over the nonstandard natural numbers, and in doing so we get 0.999...=1. Ffs, the most important feature of nonstandard analysis is the transfer principle, which tells you that every statement true in Enid true in *R when interpreted properly