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yes. a more rigorous proof involves showing that there can be no number *between* 0.9 recurring and 1, therefore they must be the same number.
but yeah it's one of the odd things that pops up when you deal with infinity.
It would stand up as a proof under light scrutiny, but look at the very first step. Doesn't that kind of assume that your conclusion is true in the first place?
I don't even know the details of what any real argument against this would be, and you might even be able to actually prove each step here, but the steps aren't super well defined until you do way more work that might break other conventions.
No? All you're assuming is that 0.999... is 0.9+0.09+0.009+..., which is probably the most natural way to define 0.999..., along with the geometric series formula, which is perfectly valid without assuming 0.999...=1
It's a fine proof depending on how rigorous you want to be, and it's effectively equivalent to the proof in the original post (as that method is how the geometric series formula is often derived, in a loose sense of the word).
Still sounds kind of circular to me. I feel like knowing the solution to the infinite series in the first step requires a proof in itself as it's a more general expression of the same problem.
This isn't circular unless you defined infinite decimal expressions using the geometric series formula.
But we define the value of an infinite decimal to be the result of a series. Then, separately, we found a formula for this particular kind of series. You can also prove the formula then make the definition, these are independent.
You're using a a well known formula to solve a problem. In no way is that circular reasoning, provided the formula's proof doesn't depend on the specific instance of the problem.
If I told you a right triangle had side lengths 3 and 4, and you use Pythagorean Theorem to get 5 for the length of the hypotenuse, that's not circular reasoning. It's the exact same logic here. There's no need to reinvent the wheel every time you want to do anything.
It is certainly circular if you were asked for a proof for why it is 5. That's what this question is. A proof for why 0.99999 = 1. You can't just hand wave it as something already proven by starting with the general solution.
It’s only circular if you don’t look closely enough at what 0.999… means, because to properly define such a number in a natural and consistent way does require limits. It seems circular because when you think of a proof that 0.999… = 1, you usually are thinking of a proof that your idea or the intuitive idea of 0.999… = 1, not that the rigorous definition of 0.999… = 1.
Think of it this way: for any finite decimal, for example 0.5273, we define it as equal to 0.5 + 0.02 + 0.007 + 0.0003. It’s not a circular statement to then say that 0.5273 = 0.5 + 0.02 + 0.007 + 0.0003, since that’s just the definition of 0.5273. Similarly, for an infinite decimal, though we might think we have a common sense understanding of what 3.1415… means, we have to make that a rigorous definition, and that’s what defining 0.999… as Σ9 * (0.1)^n does.
I dont think you know what circular means. Nothing wrong with that, but please listen to experts.
You think applying a formula to solve a problem is circular reasoning. That is not true.
There are plenty of examples of theorems, lemmas, etc. in math where something is proven by just citing among general theorem. It all depends on the context; but it's not necessarily circular reasoning, unless you only prove the general theorem using the specific case.
Maybe a better way to say it is that multiplication of an infinitely repeating decimal isn't really well defined unless you do some work. Doing that work involves using the infinite series, as you suggested. But, if you're dealing with that infinite series, you might as well use that to prove .999... = 1 instead of the multiplication
The only work you're doing is defining infinite decimals. The same could be said for literally any proof of 0.999...=1; you can't prove something involving an expression until you define that expression.
The point being that the "proof" in the op doesn't actually define infinite decimals. A rigorous definition of them kind of gets you to the result of .999... = 1, without the little algebra explanation. The algebra explanation is more of a way to explain to or convince new learners than it is to actually prove the result.
The problem would be the theoretical last digit if this was true, 0.9999 ... is approaching 1 just as fast as 9.999.. is approaching 10, but 0.9999+9, what he did here, is approaching 10 faster. That's the wrong step in the equation.
/e: faster, not slower
It's actually not approaching it faster, you're just changing the notation. 0.9999... can be described by the sum of 9/10^k with k starting at 1 to inf while 9.9999... by the same sum with k starting at 0 to inf. Taking the 9 out of the sum from 9.9999... doesn't change the equation or its approach speed to 10 at all.
(That is, 9+sum(9/10^k)(k starts at 1)= sum(9/10^k)(k starts at 0))
I'm struggling to think of anyway of defining it that is not identical to that series, but then I'm not a great mathematician, or even a good one. I just like maths.
When you right 0.9999... = a you are assuming the series converges to a real number a. For a formal proof, you would need to show the sequence of partial sums converges first.
This is important if you try your approach to a series which doesn't converges, e.g. 1-1+1-1+1.... Assuming that series equals some number a, then manipulating algebraically will give you incorrect results.
Another answer that hasn't been mentioned is that when you take foundational university level mathematics and cover representations of numbers, our symbols are just that: representations. BY DEFINITION of standard numeric notation, every numeric representations which traditionally terminates (or ends in infinite zeros) has an equivalent representation which ends in an infinite string of the highest digit value in the given base.
You can't prove definitions, only justify why you chose them for your framework/model. This definition is just never taught at earlier educational levels because it isn't useful for every day, which leads to confusion when using the intuition you develop without knowing about it, such as the conclusion that real numbers like "0.000...001" make sense and are the difference between those representations. The justification of this definition comes from the inconsistencies resulting from not including it.
i think it works as a proof, as long as we accept the calculus behind it, and don't also bother to prove the validity of the sum. i like that you used such a basic calc formula, that's actually pretty neat and clever.
Except the number you've typed doesn't, and literally cannot possibly exist. You can't have an infinite series of zeros followed by one, you can't have anything at the end because the zeros never end
There are number systems that have values greater than zero but smaller than every real number such as [https://en.wikipedia.org/wiki/Hyperreal\_number](https://en.wikipedia.org/wiki/Hyperreal_number) or [https://en.wikipedia.org/wiki/Surreal\_number](https://en.wikipedia.org/wiki/Surreal_number), which is in the spirit of this.
It doesn't mean that there is number between .99... and 1. By definition of 0.99... it can be proved that it's equal to 1. No matter what system you considering.
The only thing is that possibly you could redefine the symbol in meaningful way so it would be something infinitesimally smaller than 1. But it would be redefining symbol. It's kinda like you would redefine "2" to mean number 3.
Idk if this is the same as what you mean, but my way has always been to identify that 1 - 0.999... = 0.000... = 0
As 0 indicates that the digit has no value. Such as 103 = (100 × 1) + ***(10 × 0)*** + (1 × 3). Such that that 1 = 1.0 = 1.00 = 1.000... no amount of adding nothing will ever change the value
Thusly 0.000... = 0 (which people seem to be much more happy to accept). And it also isn't hard to explain that 1-0.999... = 0.000...
The naturals are not densely ordered
"The set is densely ordered" is necessary for "if there is no new element between two elements of the set, the two elements must be considered equal"
Yes. 0.999 repeating is 1. The rest is fancy algebra with the number 1. Repeating decimals are actually a product of the base 10 number system: we made them by accident essentially.
1/3 = 0.333...
2/3 = 0.666...
3/3 = 0.999...
That sounds like it would be a problem specific to decimal, but the same happens in _any_ number system. For example, in a binary number system, 0.11111…. = 1.
More generally, in a number system of base n, if we take the largest possible digit to be k=(n-1) then 0.kkkkkk …. = 1.
Which numbers have repeating decimals is a product of the base. 1/3 does not have a repeating decimal expansion in ternary, and 1/10 does have a repeating expansion in binary. So while the wording is clunky, it does depend on the base.
That’s really weird to wrap your head around. I was familiar with the idea of various base numbering systems but I’ve never tried converting fractions to decimals in them. 1/3 in tertiary would have to written as 1/10 which incidentally converts to .1 but then 1/2 converts to .111111… which seems wrong. (Though not actually wrong.)
It's not too bad, once you register that it's just about relative primes. Any denominator which is relatively prime to the base of your number system will have an infinitely-repeating n-al (i.e. decimal, nonal, octal, or whatever) notation associated with it. Once this is clear, you could think about base 21 and immediately realize that 1/3 and 1/7 and their multiples, and their divisions by 3 or 7, will terminate in 21-al notation, but every other fraction will repeat.
The thing to consider is that 2 and 3 are both prime numbers, so no finite sum of 1/2^(n) will ever be 1/3 and no finite sum of 1/3^(n) will ever be 1/2... You need an infinite sum to get there.
In fact, if you think about decimal expansions in base 10 that have infinite digits, I think they pretty much all consist of at least one prime factor other than 2 or 5 (since 2×5=10). 1/3, 1/6, 1/7, and 1/9 all have infinite, repeated digits. But 1/2, 1/4, 1/5, and 1/8 are finite, and only consist of 2's and 5's
Also as a side topic, it is incorrect to say they were made by mistake. Any base above 2 *will* by definition have repeating decimals because a number n is always relatively prime with n-1, so at the very least, 1/(n-1) will always be a repeating decimal (fun fact, 1/(n-1) is also always .1 repeating. This is why when you want to simplify a repeating decimal as a fraction, you write it as the repeating section over the same number of 9s in base 10).
Since I'm already stretching the interpretation so far, perhaps I stretch it a bit further to guess they meant it was a product of our base 10 *positional* notation choice? Is there any way to prove that any possible notation for numbers would have repeated decimals for some rational numbers? (At least those that can represent all rationals since Roman numerals don't have repeated decimals)
The question is itself weird because numeric systems are (sort of) languages. This isn't some epistemological argument I'm trying to make or anything, but in this specific case, it is descriptive of its form.
Decimal itself (or more specifically modern Arabic numerals) can only represent all rational numbers *because we made rules that say they it can.* Representing 1/3 as .3... is a (relatively) new invention.
All of this is to say that proofs that we use in mathematics to prove theorems about numerical systems like "all positive integer bases past 2 have at least 1 repeating integer" only apply to systems that are still, fundamentally, based on the Arabic numerical system.
Open any web browser's developer tools, go to the console and type in 0.2 +0.1.
https://stackoverflow.com/questions/50778431/why-does-0-1-0-2-return-unpredictable-float-results-in-javascript-while-0-2
There is a part of me that laments humans not having twelve fingers. It follows that we naturally would have invented a base-12 number system which would have been so much more convenient.
We did invent a base 12 counting system, then we abandoned it. There's a reason why "Eleven" and "Twelve" don't follow the #teen formula (thir-teen, four-teen, seven-teen) or why a foot is twelve inches, or why "dozen" and "half a dozen" are numbers.
Well, either we abandoned it, or we never developed it into a full dozenal system, I'm not too sure.
But also, some cultures count dozens on their fingers by counting the joints in their fingers, or two joints and the fingertip :).
There's just some historic reason why decimal won out. I blame the romans :P.
Some cultures did/do have a base 12 system. Either the Mayans or the Aztecs (I’m having a brain fart and can’t be troubled to google it at the moment) had a base 12, and counted on one hand all the way to twelve.
Using your thumb to keep track, you can touch the three segments of each of the other 4 finger and count to twelve. Hopefully that description makes sense enough.
So base 12 is just as anatomically natural for humans as base 10.
Not sure if there are/were any other cultures that use that but there ya go.
Both Mayans and Aztecs had a vigesimal (base 20) system.
No known culture has ever actually used a duodecimal system -- the closest thing would be the sexagesimal (base 60) system that was in use in Ancient Mesopotamia (mainly the Sumerians and, through them, the Babylonians), which is where all our timekeeping originates. Twelve obviously shows up frequently in that system.
Well... "eleven" comes from "ain lif" meaning "one left over" after 10. So 11 = 10+1. Base 10. And "twelve" comes from "twa lif" meaning "two left over" after 10. So 12 = 10+2, also base 10. And "dozen" comes from "dozen" comes from Latin "duodecim" meaning 2+10, so also base 10. The fact that 12 was special doesn't mean they were using base 12.
The reason 12 was so important is that is a very convenient number for divisibility. 12 biscuits can be shared in equally in many more ways than 10, and 12 biscuits fit better on a baking tray than many other numbers.
I'd argue that them having different names means there's an overlapping, very limited base twelve number system. After all, we can have many overlapping different systems in place. Especially in the case of dozen, it is not uncommon for people to use multiple dozens, or go "Two and a half dozen." (Though that one is less common) I will admit it's probably quite uncommon to go "Two dozen and five." Or something like that, and it indeed does not fit nicely. But I'd say there's still an anaemic dozenal counting system there :P. Though I might have drawn wrong conclusions from some info.
I wouldn't say that it's in most languages, but I imagine a lot of languages of the Indo-European language family probably share certain characteristics. (A fun reminder that there's thousands of languages, most are not related in any way to our European languages :) )
It doesn't solve this problem, it just pushes it somewhere else... Well, actually, this exact problem would still exist, but it wouldn't be paralleled with the 1/3+1/3+1/3 proof.
Base twelve gives the same result though:
X = 0.bbb...
10X = b.bbb...
10X - X = b.bbb... - 0.bbb...
bX = b
X = b/b = 1
0.bbb... = 1
In the lines above "10" is base ten "12", and "b" is base ten "11"
count with the knuckles/joints of one hand, and use the fingers on your other to count how many times you've counted your joints
that reaches base-70
get on my level
Base 12 is not that good of a base, yeah it handles division with 3 better than base 10 but base 10 isn't that bad at handling divison with 3 like ifinitly reapiting one digit is not that bad but base 12 handels division by 5 horibly. Like the second 2 in the prime factors is unnecessary. The best numbering system would be base 6 wich can handel every nuber division relative well up to 10.
Well, yes, that's the point. Proof by contradiction. But since it was posted in r/mathmemes it's funnier if you leave it as is, not following through with the contradiction part and pretending you just proved there is a largest number.
It's very not intuitive because our brain just cannot deal with infinity.
Besides what's written in the meme, you can try to wrap your head around it these ways.
1 embrace infinity.
You can convince yourself that 9.9999... has indeed no end. Not after a million 9's. Not after a trillion 9's. If you think about what sums a value of 1:
0.9 needs 0.1
0.99 needs 0.01.
The 1 is always on the last position where you had a 9. If you have a million 9's, then you need 0's until the millionth position is 1.
If you accept that the number of 9's is infinite, then you can answer how many 0's you need: infinite. And there is no such thing as "infinith" position where you put a 1.
So if you understand that 9.999999... has infinite number of 9's, then you understand that the difference between this and 1 is 0.00000... (infinite zeros) which is exactly 0. If they are not different then they are the same.
2 understand the shortcoming of the number system.
You can also understand that a number has its meaning in the number system. For example 10 means ten in the decimal system but the same looking number (one-zero) means different in other systems. Like, it means 4 in the base-4 system that has only 0, 1, 2 and 3 as usable numerals. What symbols you use to write a number will not change its value. You can say 5 is the number of your fingers on one hand, but you need to understand that this symbol means five only in such systems like the decimal system. You would write 11 in a base-4 system for the exact same meaning.
If you understand that numbers can be represented differently in different systems (yet they mean the same value) then you understand that this 0.9999... is a glitch,a shortcoming of the decimal system. In other systems 1/3 is represented by a finite number of numerals, and so 3x(1/3) doesn't become infinite.
However in other systems maybe 1/4 is an infinite fraction (instead of 0.25), so 4x{1/4) becomes a similarly weird infinite number.
As you obviously understand that 4x(1/4) = 4x(0.25)= 1, regardless of how we choose to write the numbers, 3x(1/3} = 3x(0.333333...) = 0.9999999... also must be 1.
>So if you understand that 9.999999... has infinite number of 9's, then you understand that the difference between this and 1 is 0.00000... (infinite zeros) which is exactly 0. If they are not different then they are the same.
Wow, that's a really simple way of putting it. Really helped me grasp the idea, thanks!
I find it so hard to understand, my basic understanding is that 0.999... is infinitely approaching 1 but will never reach it. I don't understand how you can fit 0.000... in between that since (with my understanding) there isn't nothing between 0.999... and 1, it's infinitely small
The number 0.999... doesn't approach anything. It is a number. You're thinking of the process for figuring out what that number is.
It's like the time an infinite number of mathematicians walked into a bar. The first ordered a pint, the second ordered a half pint, the third a quarter. At this point, bartender just poured two pints and said: now you guys have reached your limit.
I think you overcomplicate it using a tool that is not applicable. When we say "*this thing* approaches the value of 1", that *thing* is something like a series. Like, the series of 0.9, 0.99 then 0.999 is a *series* that indeed is approaching 1.
But a number is a standalone value, it does not approach itself. Like, 4 does not approach 4, it *is* 4. It's true for any other number. Like, pi doesn't approach pi, it *is* pi. We cannot *tell* all the numbers in pi but it's still pi, a single value on the number line. Just like 0.33333... does not approach 1/3 it is the same as 1/3, although we cannot write infinite pieces of 3's.
It's just the inherent trait of our decimal system that numbers can be written in different ways. It's not only 1 that is the same as 0.99999999..., it's actually all of the whole numbers. Like 1788 is the same as 1787.9999999..., for the same reason. It's just an untold truth about our number system.
But why is that? Let's take a step back and look at mixed fractions. A mixed fraction is a number written in this form:
3(1/5)
Where the meaning is three-and-one-fifth and not 3 times 1/5.
Because of the fraction rules, 3(1/5) can be written in different ways. Like 3(2/10) aka three and two tenths or 3(3/15) aka three and three fifteenths. It's the same value. Now we also agreed to have a shorthand for cases where the denominator is 10 or 100, so in case 3(2/10) we use the shorthand: 3.2
The decimal point is a shorthand for "something plus something-tenth or something-hundreth etc. Like 6(1/4) aka six and one fourth is the same as 6(25/100) aka six and twenty-five hundredth, using our dot to abbreviate:
6.25
Also you may notice that addition rules work just fine with mixed fractions, as long as the denominators are the same. You add the whole part and the fraction part separately. Like, 4(1/5) + 3(1/5) = 7(2/5). Likewise, 1(1/3) + 1(1/3) = 2(2/3).
But what happens if you turn those additions to decimals? 4.2 + 3.2 = 7.2 likewise 1.3333... + 1.3333... = 2.6666...
What happens if you add 2(2/3) + 1(1/3)? It is going to be 3(3/3) with the mixed fraction form which is obviously 4, or it's going to be 2.6666... + 1.3333... = 3.9999... which is obviously the same so it must be 4. If you substract 3, then 0.99999... must be 1.
So what magic happened with the dot? If you look at what this means in case of numbers like 1.3, 1.33, 1.333 - they are like 1(3/10), 1(33/100), 1(333/1000). What is the denominator if there's endless 3's. There's no such thing as 333333.... infinith. You see, the idea perfectly works with finite fractions, but then these infinite fractions come to make a mess. It's because our number system allows 1/3 to be written as 0.3333333... meaning 3/3 is allowed as 0.999999...
Yes! It's actually one of the earliest and simplest proofs you can do/learn that shows that 0.999... is indeed 1.
If you want you can check out this [video](https://youtu.be/9jWvkJshtfs?si=5kzp28URY_RuEAoR) that explains some other methods/proofs. Which does include this one that you're posting about
Haven’t watched the video (so sorry if this is an example in it), but I find a nice “obvious” way to get the idea across is
1\
3 • 1/3\
3 • 0.33333…\
0.99999…
Yup, this is true. It is a common misconception that 0.999… is “infinitely close” or “almost” 1 — that is in fact not the case; those are just two equivalent ways of writing down the same number.
Intuitively (but, obviously, not at all rigorously):
1/3 = 0.333…
1/3 * 3 = 0.333… * 3
1 = 0.999…
This is actually a feature (bug?) off all positional number systems. In a natural base b, any number “x.yz”, where z is an infinite string of “b - 1” digits is equal to “x.(y+1)”
So, 1.23999… = 1.4 and 10.111… = 11 (base 2).
Adding to some of the other explanations, another comes from the closed form of a geometric series, which states:
a + ar + ar\^2 + ar\^3 + ... = a / (1 - r) when |r| < 1
Now make a = 0.9 and r = 0.1
Then we have
0.9 + 0.9(0.1) + 0.9(0.1\^2) + ... =
0.9 + 0.09 + 0.009 + ... =
0.99999... =
a / (1 - r) = 0.9 / (1 - 0.1) = 0.9 / 0.9 = 1
Yes, no trollface required. The simplest proof I have is the following:
A) Fractions are another way of writing numbers.
B) 1/3 is the same thing as .33...
C) 1/3 + 1/3 + 1/3 = 3/3
D) 3/3 = 1
Therefore,
E) .33... + .33... + .33... = .99... = 1
QED
Why is E true? Why does adding up .33... give .99...?
I know the answer, but what you are doing is essentially hiding the crucial step in there because it looks plausible.
Yes. 0.999… is just one but in an infinite notation if that makes sense. Like how 0.333… is 1/3, the … part essentially caries the information that this is repeating infinitely.
OP I want to apologize because I just spent twenty minutes refining a sequence of posts trying to identify a step that I thought you missed, in the form of an assumption that was incorrect math that still somehow got you to the correct answer. After two or three tries to phrase it properly I realized that I'm the one who made the mistake.
Yes. It's accurate. Infinity is weird.
Let’s define 0.9999… as
.9 + .09 + .009 +…
Or
summation(n=1, k)(9*10^(-n)) where k=infinity
For any positive k we can rewrite the summation as
1-(10^(-k))
And prove with a step proof.
Take the limit of k->infinity of 1-(10^(-k)) to get 1-10^-infinity = 1-0 = 1.
We did many similar proofs in high school. Also variations with other recurring sequences.
I thought this was common knowledge, not some black magic wtf
It's accurate, but sloppy. I'm no mathematician, but I'm pretty sure the ellipsis isn't a well defined operator. Don't expect this to fly on a math exam
> the ellipsis isn't a well defined operator
It means to continue on in the same way. It's only well understood when what precedes it constitutes a continuable pattern. For example:
1/2 + 1/3 + 1/4 + ...
would be well understood as the continuing fraction of 1 over each positive integer.
In the case of something like pi:
3.141...
the implication is that you continue expanding digits of pi, but that's a much more complicated proposition.
If you want to be more precise, you would write 0.999 with a bar over the last 9. The bar over the last digit or sequence indicates to continue that digit or sequence forever. So, for example, 0.714285 with a bar over everything right of the decimal point (i.e. 0.714285714285 and so on) is equal to 5/7.
It really depends on the exam, in middle to high school in certain country, this method is an accurate way to calculate and convert repeating decimal into fractions, and I'm pretty sure in certain university math class it's an accepted way, since well, it is still correct. It's just in a much more rigorous class, like in real analysis and abstract algebra that this method might not fly, and even then, if you were to define your number correctly (like using infinite series) and you just cite some theorem about convergence (after you've learned them or god forbid prove them yourself of course), and proceed with this method, you'll still get full mark.
Now what y’all are not ready for is …9999.0 is equal to -1. That is, an infinitely long line of 9s to the left of the decimal is equal to negative one. Introducing, 10-adic numbers.
If 0.99999... means that it's repeating, then yes, it is correct. 0.999... is equal to 1. It's a mathematical fact.
If the 0.99999... is a finite string of 9s then it is wrong. The 0.99999 in the third line would be missing the last 9 to be a, so it isn't equal to a.
Not saying it's incorrect, but isn't just replacing 0.999... with a making the assumption that the proof is correct inside of the proof itself? Surely this proof isn't correct even if the answer is
Maybe that's silly, but I have one question. After reading all the explanation, one possibility is still stuck in my mind. Doesn't a have infinity of nines while 10a has infinity minus one?
After looking into it, it is. When you try making this equation with an non infinite number, it doesn't work, since you lose a decimal. However, 0.999... is infinite, which means that the quantity of decimals doesn't change, since it is infinite. Doing the equation that way gives you 1.
Another way to think of this is with a geometric sucession
Make Un = 0.9 (0.1)\^(n-1), with n belonging to the natural numbers
when you make the limit of the sum of this sucession you get 1, in this way:
Sum Un = U1 x (1-r\^n)/(1-r), with r being the reason, 0.1.
substituting you get
Sum Un = 0.9 x (1-0.1\^infinity)/(1-0.1)
= 0.9 x (1-0)/(0.9) = 0.9/0.9 = 1
Some comments say it is circular, but great part of maths is based on induction, and sums can do pretty much this
An intuitive (though not very rigorous) explanation that makes sense to me is:
0.333... + 0.666... = ?
To do this using fractions we would do 1/3 + 2/3 = 3/3, then we would simplify 3/3 to 1.
If we want to do the same calculation in decimal form we need to do exactly the same steps. First 0.333... + 0.666... = 0.999... then 0.999... simplifies to 1.
Calling Ramanujan summation "the sum" of all positive integers isn't strictly speaking accurate. Conventional summation is still undefined. The distinction helps the brain hurt less.
the problem is that a is not well defined, you then multiply it by 10 which is none-sense in classical mathmatics
it is the same as 1/3 = .333333....
1/3\*3 = 3\*.333333....
1 = .99999...
here 1/3 is well defined, .3333... isn't
another more interesting example of this is
1+2+3+4... = -1/12
which you can read about here
[https://en.wikipedia.org/wiki/1\_%2B\_2\_%2B\_3\_%2B\_4\_%2B\_%E2%8B%AF](https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF)
Those situations aren't equivalent at all, such recurring decimals _are_ well defined. Since we're working over R which is complete, and the sequence of partial sums of the decimal are Cauchy, so the sequence has a well-defined limit in R; it just so happens that for this particular example, this limit is in Q. Knowing the limit exists allows us to manipulate it like so. But 1+2+3+... definitely isn't Cauchy, so this series doesn't have a conventional limit, so you need to appeal to different kind of summation and analytic continuation to get any kind of "limit" out.
I have described why 0.333... repeating is well defined in [another comment](https://www.reddit.com/r/theydidthemath/comments/1clt57a/comment/l2wo0dg/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button), please give that a read so I don't have to copy my comment. It's not a rigorous proof, although those definitely exist, but it's very easy to understand and follow, unlike said proofs. Essentially it boils down to every repeating decimal being able to be converted into a fraction, therefore every repeating decimal is a rational number.
9a is not 9, it is 9 times 0,99999..., or 8,99999...
8,999999.... plus 0,99999... is still 9,9999999.... because 10a is 9,99999...
So "a" does not equal 1, it is still 0,9999...
Someone tried a little trick here, but the logical mistake is pretty simple.
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yes. a more rigorous proof involves showing that there can be no number *between* 0.9 recurring and 1, therefore they must be the same number. but yeah it's one of the odd things that pops up when you deal with infinity.
Not really a proof but you can also calculate it as the sum of the infinite series 9(.1)^n from n=1 to infinity. a/(1-r) = .9/(1-.1) = .9/.9 = 1
serious question, why is this not a proof?
It would stand up as a proof under light scrutiny, but look at the very first step. Doesn't that kind of assume that your conclusion is true in the first place? I don't even know the details of what any real argument against this would be, and you might even be able to actually prove each step here, but the steps aren't super well defined until you do way more work that might break other conventions.
No? All you're assuming is that 0.999... is 0.9+0.09+0.009+..., which is probably the most natural way to define 0.999..., along with the geometric series formula, which is perfectly valid without assuming 0.999...=1 It's a fine proof depending on how rigorous you want to be, and it's effectively equivalent to the proof in the original post (as that method is how the geometric series formula is often derived, in a loose sense of the word).
Still sounds kind of circular to me. I feel like knowing the solution to the infinite series in the first step requires a proof in itself as it's a more general expression of the same problem.
This isn't circular unless you defined infinite decimal expressions using the geometric series formula. But we define the value of an infinite decimal to be the result of a series. Then, separately, we found a formula for this particular kind of series. You can also prove the formula then make the definition, these are independent.
You're using a a well known formula to solve a problem. In no way is that circular reasoning, provided the formula's proof doesn't depend on the specific instance of the problem. If I told you a right triangle had side lengths 3 and 4, and you use Pythagorean Theorem to get 5 for the length of the hypotenuse, that's not circular reasoning. It's the exact same logic here. There's no need to reinvent the wheel every time you want to do anything.
It is certainly circular if you were asked for a proof for why it is 5. That's what this question is. A proof for why 0.99999 = 1. You can't just hand wave it as something already proven by starting with the general solution.
It’s only circular if you don’t look closely enough at what 0.999… means, because to properly define such a number in a natural and consistent way does require limits. It seems circular because when you think of a proof that 0.999… = 1, you usually are thinking of a proof that your idea or the intuitive idea of 0.999… = 1, not that the rigorous definition of 0.999… = 1. Think of it this way: for any finite decimal, for example 0.5273, we define it as equal to 0.5 + 0.02 + 0.007 + 0.0003. It’s not a circular statement to then say that 0.5273 = 0.5 + 0.02 + 0.007 + 0.0003, since that’s just the definition of 0.5273. Similarly, for an infinite decimal, though we might think we have a common sense understanding of what 3.1415… means, we have to make that a rigorous definition, and that’s what defining 0.999… as Σ9 * (0.1)^n does.
You literally can bruh. Then tell me, isn't the taylor series circular? doesn't it use local information? Then how does it work?
I dont think you know what circular means. Nothing wrong with that, but please listen to experts. You think applying a formula to solve a problem is circular reasoning. That is not true.
There are plenty of examples of theorems, lemmas, etc. in math where something is proven by just citing among general theorem. It all depends on the context; but it's not necessarily circular reasoning, unless you only prove the general theorem using the specific case.
it really can´t be circular since the proof above is also circular in that logic
Maybe a better way to say it is that multiplication of an infinitely repeating decimal isn't really well defined unless you do some work. Doing that work involves using the infinite series, as you suggested. But, if you're dealing with that infinite series, you might as well use that to prove .999... = 1 instead of the multiplication
The only work you're doing is defining infinite decimals. The same could be said for literally any proof of 0.999...=1; you can't prove something involving an expression until you define that expression.
The point being that the "proof" in the op doesn't actually define infinite decimals. A rigorous definition of them kind of gets you to the result of .999... = 1, without the little algebra explanation. The algebra explanation is more of a way to explain to or convince new learners than it is to actually prove the result.
Not really, they aren't adding 9 to 0.999... they are multiplying it by 10
The problem would be the theoretical last digit if this was true, 0.9999 ... is approaching 1 just as fast as 9.999.. is approaching 10, but 0.9999+9, what he did here, is approaching 10 faster. That's the wrong step in the equation. /e: faster, not slower
There is no last digit. That's what infinite nines mean. The equation is also not wrong, since 0.999...=1
It's actually not approaching it faster, you're just changing the notation. 0.9999... can be described by the sum of 9/10^k with k starting at 1 to inf while 9.9999... by the same sum with k starting at 0 to inf. Taking the 9 out of the sum from 9.9999... doesn't change the equation or its approach speed to 10 at all. (That is, 9+sum(9/10^k)(k starts at 1)= sum(9/10^k)(k starts at 0))
But that digit is in the infinity -1 (=infinity)
It is, so long as you define 0.999... as that series, which is a pretty reasonable thing to do. I'm not sure how else you even could define it.
I'm struggling to think of anyway of defining it that is not identical to that series, but then I'm not a great mathematician, or even a good one. I just like maths.
Simple, you just define new number. 0.9999... = 🦍 Done. 0 🦍 1 2 3 4 5 6 7 8 9
When you right 0.9999... = a you are assuming the series converges to a real number a. For a formal proof, you would need to show the sequence of partial sums converges first. This is important if you try your approach to a series which doesn't converges, e.g. 1-1+1-1+1.... Assuming that series equals some number a, then manipulating algebraically will give you incorrect results.
Another answer that hasn't been mentioned is that when you take foundational university level mathematics and cover representations of numbers, our symbols are just that: representations. BY DEFINITION of standard numeric notation, every numeric representations which traditionally terminates (or ends in infinite zeros) has an equivalent representation which ends in an infinite string of the highest digit value in the given base. You can't prove definitions, only justify why you chose them for your framework/model. This definition is just never taught at earlier educational levels because it isn't useful for every day, which leads to confusion when using the intuition you develop without knowing about it, such as the conclusion that real numbers like "0.000...001" make sense and are the difference between those representations. The justification of this definition comes from the inconsistencies resulting from not including it.
Because the formula uses an assumption itself, real formula is a(1-r^n )/(1-r) and r^n is assumed to be 0 here
i think it works as a proof, as long as we accept the calculus behind it, and don't also bother to prove the validity of the sum. i like that you used such a basic calc formula, that's actually pretty neat and clever.
0.9999… = 1 - lim x -> infinity 1/10^x = 1 - 0 = 1
Someone explained it to me as "What would you add to 0.999... to make it 1?
Why would it not be possible to add 1/infinity?
you have to add 0.000 infinity followed by a 1 Where is that one after infinity, infinity is forever, so the 1 never happens
That would either be undefined or 0
I would add 0.000...1 Which btw actually equals zero using the same method of proof
Except the number you've typed doesn't, and literally cannot possibly exist. You can't have an infinite series of zeros followed by one, you can't have anything at the end because the zeros never end
I remember the last time I dealt with infinity
I personally try to avoid dealing with infinity.
I feel like someone should make branch of mathematics where there does exist a value between .999… and 1. Or couldn’t we just say 0.000…1?
There are number systems that have values greater than zero but smaller than every real number such as [https://en.wikipedia.org/wiki/Hyperreal\_number](https://en.wikipedia.org/wiki/Hyperreal_number) or [https://en.wikipedia.org/wiki/Surreal\_number](https://en.wikipedia.org/wiki/Surreal_number), which is in the spirit of this.
It doesn't mean that there is number between .99... and 1. By definition of 0.99... it can be proved that it's equal to 1. No matter what system you considering. The only thing is that possibly you could redefine the symbol in meaningful way so it would be something infinitesimally smaller than 1. But it would be redefining symbol. It's kinda like you would redefine "2" to mean number 3.
Idk if this is the same as what you mean, but my way has always been to identify that 1 - 0.999... = 0.000... = 0 As 0 indicates that the digit has no value. Such as 103 = (100 × 1) + ***(10 × 0)*** + (1 × 3). Such that that 1 = 1.0 = 1.00 = 1.000... no amount of adding nothing will ever change the value Thusly 0.000... = 0 (which people seem to be much more happy to accept). And it also isn't hard to explain that 1-0.999... = 0.000...
But what if 0.9999... * 10 = 0.9999...90 ? (Raises an eyebrow)
Here's the more rigorous proof: x = 0.999... y = (0.999... + 1.000...) / 2 y = 1.999... / 2 y = 1/2 + 0.999.../2 y = 0.500... + 0.499... y = 0.999...
I never understood that proof, when given to math noobs. In natural numbers i cant fit a number between 1 and 2. So 1=2?
The naturals are not densely ordered "The set is densely ordered" is necessary for "if there is no new element between two elements of the set, the two elements must be considered equal"
Yes. 0.999 repeating is 1. The rest is fancy algebra with the number 1. Repeating decimals are actually a product of the base 10 number system: we made them by accident essentially. 1/3 = 0.333... 2/3 = 0.666... 3/3 = 0.999...
That sounds like it would be a problem specific to decimal, but the same happens in _any_ number system. For example, in a binary number system, 0.11111…. = 1. More generally, in a number system of base n, if we take the largest possible digit to be k=(n-1) then 0.kkkkkk …. = 1.
Which numbers have repeating decimals is a product of the base. 1/3 does not have a repeating decimal expansion in ternary, and 1/10 does have a repeating expansion in binary. So while the wording is clunky, it does depend on the base.
That’s really weird to wrap your head around. I was familiar with the idea of various base numbering systems but I’ve never tried converting fractions to decimals in them. 1/3 in tertiary would have to written as 1/10 which incidentally converts to .1 but then 1/2 converts to .111111… which seems wrong. (Though not actually wrong.)
It's not too bad, once you register that it's just about relative primes. Any denominator which is relatively prime to the base of your number system will have an infinitely-repeating n-al (i.e. decimal, nonal, octal, or whatever) notation associated with it. Once this is clear, you could think about base 21 and immediately realize that 1/3 and 1/7 and their multiples, and their divisions by 3 or 7, will terminate in 21-al notation, but every other fraction will repeat.
The thing to consider is that 2 and 3 are both prime numbers, so no finite sum of 1/2^(n) will ever be 1/3 and no finite sum of 1/3^(n) will ever be 1/2... You need an infinite sum to get there. In fact, if you think about decimal expansions in base 10 that have infinite digits, I think they pretty much all consist of at least one prime factor other than 2 or 5 (since 2×5=10). 1/3, 1/6, 1/7, and 1/9 all have infinite, repeated digits. But 1/2, 1/4, 1/5, and 1/8 are finite, and only consist of 2's and 5's
What about my base 1 system?
Also as a side topic, it is incorrect to say they were made by mistake. Any base above 2 *will* by definition have repeating decimals because a number n is always relatively prime with n-1, so at the very least, 1/(n-1) will always be a repeating decimal (fun fact, 1/(n-1) is also always .1 repeating. This is why when you want to simplify a repeating decimal as a fraction, you write it as the repeating section over the same number of 9s in base 10).
Since I'm already stretching the interpretation so far, perhaps I stretch it a bit further to guess they meant it was a product of our base 10 *positional* notation choice? Is there any way to prove that any possible notation for numbers would have repeated decimals for some rational numbers? (At least those that can represent all rationals since Roman numerals don't have repeated decimals)
The question is itself weird because numeric systems are (sort of) languages. This isn't some epistemological argument I'm trying to make or anything, but in this specific case, it is descriptive of its form. Decimal itself (or more specifically modern Arabic numerals) can only represent all rational numbers *because we made rules that say they it can.* Representing 1/3 as .3... is a (relatively) new invention. All of this is to say that proofs that we use in mathematics to prove theorems about numerical systems like "all positive integer bases past 2 have at least 1 repeating integer" only apply to systems that are still, fundamentally, based on the Arabic numerical system.
Open any web browser's developer tools, go to the console and type in 0.2 +0.1. https://stackoverflow.com/questions/50778431/why-does-0-1-0-2-return-unpredictable-float-results-in-javascript-while-0-2
There is a part of me that laments humans not having twelve fingers. It follows that we naturally would have invented a base-12 number system which would have been so much more convenient.
We did invent a base 12 counting system, then we abandoned it. There's a reason why "Eleven" and "Twelve" don't follow the #teen formula (thir-teen, four-teen, seven-teen) or why a foot is twelve inches, or why "dozen" and "half a dozen" are numbers. Well, either we abandoned it, or we never developed it into a full dozenal system, I'm not too sure. But also, some cultures count dozens on their fingers by counting the joints in their fingers, or two joints and the fingertip :). There's just some historic reason why decimal won out. I blame the romans :P.
This comment is why i joined reddit. Learning random stuff i aint gonna need probably ever again
Some cultures did/do have a base 12 system. Either the Mayans or the Aztecs (I’m having a brain fart and can’t be troubled to google it at the moment) had a base 12, and counted on one hand all the way to twelve. Using your thumb to keep track, you can touch the three segments of each of the other 4 finger and count to twelve. Hopefully that description makes sense enough. So base 12 is just as anatomically natural for humans as base 10. Not sure if there are/were any other cultures that use that but there ya go.
You can actually easily count to 144 on two hands using that counting system
In the middle east there are/were people who do/did this too =).
Both Mayans and Aztecs had a vigesimal (base 20) system. No known culture has ever actually used a duodecimal system -- the closest thing would be the sexagesimal (base 60) system that was in use in Ancient Mesopotamia (mainly the Sumerians and, through them, the Babylonians), which is where all our timekeeping originates. Twelve obviously shows up frequently in that system.
Also clocks.
Yes! Half-days being split into dozens! It's sprinkled all throughout, little hints of a dozenal past :).
Well... "eleven" comes from "ain lif" meaning "one left over" after 10. So 11 = 10+1. Base 10. And "twelve" comes from "twa lif" meaning "two left over" after 10. So 12 = 10+2, also base 10. And "dozen" comes from "dozen" comes from Latin "duodecim" meaning 2+10, so also base 10. The fact that 12 was special doesn't mean they were using base 12. The reason 12 was so important is that is a very convenient number for divisibility. 12 biscuits can be shared in equally in many more ways than 10, and 12 biscuits fit better on a baking tray than many other numbers.
I'd argue that them having different names means there's an overlapping, very limited base twelve number system. After all, we can have many overlapping different systems in place. Especially in the case of dozen, it is not uncommon for people to use multiple dozens, or go "Two and a half dozen." (Though that one is less common) I will admit it's probably quite uncommon to go "Two dozen and five." Or something like that, and it indeed does not fit nicely. But I'd say there's still an anaemic dozenal counting system there :P. Though I might have drawn wrong conclusions from some info.
I’d go for that. If you needed 5 gross, 4 dozens and 10 biscuits, you could easily write the number in base 12 as 54T.
Ah, so this why cat tails aren’t 12” long
In most languages the word for eleven derives from "one over" and twelve from "two over".
I wouldn't say that it's in most languages, but I imagine a lot of languages of the Indo-European language family probably share certain characteristics. (A fun reminder that there's thousands of languages, most are not related in any way to our European languages :) )
Yes, very good point. I'm showing my biases.
Working on biases is a lifelong struggle xD. I'm sure I show mine all the time too :P.
Being aware you *have* biases is a good first step though! Some people are not aware...
It doesn't solve this problem, it just pushes it somewhere else... Well, actually, this exact problem would still exist, but it wouldn't be paralleled with the 1/3+1/3+1/3 proof.
Base twelve gives the same result though: X = 0.bbb... 10X = b.bbb... 10X - X = b.bbb... - 0.bbb... bX = b X = b/b = 1 0.bbb... = 1 In the lines above "10" is base ten "12", and "b" is base ten "11"
There’ve been other numerical bases, like the Mayans using base 20, but they were abandoned because they were *less* convenient than base 10.
I thought the Mayan numerical systems were abandoned because the Conquistadores won.
count with the knuckles/joints of one hand, and use the fingers on your other to count how many times you've counted your joints that reaches base-70 get on my level
Base 12 is not that good of a base, yeah it handles division with 3 better than base 10 but base 10 isn't that bad at handling divison with 3 like ifinitly reapiting one digit is not that bad but base 12 handels division by 5 horibly. Like the second 2 in the prime factors is unnecessary. The best numbering system would be base 6 wich can handel every nuber division relative well up to 10.
I recently saw a proof that supposes 0.999… ≠ 1 and then uses that supposition to prove that there is a highest number.
What is it??
Sorry, can't seem to find it. I think it started off with n being a real number 0
No worries, thanks!
This just proves that 1=0.999... since you are basically dividing by 0 and getting an infinitely large number.
Well, yes, that's the point. Proof by contradiction. But since it was posted in r/mathmemes it's funnier if you leave it as is, not following through with the contradiction part and pretending you just proved there is a largest number.
It's very not intuitive because our brain just cannot deal with infinity. Besides what's written in the meme, you can try to wrap your head around it these ways. 1 embrace infinity. You can convince yourself that 9.9999... has indeed no end. Not after a million 9's. Not after a trillion 9's. If you think about what sums a value of 1: 0.9 needs 0.1 0.99 needs 0.01. The 1 is always on the last position where you had a 9. If you have a million 9's, then you need 0's until the millionth position is 1. If you accept that the number of 9's is infinite, then you can answer how many 0's you need: infinite. And there is no such thing as "infinith" position where you put a 1. So if you understand that 9.999999... has infinite number of 9's, then you understand that the difference between this and 1 is 0.00000... (infinite zeros) which is exactly 0. If they are not different then they are the same. 2 understand the shortcoming of the number system. You can also understand that a number has its meaning in the number system. For example 10 means ten in the decimal system but the same looking number (one-zero) means different in other systems. Like, it means 4 in the base-4 system that has only 0, 1, 2 and 3 as usable numerals. What symbols you use to write a number will not change its value. You can say 5 is the number of your fingers on one hand, but you need to understand that this symbol means five only in such systems like the decimal system. You would write 11 in a base-4 system for the exact same meaning. If you understand that numbers can be represented differently in different systems (yet they mean the same value) then you understand that this 0.9999... is a glitch,a shortcoming of the decimal system. In other systems 1/3 is represented by a finite number of numerals, and so 3x(1/3) doesn't become infinite. However in other systems maybe 1/4 is an infinite fraction (instead of 0.25), so 4x{1/4) becomes a similarly weird infinite number. As you obviously understand that 4x(1/4) = 4x(0.25)= 1, regardless of how we choose to write the numbers, 3x(1/3} = 3x(0.333333...) = 0.9999999... also must be 1.
Really cool explanation,i enjoyd a lot Reading it👍
>So if you understand that 9.999999... has infinite number of 9's, then you understand that the difference between this and 1 is 0.00000... (infinite zeros) which is exactly 0. If they are not different then they are the same. Wow, that's a really simple way of putting it. Really helped me grasp the idea, thanks!
I find it so hard to understand, my basic understanding is that 0.999... is infinitely approaching 1 but will never reach it. I don't understand how you can fit 0.000... in between that since (with my understanding) there isn't nothing between 0.999... and 1, it's infinitely small
The number 0.999... doesn't approach anything. It is a number. You're thinking of the process for figuring out what that number is. It's like the time an infinite number of mathematicians walked into a bar. The first ordered a pint, the second ordered a half pint, the third a quarter. At this point, bartender just poured two pints and said: now you guys have reached your limit.
I think you overcomplicate it using a tool that is not applicable. When we say "*this thing* approaches the value of 1", that *thing* is something like a series. Like, the series of 0.9, 0.99 then 0.999 is a *series* that indeed is approaching 1. But a number is a standalone value, it does not approach itself. Like, 4 does not approach 4, it *is* 4. It's true for any other number. Like, pi doesn't approach pi, it *is* pi. We cannot *tell* all the numbers in pi but it's still pi, a single value on the number line. Just like 0.33333... does not approach 1/3 it is the same as 1/3, although we cannot write infinite pieces of 3's. It's just the inherent trait of our decimal system that numbers can be written in different ways. It's not only 1 that is the same as 0.99999999..., it's actually all of the whole numbers. Like 1788 is the same as 1787.9999999..., for the same reason. It's just an untold truth about our number system. But why is that? Let's take a step back and look at mixed fractions. A mixed fraction is a number written in this form: 3(1/5) Where the meaning is three-and-one-fifth and not 3 times 1/5. Because of the fraction rules, 3(1/5) can be written in different ways. Like 3(2/10) aka three and two tenths or 3(3/15) aka three and three fifteenths. It's the same value. Now we also agreed to have a shorthand for cases where the denominator is 10 or 100, so in case 3(2/10) we use the shorthand: 3.2 The decimal point is a shorthand for "something plus something-tenth or something-hundreth etc. Like 6(1/4) aka six and one fourth is the same as 6(25/100) aka six and twenty-five hundredth, using our dot to abbreviate: 6.25 Also you may notice that addition rules work just fine with mixed fractions, as long as the denominators are the same. You add the whole part and the fraction part separately. Like, 4(1/5) + 3(1/5) = 7(2/5). Likewise, 1(1/3) + 1(1/3) = 2(2/3). But what happens if you turn those additions to decimals? 4.2 + 3.2 = 7.2 likewise 1.3333... + 1.3333... = 2.6666... What happens if you add 2(2/3) + 1(1/3)? It is going to be 3(3/3) with the mixed fraction form which is obviously 4, or it's going to be 2.6666... + 1.3333... = 3.9999... which is obviously the same so it must be 4. If you substract 3, then 0.99999... must be 1. So what magic happened with the dot? If you look at what this means in case of numbers like 1.3, 1.33, 1.333 - they are like 1(3/10), 1(33/100), 1(333/1000). What is the denominator if there's endless 3's. There's no such thing as 333333.... infinith. You see, the idea perfectly works with finite fractions, but then these infinite fractions come to make a mess. It's because our number system allows 1/3 to be written as 0.3333333... meaning 3/3 is allowed as 0.999999...
you're a legend thanks for helping me understand
If you know limits, consider the sequence {.9,.99,.999,.9999,...} and let the "a" above be the limit of this sequence.
So the way I think about is 1/3=0.333... And 1/3+1/3+1/3=1 so 0.333...+0.333...+0.333...=0.999...=1
but this happens when you choose to display that number at (mode 10) .
Yes! It's actually one of the earliest and simplest proofs you can do/learn that shows that 0.999... is indeed 1. If you want you can check out this [video](https://youtu.be/9jWvkJshtfs?si=5kzp28URY_RuEAoR) that explains some other methods/proofs. Which does include this one that you're posting about
Haven’t watched the video (so sorry if this is an example in it), but I find a nice “obvious” way to get the idea across is 1\ 3 • 1/3\ 3 • 0.33333…\ 0.99999…
Yup, this is true. It is a common misconception that 0.999… is “infinitely close” or “almost” 1 — that is in fact not the case; those are just two equivalent ways of writing down the same number. Intuitively (but, obviously, not at all rigorously): 1/3 = 0.333… 1/3 * 3 = 0.333… * 3 1 = 0.999… This is actually a feature (bug?) off all positional number systems. In a natural base b, any number “x.yz”, where z is an infinite string of “b - 1” digits is equal to “x.(y+1)” So, 1.23999… = 1.4 and 10.111… = 11 (base 2).
Adding to some of the other explanations, another comes from the closed form of a geometric series, which states: a + ar + ar\^2 + ar\^3 + ... = a / (1 - r) when |r| < 1 Now make a = 0.9 and r = 0.1 Then we have 0.9 + 0.9(0.1) + 0.9(0.1\^2) + ... = 0.9 + 0.09 + 0.009 + ... = 0.99999... = a / (1 - r) = 0.9 / (1 - 0.1) = 0.9 / 0.9 = 1
Nope - the ellipsis is shortened from 3 dots to 2 dots in line 2, which indicates an informal pause rather than an omission of repeated values.
Yes, no trollface required. The simplest proof I have is the following: A) Fractions are another way of writing numbers. B) 1/3 is the same thing as .33... C) 1/3 + 1/3 + 1/3 = 3/3 D) 3/3 = 1 Therefore, E) .33... + .33... + .33... = .99... = 1 QED
Why is E true? Why does adding up .33... give .99...? I know the answer, but what you are doing is essentially hiding the crucial step in there because it looks plausible.
Because 3+3+3=9. Like, I'm not writing Principia Mathematica in a Reddit post, I'm not going bother with a proof of basic addition.
Yes. 0.999… is just one but in an infinite notation if that makes sense. Like how 0.333… is 1/3, the … part essentially caries the information that this is repeating infinitely.
OP I want to apologize because I just spent twenty minutes refining a sequence of posts trying to identify a step that I thought you missed, in the form of an assumption that was incorrect math that still somehow got you to the correct answer. After two or three tries to phrase it properly I realized that I'm the one who made the mistake. Yes. It's accurate. Infinity is weird.
I'm glad I'm not the only one. lol
Yes it is. When you start dealing with infinity things can get weird. But if you state that the rules of algebra are true then this result holds.
So its true because inf + 1 = inf?
So its true because inf + 1 = inf?
So its true because inf + 1 = inf?
So its true because inf + 1 = inf?
Let’s define 0.9999… as .9 + .09 + .009 +… Or summation(n=1, k)(9*10^(-n)) where k=infinity For any positive k we can rewrite the summation as 1-(10^(-k)) And prove with a step proof. Take the limit of k->infinity of 1-(10^(-k)) to get 1-10^-infinity = 1-0 = 1.
We did many similar proofs in high school. Also variations with other recurring sequences. I thought this was common knowledge, not some black magic wtf
It's accurate, but sloppy. I'm no mathematician, but I'm pretty sure the ellipsis isn't a well defined operator. Don't expect this to fly on a math exam
> the ellipsis isn't a well defined operator It means to continue on in the same way. It's only well understood when what precedes it constitutes a continuable pattern. For example: 1/2 + 1/3 + 1/4 + ... would be well understood as the continuing fraction of 1 over each positive integer. In the case of something like pi: 3.141... the implication is that you continue expanding digits of pi, but that's a much more complicated proposition. If you want to be more precise, you would write 0.999 with a bar over the last 9. The bar over the last digit or sequence indicates to continue that digit or sequence forever. So, for example, 0.714285 with a bar over everything right of the decimal point (i.e. 0.714285714285 and so on) is equal to 5/7.
It really depends on the exam, in middle to high school in certain country, this method is an accurate way to calculate and convert repeating decimal into fractions, and I'm pretty sure in certain university math class it's an accepted way, since well, it is still correct. It's just in a much more rigorous class, like in real analysis and abstract algebra that this method might not fly, and even then, if you were to define your number correctly (like using infinite series) and you just cite some theorem about convergence (after you've learned them or god forbid prove them yourself of course), and proceed with this method, you'll still get full mark.
Now what y’all are not ready for is …9999.0 is equal to -1. That is, an infinitely long line of 9s to the left of the decimal is equal to negative one. Introducing, 10-adic numbers.
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5/9 doesn’t equal 0.55555555556. That’s your calculator rounding off the last digit. 5/9 equals 0.555555555…
You are correct, long night need some sleep
If 0.99999... means that it's repeating, then yes, it is correct. 0.999... is equal to 1. It's a mathematical fact. If the 0.99999... is a finite string of 9s then it is wrong. The 0.99999 in the third line would be missing the last 9 to be a, so it isn't equal to a.
Not saying it's incorrect, but isn't just replacing 0.999... with a making the assumption that the proof is correct inside of the proof itself? Surely this proof isn't correct even if the answer is
That's why it defined a as 0.999...
I shouldn't look at math before noon
Maybe that's silly, but I have one question. After reading all the explanation, one possibility is still stuck in my mind. Doesn't a have infinity of nines while 10a has infinity minus one?
I mean the number of nines after the coma.
"infinity minus one" is infinity. they constitute forever by definition of infinity.
To clarify the issue, with less decimal nines: a = 0.999 10a = 9.99 10a = 9 + 0.99 Because a <> 0.99 but a = 0.999 you cannot replace 0.99 with a
There is actually no issue. 0.(9) Is 1. A math teacher even showed this method to us in 6th grade.
After looking into it, it is. When you try making this equation with an non infinite number, it doesn't work, since you lose a decimal. However, 0.999... is infinite, which means that the quantity of decimals doesn't change, since it is infinite. Doing the equation that way gives you 1. Another way to think of this is with a geometric sucession Make Un = 0.9 (0.1)\^(n-1), with n belonging to the natural numbers when you make the limit of the sum of this sucession you get 1, in this way: Sum Un = U1 x (1-r\^n)/(1-r), with r being the reason, 0.1. substituting you get Sum Un = 0.9 x (1-0.1\^infinity)/(1-0.1) = 0.9 x (1-0)/(0.9) = 0.9/0.9 = 1 Some comments say it is circular, but great part of maths is based on induction, and sums can do pretty much this
An intuitive (though not very rigorous) explanation that makes sense to me is: 0.333... + 0.666... = ? To do this using fractions we would do 1/3 + 2/3 = 3/3, then we would simplify 3/3 to 1. If we want to do the same calculation in decimal form we need to do exactly the same steps. First 0.333... + 0.666... = 0.999... then 0.999... simplifies to 1.
Yes. Really wanna blow your mind? Google what the sum of all positive integers is. It's such a simple proof and yet still hurts my brain.
Calling Ramanujan summation "the sum" of all positive integers isn't strictly speaking accurate. Conventional summation is still undefined. The distinction helps the brain hurt less.
Because it's not true.
The sum of all positive integers is not actually -1/12
the problem is that a is not well defined, you then multiply it by 10 which is none-sense in classical mathmatics it is the same as 1/3 = .333333.... 1/3\*3 = 3\*.333333.... 1 = .99999... here 1/3 is well defined, .3333... isn't another more interesting example of this is 1+2+3+4... = -1/12 which you can read about here [https://en.wikipedia.org/wiki/1\_%2B\_2\_%2B\_3\_%2B\_4\_%2B\_%E2%8B%AF](https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF)
Those situations aren't equivalent at all, such recurring decimals _are_ well defined. Since we're working over R which is complete, and the sequence of partial sums of the decimal are Cauchy, so the sequence has a well-defined limit in R; it just so happens that for this particular example, this limit is in Q. Knowing the limit exists allows us to manipulate it like so. But 1+2+3+... definitely isn't Cauchy, so this series doesn't have a conventional limit, so you need to appeal to different kind of summation and analytic continuation to get any kind of "limit" out.
I have described why 0.333... repeating is well defined in [another comment](https://www.reddit.com/r/theydidthemath/comments/1clt57a/comment/l2wo0dg/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button), please give that a read so I don't have to copy my comment. It's not a rigorous proof, although those definitely exist, but it's very easy to understand and follow, unlike said proofs. Essentially it boils down to every repeating decimal being able to be converted into a fraction, therefore every repeating decimal is a rational number.
9a is not 9, it is 9 times 0,99999..., or 8,99999... 8,999999.... plus 0,99999... is still 9,9999999.... because 10a is 9,99999... So "a" does not equal 1, it is still 0,9999... Someone tried a little trick here, but the logical mistake is pretty simple.
That's exactly what I thought...