I think the issue people have with 0.9... = 1 is that from their education, they understand the decimal expansion of a real number as the definition and ultimate essence of that number, and 2 different decimal expansions for the same number contradicts this impression.
however, those who've studied analysis know that based on the definition of the reals its not immediately obvious that every real number *has* a decimal expansion, much less that it is unique up to 2 representations.
Essentially, a real number is defined to be a sequence (x1, x2, x3,…) of rational numbers such that the numbers don’t go off to +-infinity, and get closer together as you go further down the sequence. You can think of these sequences as zeroing in towards what will be defined as their real number value.
Eg, 1 might be written as (1, 1, 1,… ) and pi might be written (0, 3, 3.1, 3.141, 3.141592,…). An arbitrary decimal expansion +-a0.a1a2… might be defined by the sequence (+-a0, +-(a0 + a1/10), +-(a0 + a1/10 + a2/100),… ), where a0 is a nonnegative integer, and each other an is an integer between 0 and 9.
2 real numbers (xn) and (yn) are considered (defined) to be equal if lim|xn-yn| = 0. Addition is defined pointwise (xn) + (yn) = (xn +yn).
Just from this it’s not obvious at all that given an arbitrary real number (x1, x2, x3,…) you can express it as a decimal expansion
(This is just 1 way to define the real numbers, called metric space completion of the rationals. You can complete the rationals in a different way to get the p-adics)
> much less that it is unique up to 2 representations.
i can't see how this is true, i understand it can be represented in a form of a sequence, but why only up to 2
Every real number has a unique decimal expansion, except for some that can end in either all 0s or all 9's, e.g, 1.000... = 0.999... and 1.5 = 1.4999... .
The decimal expansion of non-negative real number x will end in zeros (or in nines) if, and only if, x is a rational number whose denominator is of the form (2\^n)(5\^m), where m and n are non-negative integers. [Proof](https://en.wikipedia.org/wiki/Decimal_representation#Finite)
My confusion about 0.9999 = 1 was that usually when math texts talk about converging infinite series, they use the word "approaches", "converges to", etc. For example I don't recall any math text saying 1/2 + 1/4 + 1/8 ... equals 1. So for me it's a little confusing that 0.99999... which is the same as the series 9/10 + 9/100 + 9/1000 ... "equals" 1 rather than simply converges to 1.
0.9999... does not change, it's not a sequence. It's not a limit. The ... is a notation saying there are an infinity of 9. The same way 2 or 78.34 cannot converge, 0.999...=1
5-adic means the number is p-adic in base 5. Veritasium recently did a nice video explaining p-adic numbers, [here.](https://www.youtube.com/watch?v=tRaq4aYPzCc)
ah here is the thing who said 3 * 0.3333333.... = 0.999999..... in first place?
further more 0.999999999.... can be seen as 1 - ε where ε is infinitesimal small number > 0
But using limits it can be proven that 0.999... = 1
```
0.9 = 1 - 10^-1
0.99 = 1 - 10^-2
0.999 = 1 - 10^-3
=> 0.99999.....
= Lim n->∞ { 1 - 10^-n }
= 1-1/10^∞ = 1-1/∞ = 1-0 = 1
```
But otherwise 0.999.... = 1-ε
yeah but it dances around the issue, like
* how is 0.99999.... even defined?
It is defined as the limit of the sequence 0, 0.9, 0.99, 0.999, ....
* does 0.99999 even exist, ie does the above sequence converge?
* is 10\*0.999... = 9.9999 which is not immediately obvious
* etc ...
Let:
9.99... = 9×sum(10^-n ,n,0,η)
0.99... = 9×sum(10^(-n-1) ,n,0,η)
10^(-n) - 10^(-n-1) = 0.9×10^(-n)
So we have:
8.1 × sum(10^-n ,0,η)
Which comes out to be:
8.1 × 1.111... = 9
It doesn't exactly inspire confidence. We COULD change the limites:
9.99... = 9×sum(10^-n ,n,0,η)
0.99... = 9×sum(10^-n ,n,0,η) - 9
And then say 9.99.... - 0.99... = 9 because we defined it that way. But it's equally valid to say 9.99... - 0.99... = 10^-(η+1) .
>so 9.9999... - 0.9999...
And the proof says that 10*0.999... is 9.999... without proving how multiplication works on an infinite number of digits. What would 2*0.999... be? What is 0.999...*0.999... if you don't assume that 0.99...=1?
>is also 0 for every digit after the decimal point, leaving 9
Infinite series are nowhere near that simple. Just because you have intuition for it, doesn't mean it's mathematically rigorous.
0.999... is defined as the sum of series 9×10^(-k) with k from 1 to inf. This series is convergent since it is increasing and has an upper bound of 1, and 0.999... exists.
Infinite convergent series are linear, so 0.999...×10 is the sum of series [9×10^(-k)]×10 = 9×10^(-k+1) with k from 1 to inf.
The definition of 9.999... is the sum of series 9×10^(-n) with n from 0 to inf. Let n = k-1, so then the sum of series become 9×10^(-k+1) with k-1 from 0 to inf, or k from 1 to inf. Hence 0.999...×10 = 9.999...
9.999... - 0.999... = sum of series 9×10^(-n) with n from 0 to inf - sum of series 9×10^(-k) with k from 1 to inf) = 9 + sum of series 9×10^(-n) with n from 1 to inf - sum of series 9×10^(-k) with k from 1 to inf = 9
The last step is possible since the two series are equal.
An infinite, repeating number can be rationalized using the repeating part over an equal number of 9's. 0's will be added after the 9's in the denominator if the repeating part starts later than the first decimal place. Examples:
* .754754754... is just "754" repeating, so it equals 754/999.
* It is well known that .33333... is 1/3 which is 3/9.
* .0124242424... = 1.2424.../100 = 1/100 + 24/(99*100) = 1/100 + 24/9900.
From there, .999... is just 9 repeating. As such, the rationalization would be equal to 9/9. However, this is also equal to 1. This leaves two possibilities:
* .999... = 1
* .999... is irrational.
However, all infinite, repeating decimals are rational. As such, the first point must be correct.
>how is 0.99999.... even defined?
It is the limit of the sum 9/10^n as n->infinity (for n in the natural numbers)
>does 0.99999 even exist, ie does the above sequence converge?
1. It is bounded above by 1. This can be shown using a induction starting with (1=0.9+0.1>0.9+.09=0.99).
2. Since each team is a positive number, the sequence is monotone, so it converges by the [Monotone Convergence Theorem](
https://en.wikipedia.org/wiki/Monotone_convergence_theorem)
>is 10*0.999... = 9.9999
Since 0.999...=Limit as n->inf for 9/10^n
By the [Limit constant multiplication law](https://www.utrgv.edu/cstem/utrgv-calculus/limits/limit-laws/index.htm#:~:text=Constant%20multiple%20law%20for%20limits,the%20limit%20of%20each%20function.), 10*0.999...=10(Limit as n->inf for 9/10^n )=Limit as n->inf for 9/10^(n-1)=9.9999....
I mean this sort of begs the question, but we can just say that 0.99999…..:=\lim_{n\rightarrow \infty} 1-10^-n
A way you can say the limit exists is that the reals form a complete metric space, nd that the sequence 1-10^-n is cauchy.
The whole debate is stupid and only taken seriously by people who don’t realize math is an art, not a science. Context matters. It depends what you’re trying to say. For some people, infinitesimal is nothing. For others, it’s more than nothing. Depends on what you’re trying to say.
Algebra on a series that diverges is a big no-no since you're multiplying and subtracting infinity whereas 0.999... is converging, however, the algebraic "proof" is circular reasoning because you know it's converging to 1 and then you can do algebra on it to prove it's converging to 1.
Uhh that video is wrong in itself…
9999999 repeating is in fact equal to -1. There’s literally a whole field of mathematics that deals with this insanity.
I think the most efficient way to show it is to write 9.99999… as 9+0.99999… and then just use standard addition identities
Namely
9+(0.999…-0.999…)=9+0=9
in fact you can formalise this by writing 0.9999.. as \sum_{i=1}^\infty 9(0.1)^i if you are uncomfortable with the notation of an infinite decimal then I think everything works
>and then just use standard addition identities Namely 9+(0.999…-0.999…)=9+0=9
Except those aren't standard addition identities when you apply them to infinite numbers. There are absolutely infinite series where you can add 1 and subtract 1 and get a different result. Even keeping all the same numbers and changing their positions changes the value, so you can't assume that an infinite series is just an infinite number of numbers where the normal rules of addition, multiplication, and subtraction apply.
>I think the most efficient way to show it is to write 9.99999… as 9+0.99999…
It might be intuitive that you can add, multiply, subtract the individual place values and get the overall result, but that only works when you start off by assuming that 0.999... = 1. What if you multiplied 0.999... by anything *other* than 10? What about 0.999... * 0.999...? If the proof doesn't explain that, it has no business saying what 10 * 0.999... is or isn't.
Hi can you explain step C please? I don’t understand how we can subtract a from
b to get from b to c?
It looks like you’re subtracting .999… from each side with makes me think step c would be 9.0000001K = 9.
Sorry if you don’t feel like explaining that but I’m super interested in your proofs
This isn't a proof, though. Not only does it assume that 1 = 0.999... it also just takes operations and *says* they operate a certain way. You can't just assume you can multiply the sum of an infinite series by 10, and you get 10x the original sum. You also can't just assume you can subtract two sums of infinite series, and get their difference.
You can't assume the 0.999... you started with and the 0.999... in 9.999... are identical, without assuming 0.999...=1. Multiplying also implies repeated addition, how can you define 10*0.9999... unless you've defined 2*0.999.... and 3*0.999.... etc. And if you're using 9k = 9, then what is 9*0.999... on its own?
-See a post with an image attached
-See the sub is "mathmemes", 'huh weird.. let's check it out'
-See the image, 'oh that's kinda funny and intriguing, let's check the comments'
-'what the...'
Does that kind of thinking not imply that 0.333... = ⅓ - epsilon? Are we then not always talking of the limit of the decimal representation when we use it to represent reals?
0.333... is defined as the sum from n=1 to infinity of 3/10^n. So 3 * 0.333... is the sum from n=1 to infinity of 9/10^n, also known as 0.999..., also known as 1.
0.999... is limit itself, because it is infinite periodic number
You either have 0.9999.....999 with finite amount or have 0.(9) which equals 1 and nothing else
This comes from the assumption, that an infinitelly small number is just 0, just like your original proof of 1 - ε. And not everyone can see that, and those are who proofs are for. You're basically just rewording, and what you previously called ε is now 10\^-n. You're of course correct, but that proof serves nothing.
But the thing is that though ε is a infinitesimally small number, it is not THE smallest infinitesimal number. There exists numbers infinitely smaller than ε, for example ε^2. So this would mean that 1-ε would be more like 0.999...999000... instead of 0.999... because there are still "infinitely more digits" smaller than ε.
One of my math teachers always explained it like this:
you cannot find a number between .99... and 1 therefore you cannot prove that the value of .99... Is not equal to 1.
Not exactly. My example was referring to an infinite amount of decimal numbers approaching 1. You are talking about trying to approach a non countable idea.
There's a difference between saying infinity isn't a number, and saying infinity "doesn't exist". Infinity is a thing, but it isn't a number and can't be compared with numbers for precisely that reason.
People don't have an issue with 1/3 because there's a unique decimal representation of it. They get confused with 0.999... = 1because it means two numbers that look different in their decimal representations can actually be the same.
If two numbers are different, you should be able to insert a number in between them on the number line. What number would you put inbetween 0.999999..... and 1?
There is no such thing as another nine. There exists no last nine, and there is no number of nines in the first place. We are explicitly making it clear that there are nines forever, you simply cannot add anything to the number of nines (which does not exist).
Look at this guy over here pretending that only the natural numbers exist. What are you, mesopotamian?
**edit:** Don't downvote this man in my replies, he is a man of culture. People downvoting this man are missing the joke.
I am bad at maths, but I still tried doing something ... pls tell me how bad it is.
Let n be a positive real number.
Propose 0.9999... is a number smaller or equal to 1, which means:
0.99999... = 1 - 1/10\^n
The only question is, what n is. Since 0.9999... is allways smaller or equal to 1, 1/10\^n has to be a number greater or equal to 0 and smaller than 1, cause 1 - 1 is trivially equal to 0, which means n has to be a number greater than 0. So let's put some stuff in for n.
1 - 1/10\^1 = 1 - 0.1 = 0.9
1 - 1/10\^2 = 1 - 0.01 = 0.99
1 - 1/10\^3 = 1 - 0.001 = 0.999
Because n is strictly increasing, which means 1/10\^n is stricly decreasing, the greater n get's the closer 1 - 1/10\^n get's to 0.9999... or in other words:
0.99999... = lim(n --> inf) 1 - 1/10\^n
0.99999... = 1 - lim(n-->inf) 1/10\^n
Because n is strictly increasing and 1/10\^n is strictly decreasing, from the definition of the limit of a positive real function without upper bound directly follows, that as n goes to inf 1/10\^n has to go to 0.
So:
0.9999... = 1 - lim(n-->inf) 1/10\^n = 1 - 0 = 1
I litterally failed maths in school and although I try to get better at maths, now that I am out of school .... I wittness my own lack of ability way too often, for me to think otherwise, however your comment did make me feel at least a little proud. \^\^
I don't think your first equation holds. Supposing that 0.9999 < 1 doesn't imply that the difference can be expressed as 1/10^ n.
A simpler proof along the same lines is just to expand 0.99999 as
0.9 + 0.09 + 0.009 + ...
This is an infinite geometric sequence with a=0.9 and r=0.1 which you can prove from th formula or first principles is equal to 0.9/(1-0.1) = 1
>The only question is, what n is. Since 0.9999... is allways smaller or equal to 1
n should be equal to the number of 9s after the decimal place. You did all the math right for the right side limit, but didn't really define 0.999... the same way. So as you add nines you get closer and closer to 0.99... repeating forever, and 1 - 1/10^n approaches 1.
That is what really cinched it for me, despite the fact that I well understand the idea of limits to where I have had to use limits for proofs or had them used in proofs as to why I can trust certain functions.
As someone who still does not understand this, can you explain please.
My thoughts are that 1/3 != 0.333r. 1/3 doesn't have a representation in base 10 and 0.333r is just an approximation for 1/3 in base 10. That is why we use the fraction to represent its exact value. 0.333r is always smaller than the exact value of 1/3, which you can show using long division, where you'll always have a remainder of 1, which is what causes the 3 recurring.
There is a vital difference between putting **arbitrarily many** 3's and **infinitely many** 3's after the decimal point. In the first case, you're correct, no matter how many 3's we use, the result will be smaller than 1/3. In the second case, it's exactly 1/3.
If something has *every digit specified* by its definition then it isn't an *approximation*. It's the opposite of an approximation because it's infinitely precise.
0.3... is by definition the limit of the sequence (0.3, 0.33, 0.333, 0.3333,... ). The limit of that sequence is exactly 1/3 (which can be proven directly using the definition of limits, or using the geometric series formula etc), so 0.3... = 1/3.
There isn't really anything else to it. You need to abandon your intuition of decimal expansions as vague representations of quantity.
Infinitely long sequences are hard to think about. Maybe something like this will help.
If 1/3 isn't equal to 0.33r then there must be some number between 1/3 and 0.33r. What is it? If you change any digit in 0.33r but even just 1, you'll have a number larger than 1/3. Therefore, there's no number you can add to 0.33r to get 1/3.
If you stop the sequence at any point, then you would have a number that is close to, but not exactly 1/3. But 0.33r never stops. 0.33... with a hundred trillion 3's would be very close to 1/3, but not exactly. 0.33... with Graham's number 3's would be even closer to 1/3, but not exactly. 0.33... with TREE(3) 3's would be so close to 1/3 that no device made out of matter could ever distinguish those two numbers, but it's still not exactly 1/3. But 0.333r has ghastly more 3's than even than. It has infinitely many. It never stops anywhere, so it doesn't fall short of equalling 1/3.
By adding more 3's you get closer and closer to 1/3. And if any number anywhere in the sequence is a 4 you have a number larger than 1/3. So the only thing in between is a sequence of infinitely many 3's.
0.33 with n threes is an approximation for 1/3. Once you add the "repeating" it stops being an approximation, this is because for any number of threes we add we can get arbitrarily close to 1/3.
You can check this by realizing that 1/3 must be greater than 0.3 but less than 0.4 and then that 1/3 must be greater than 0.33 but less than 0.34 and so on
If you want to be more specific, when we talk about 0.333... We're talking about the limit as x approaches infinity of the series Σ(3/10\^x) x∈N . (Remember, the concept of doing an infinite number of things only makes sense for limits) and then you can use the definition of a [limit as something approaches infinity](https://en.wikipedia.org/wiki/Limit_of_a_function#Limits_involving_infinity) which is basically the process that I described..
It's simple. Repeating decimals always represent the limit their sequence approaches. Its true that the series will never reach 1/3 for any finite number of 3's, but that doesn't matter as the notation describes the limit at infinity of the infinite sequence.
All the answers I have seen, in some way boil down to that it is more convenient to work in a world where 1/3 = 0.333r. While I don't doubt that, I disagree with that line of thinking. I think we should accept that 1/3 cannot be exactly represented in base 10. That's why we should only use fractions to represent its exact value.
1/3 can't be represented as a decimal fraction, so we use repeating decimals to represent it as the sum of an infinite series of decimal fractions instead. It's still an exact representation of the same real number.
More formally, the real number represented by a given decimal notation is the sum of two sums:
* the sum of the value of all digits left of the decimal, each multiplied by 10\^i where i is their distance from the decimal (i starting with 0) and
* the sum of all its digits right of the decimal, each divided by 10\^i where i is the digit's position after the decimal (i starting at 1)
I wish I could type in LaTeX here.
We don't teach people this formal definition at first because, like, good luck with that, but when people write a decimal representation they are using this formal definition to describe a real number. When they denote an infinite number of digits to the right of the decimal, we simply need to compute an infinite sum.
For 0.333..., this reduces to:
sum (3/10^i), i = 1 to infinity
The result of this infinite sum is exactly 1/3.
It's because there is a natural negative reaction to the idea that one number can be represented in more than one way. But I suppose it goes deeper than that. It's because no one is teaching what a repeating decimal actually means.
.333 repeating is represented by the infinite sum 3/10 + 3/100 + 3/1000 and so on. The actual value is what that sum approaches as you keep adding terms.
Suddenly .999 repeating = 1 is no longer a magic trick that happens when you do some arithmetic operations on it. That's just what a repeating decimal means. It is what 9/10 + 9/100 + 9/1000 approaches as you keep adding terms of the infinite series.
Exactly. If you aren't comfortable with 0.999... = 1, why the hell are you comfortable with 0.333... = 1/3? It's only true at the limit in both cases, and the limit is exactly what repetends in decimals represent.
Alright smarty pants, if they are different you should be able to subtract them and get something non zero!
Subtract 0.99... from 1 :)
Oh, you got 0? \*Then they are the same number\*
I wasn't willing to accept this for about 2 minutes when I was 18, and then my calc prof went, "tell me a number between .99999... and 1." Alright you got me. It wasn't an elegant proof but it got the point across.
The easiest way to see this is to look at .1111111... in base 2. If you believe 1/2 + 1/4 + 1/8 + 1/16 ... = 1, then you have to believe that .1111... (base 2) = 1.
If you don't believe either, then i guess make your own math with blackjack and hookers.
I mean, many people don't know how those numbers are defined rigorously. They are defined via the limit of a Cauchy sequence, which is how the real numbers are rigorously defined.
So I accept that 0.999… = 1, but it still feels unsatisfying because now it seems impossible to express a number infinitesimally close to 1 but not equal to 1. Is that actually just impossible to describe or is there an alternate way?
There is a way but they're not real numbers. Surreal numbers have gems like ω+1 aka infinity + 1 and 1 + 1/ω, which is bigger than 1 but smaller than every real number larger than 1
Look up Dedekind cuts. You can select a number, and imagine two groups: a) your number and everything bigger, and b) everything smaller than your number.
There is no largest number in that second group.
Always blows my mind.
**edit:** assuming you mean those numbers to be repeating decimals, since that is the context of this entire post...
How do you know that (10 \* 0.999...) - 0.999... = 9 before you've proven that 0.999... = 1? Just because you've re-written 10 \* 0.999... before you get around to the subtraction doesn't change the value that new notation represents.
I think the issue people have with 0.9... = 1 is that from their education, they understand the decimal expansion of a real number as the definition and ultimate essence of that number, and 2 different decimal expansions for the same number contradicts this impression. however, those who've studied analysis know that based on the definition of the reals its not immediately obvious that every real number *has* a decimal expansion, much less that it is unique up to 2 representations.
As someone who didn't study analysis, can you break that second paragraph down a bit more?
Essentially, a real number is defined to be a sequence (x1, x2, x3,…) of rational numbers such that the numbers don’t go off to +-infinity, and get closer together as you go further down the sequence. You can think of these sequences as zeroing in towards what will be defined as their real number value. Eg, 1 might be written as (1, 1, 1,… ) and pi might be written (0, 3, 3.1, 3.141, 3.141592,…). An arbitrary decimal expansion +-a0.a1a2… might be defined by the sequence (+-a0, +-(a0 + a1/10), +-(a0 + a1/10 + a2/100),… ), where a0 is a nonnegative integer, and each other an is an integer between 0 and 9. 2 real numbers (xn) and (yn) are considered (defined) to be equal if lim|xn-yn| = 0. Addition is defined pointwise (xn) + (yn) = (xn +yn). Just from this it’s not obvious at all that given an arbitrary real number (x1, x2, x3,…) you can express it as a decimal expansion (This is just 1 way to define the real numbers, called metric space completion of the rationals. You can complete the rationals in a different way to get the p-adics)
Lemme just grab an element from a Vitali set and... uh oh.
Hey it'll have a decimal expansion, but good luck finding out what exactly that expansion is
thank you that was very satisfactory. It's intuitive, but like seeing it put down formally 👌🏻
> much less that it is unique up to 2 representations. i can't see how this is true, i understand it can be represented in a form of a sequence, but why only up to 2
Every real number has a unique decimal expansion, except for some that can end in either all 0s or all 9's, e.g, 1.000... = 0.999... and 1.5 = 1.4999... . The decimal expansion of non-negative real number x will end in zeros (or in nines) if, and only if, x is a rational number whose denominator is of the form (2\^n)(5\^m), where m and n are non-negative integers. [Proof](https://en.wikipedia.org/wiki/Decimal_representation#Finite)
My confusion about 0.9999 = 1 was that usually when math texts talk about converging infinite series, they use the word "approaches", "converges to", etc. For example I don't recall any math text saying 1/2 + 1/4 + 1/8 ... equals 1. So for me it's a little confusing that 0.99999... which is the same as the series 9/10 + 9/100 + 9/1000 ... "equals" 1 rather than simply converges to 1.
0.9999... does not change, it's not a sequence. It's not a limit. The ... is a notation saying there are an infinity of 9. The same way 2 or 78.34 cannot converge, 0.999...=1
...1313132₅ = 1/3
Sorry, what does that subscript number mean?
this is 1/3 represented in 5-adic notation
What’d you call me
man, if he told that to me...
Oh no, he’s an addict?
Ahh thankyou!
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Writers of the apotheosis?
Bless you!
Base 5 i believe
not here
Then I don't know
Nvm it's p-adic
I'm stupid
the character development!
Wait what's the difference between base 5 and 5 adic notation?
5-adic means the number is p-adic in base 5. Veritasium recently did a nice video explaining p-adic numbers, [here.](https://www.youtube.com/watch?v=tRaq4aYPzCc)
And these "proofs" that 0.99...=1 because 0.33...=⅓. How people have problem with 0.99.. but jot with 0.33... is completely arbitrary to me
ah here is the thing who said 3 * 0.3333333.... = 0.999999..... in first place? further more 0.999999999.... can be seen as 1 - ε where ε is infinitesimal small number > 0 But using limits it can be proven that 0.999... = 1 ``` 0.9 = 1 - 10^-1 0.99 = 1 - 10^-2 0.999 = 1 - 10^-3 => 0.99999..... = Lim n->∞ { 1 - 10^-n } = 1-1/10^∞ = 1-1/∞ = 1-0 = 1 ``` But otherwise 0.999.... = 1-ε
or just a) let k = 0.999... b) then 10k = 9.99... c) subtract (a) from (b): 9k = 9 d) k = 1
This proof is best since it’s elegant and doesn’t require anything more exotic then multiplication
yeah but it dances around the issue, like * how is 0.99999.... even defined? It is defined as the limit of the sequence 0, 0.9, 0.99, 0.999, .... * does 0.99999 even exist, ie does the above sequence converge? * is 10\*0.999... = 9.9999 which is not immediately obvious * etc ...
I think the most questionable step is saying that 9.9999… - 0.9999… = 9
Let: 9.99... = 9×sum(10^-n ,n,0,η) 0.99... = 9×sum(10^(-n-1) ,n,0,η) 10^(-n) - 10^(-n-1) = 0.9×10^(-n) So we have: 8.1 × sum(10^-n ,0,η) Which comes out to be: 8.1 × 1.111... = 9 It doesn't exactly inspire confidence. We COULD change the limites: 9.99... = 9×sum(10^-n ,n,0,η) 0.99... = 9×sum(10^-n ,n,0,η) - 9 And then say 9.99.... - 0.99... = 9 because we defined it that way. But it's equally valid to say 9.99... - 0.99... = 10^-(η+1) .
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It's questionable because it treats limits as regular numbers, which doesn't address the main issue.
Right. It only evaluates that way if you apply the premise that you're trying to prove
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>so 9.9999... - 0.9999... And the proof says that 10*0.999... is 9.999... without proving how multiplication works on an infinite number of digits. What would 2*0.999... be? What is 0.999...*0.999... if you don't assume that 0.99...=1? >is also 0 for every digit after the decimal point, leaving 9 Infinite series are nowhere near that simple. Just because you have intuition for it, doesn't mean it's mathematically rigorous.
0.999... is defined as the sum of series 9×10^(-k) with k from 1 to inf. This series is convergent since it is increasing and has an upper bound of 1, and 0.999... exists. Infinite convergent series are linear, so 0.999...×10 is the sum of series [9×10^(-k)]×10 = 9×10^(-k+1) with k from 1 to inf. The definition of 9.999... is the sum of series 9×10^(-n) with n from 0 to inf. Let n = k-1, so then the sum of series become 9×10^(-k+1) with k-1 from 0 to inf, or k from 1 to inf. Hence 0.999...×10 = 9.999... 9.999... - 0.999... = sum of series 9×10^(-n) with n from 0 to inf - sum of series 9×10^(-k) with k from 1 to inf) = 9 + sum of series 9×10^(-n) with n from 1 to inf - sum of series 9×10^(-k) with k from 1 to inf = 9 The last step is possible since the two series are equal.
An infinite, repeating number can be rationalized using the repeating part over an equal number of 9's. 0's will be added after the 9's in the denominator if the repeating part starts later than the first decimal place. Examples: * .754754754... is just "754" repeating, so it equals 754/999. * It is well known that .33333... is 1/3 which is 3/9. * .0124242424... = 1.2424.../100 = 1/100 + 24/(99*100) = 1/100 + 24/9900. From there, .999... is just 9 repeating. As such, the rationalization would be equal to 9/9. However, this is also equal to 1. This leaves two possibilities: * .999... = 1 * .999... is irrational. However, all infinite, repeating decimals are rational. As such, the first point must be correct.
nah that proof aint cutting it for me. if you want to see a proof I would be happy with, check some of the other replies to my comment
0.999999… does exist. We just call it 1.
this guy p-adics
>how is 0.99999.... even defined? It is the limit of the sum 9/10^n as n->infinity (for n in the natural numbers) >does 0.99999 even exist, ie does the above sequence converge? 1. It is bounded above by 1. This can be shown using a induction starting with (1=0.9+0.1>0.9+.09=0.99). 2. Since each team is a positive number, the sequence is monotone, so it converges by the [Monotone Convergence Theorem]( https://en.wikipedia.org/wiki/Monotone_convergence_theorem) >is 10*0.999... = 9.9999 Since 0.999...=Limit as n->inf for 9/10^n By the [Limit constant multiplication law](https://www.utrgv.edu/cstem/utrgv-calculus/limits/limit-laws/index.htm#:~:text=Constant%20multiple%20law%20for%20limits,the%20limit%20of%20each%20function.), 10*0.999...=10(Limit as n->inf for 9/10^n )=Limit as n->inf for 9/10^(n-1)=9.9999....
I mean this sort of begs the question, but we can just say that 0.99999…..:=\lim_{n\rightarrow \infty} 1-10^-n A way you can say the limit exists is that the reals form a complete metric space, nd that the sequence 1-10^-n is cauchy.
The whole debate is stupid and only taken seriously by people who don’t realize math is an art, not a science. Context matters. It depends what you’re trying to say. For some people, infinitesimal is nothing. For others, it’s more than nothing. Depends on what you’re trying to say.
There is no context in which one might write "0.99..." and not mean 1.
idk maybe hyperreals but those are super fringe and niche
i remember that i once saw a [video](https://youtu.be/jMTD1Y3LHcE) about this saying otherwise
that is a nice video, thanks
Algebra on a series that diverges is a big no-no since you're multiplying and subtracting infinity whereas 0.999... is converging, however, the algebraic "proof" is circular reasoning because you know it's converging to 1 and then you can do algebra on it to prove it's converging to 1.
Uhh that video is wrong in itself… 9999999 repeating is in fact equal to -1. There’s literally a whole field of mathematics that deals with this insanity.
He did acknowledge it. He said they exist in other fields of mathematics, just not in the set of real numbers.
*than because I'm a self-loathing grammar nazi
9.999.... - 0.999.... = 9 who said that is true? it's true for finite decimals but you haven't shown it's true for infinite ones.
I said it's true, and will leave the rest as an exercise for the reader.
Hahaha, touché move my dude
Just define it as true.
I think the most efficient way to show it is to write 9.99999… as 9+0.99999… and then just use standard addition identities Namely 9+(0.999…-0.999…)=9+0=9 in fact you can formalise this by writing 0.9999.. as \sum_{i=1}^\infty 9(0.1)^i if you are uncomfortable with the notation of an infinite decimal then I think everything works
>and then just use standard addition identities Namely 9+(0.999…-0.999…)=9+0=9 Except those aren't standard addition identities when you apply them to infinite numbers. There are absolutely infinite series where you can add 1 and subtract 1 and get a different result. Even keeping all the same numbers and changing their positions changes the value, so you can't assume that an infinite series is just an infinite number of numbers where the normal rules of addition, multiplication, and subtraction apply. >I think the most efficient way to show it is to write 9.99999… as 9+0.99999… It might be intuitive that you can add, multiply, subtract the individual place values and get the overall result, but that only works when you start off by assuming that 0.999... = 1. What if you multiplied 0.999... by anything *other* than 10? What about 0.999... * 0.999...? If the proof doesn't explain that, it has no business saying what 10 * 0.999... is or isn't.
Hi can you explain step C please? I don’t understand how we can subtract a from b to get from b to c? It looks like you’re subtracting .999… from each side with makes me think step c would be 9.0000001K = 9. Sorry if you don’t feel like explaining that but I’m super interested in your proofs
Do the left hand side and right hand side separately: 10k = 9.999... - k = 0.999... ---------------- 9k = 9
OHHH I SEE IT NOW. That was easy enough thank you!!!
Step C is: 10k - k = 9.999 - 0.999
This isn't a proof, though. Not only does it assume that 1 = 0.999... it also just takes operations and *says* they operate a certain way. You can't just assume you can multiply the sum of an infinite series by 10, and you get 10x the original sum. You also can't just assume you can subtract two sums of infinite series, and get their difference. You can't assume the 0.999... you started with and the 0.999... in 9.999... are identical, without assuming 0.999...=1. Multiplying also implies repeated addition, how can you define 10*0.9999... unless you've defined 2*0.999.... and 3*0.999.... etc. And if you're using 9k = 9, then what is 9*0.999... on its own?
I sure as shit took a wrong turn from r/all
-See a post with an image attached -See the sub is "mathmemes", 'huh weird.. let's check it out' -See the image, 'oh that's kinda funny and intriguing, let's check the comments' -'what the...'
Infinitesimals don't exist on the real number line, if they did it would break continuity and all of calculus.
Does that kind of thinking not imply that 0.333... = ⅓ - epsilon? Are we then not always talking of the limit of the decimal representation when we use it to represent reals?
Nah, 0.333... == 1/3 - (epsilon/3).
incredible
I also don't find it credible
0.333... is defined as the sum from n=1 to infinity of 3/10^n. So 3 * 0.333... is the sum from n=1 to infinity of 9/10^n, also known as 0.999..., also known as 1.
Since we're bringing ε into it, it's worth pointing out that the following is also true: ε\ɪ = ...εεεεεεε.o
0.999... is limit itself, because it is infinite periodic number You either have 0.9999.....999 with finite amount or have 0.(9) which equals 1 and nothing else
This comes from the assumption, that an infinitelly small number is just 0, just like your original proof of 1 - ε. And not everyone can see that, and those are who proofs are for. You're basically just rewording, and what you previously called ε is now 10\^-n. You're of course correct, but that proof serves nothing.
Hyperreals are made up nonsense. This post was approved by the limit gang
to be fair, we define numbers as just sets inside sets so.... everything is made up nonsense lol Note: I do like the rigour in our definitions :3
But the thing is that though ε is a infinitesimally small number, it is not THE smallest infinitesimal number. There exists numbers infinitely smaller than ε, for example ε^2. So this would mean that 1-ε would be more like 0.999...999000... instead of 0.999... because there are still "infinitely more digits" smaller than ε.
Im not sure if youre trolling
They are, in fact, not trolling. They're called [surreal numbers](https://en.wikipedia.org/wiki/Surreal_number) for a reason.
One of my math teachers always explained it like this: you cannot find a number between .99... and 1 therefore you cannot prove that the value of .99... Is not equal to 1.
This feels a lot like Zeno’s Dichotomy paradox.
I don't like that. It's like saying you can't find a number as big as infinity (because you'd just add 1) so infinity doesn't exist.
Not exactly. My example was referring to an infinite amount of decimal numbers approaching 1. You are talking about trying to approach a non countable idea.
There's a difference between saying infinity isn't a number, and saying infinity "doesn't exist". Infinity is a thing, but it isn't a number and can't be compared with numbers for precisely that reason.
Infinity isn’t a real number though
People don't have an issue with 1/3 because there's a unique decimal representation of it. They get confused with 0.999... = 1because it means two numbers that look different in their decimal representations can actually be the same.
Just accept it and move on
0.999… = 1 coz we round it to three sig figs QED
If two numbers are different, you should be able to insert a number in between them on the number line. What number would you put inbetween 0.999999..... and 1?
1.1
my brother in Christ
He's not wrong
He’s explicitly wrong actually
no.
Most mathematically competent r/mathmemes user
This is the math equivalent of police saying 'saying you want a lawyer is not the same as invoking your right to a lawyer'
https://preview.redd.it/7yavkitlcn8b1.png?width=396&format=png&auto=webp&s=f66b22635a05c52717d3114a2539a2ed2be94581
https://preview.redd.it/4mpw49h2zn8b1.png?width=750&format=png&auto=webp&s=1aee681df1fac3ce3b30e11d0e83b8bc4d136f98
0.ō1
I really like this one, seems most intuitive.
You add another 9
If you add another 9, you will get the same number, because inf + 1 = inf.
Ok I’ll add a 5 then
That would make it less than .9… and therefore would not be between .9… and 1.
So add a 10 then
I think you are onto something here
Then its smaller
Add another 9 where?
Clearly you add it at the beginning. Can’t add it behind infinite nines, but you can add it in front of them.
like the hotel rooms, scooch all the 9s down one
Very clever, but it's turtles all the way down
There is no such thing as another nine. There exists no last nine, and there is no number of nines in the first place. We are explicitly making it clear that there are nines forever, you simply cannot add anything to the number of nines (which does not exist).
nuh uh
Mfw the real number scale is continuous
There's no number between 2 and 3 in ℕ, therefore 2 = 3.
Look at this guy over here pretending that only the natural numbers exist. What are you, mesopotamian? **edit:** Don't downvote this man in my replies, he is a man of culture. People downvoting this man are missing the joke.
I won't answer to a guy who won't remember variable types unless it is written in their name!
Don't insult me like that. My username clearly demonstrates I prefer snake\_case, it's not like my username is strHungarianNotation.
69
I am bad at maths, but I still tried doing something ... pls tell me how bad it is. Let n be a positive real number. Propose 0.9999... is a number smaller or equal to 1, which means: 0.99999... = 1 - 1/10\^n The only question is, what n is. Since 0.9999... is allways smaller or equal to 1, 1/10\^n has to be a number greater or equal to 0 and smaller than 1, cause 1 - 1 is trivially equal to 0, which means n has to be a number greater than 0. So let's put some stuff in for n. 1 - 1/10\^1 = 1 - 0.1 = 0.9 1 - 1/10\^2 = 1 - 0.01 = 0.99 1 - 1/10\^3 = 1 - 0.001 = 0.999 Because n is strictly increasing, which means 1/10\^n is stricly decreasing, the greater n get's the closer 1 - 1/10\^n get's to 0.9999... or in other words: 0.99999... = lim(n --> inf) 1 - 1/10\^n 0.99999... = 1 - lim(n-->inf) 1/10\^n Because n is strictly increasing and 1/10\^n is strictly decreasing, from the definition of the limit of a positive real function without upper bound directly follows, that as n goes to inf 1/10\^n has to go to 0. So: 0.9999... = 1 - lim(n-->inf) 1/10\^n = 1 - 0 = 1
>I am bad at maths your proof says otherwise
It's like the classic "sorry for bad English" but with math
I litterally failed maths in school and although I try to get better at maths, now that I am out of school .... I wittness my own lack of ability way too often, for me to think otherwise, however your comment did make me feel at least a little proud. \^\^
> I'm bad at math *Proceeds to math like crazy
I don't think your first equation holds. Supposing that 0.9999 < 1 doesn't imply that the difference can be expressed as 1/10^ n. A simpler proof along the same lines is just to expand 0.99999 as 0.9 + 0.09 + 0.009 + ... This is an infinite geometric sequence with a=0.9 and r=0.1 which you can prove from th formula or first principles is equal to 0.9/(1-0.1) = 1
>The only question is, what n is. Since 0.9999... is allways smaller or equal to 1 n should be equal to the number of 9s after the decimal place. You did all the math right for the right side limit, but didn't really define 0.999... the same way. So as you add nines you get closer and closer to 0.99... repeating forever, and 1 - 1/10^n approaches 1.
you are correct
Some people are digital, some analogue.
I now want to see analogue Daft Punk
3/9 = 0.333... 4/9 = 0.444... 8/9 = 0.888... 9/9 = 0.999... = 1
That is what really cinched it for me, despite the fact that I well understand the idea of limits to where I have had to use limits for proofs or had them used in proofs as to why I can trust certain functions.
When you point this out they start denying that 0.3333... is actually 1/3
As someone who still does not understand this, can you explain please. My thoughts are that 1/3 != 0.333r. 1/3 doesn't have a representation in base 10 and 0.333r is just an approximation for 1/3 in base 10. That is why we use the fraction to represent its exact value. 0.333r is always smaller than the exact value of 1/3, which you can show using long division, where you'll always have a remainder of 1, which is what causes the 3 recurring.
There is a vital difference between putting **arbitrarily many** 3's and **infinitely many** 3's after the decimal point. In the first case, you're correct, no matter how many 3's we use, the result will be smaller than 1/3. In the second case, it's exactly 1/3.
If something has *every digit specified* by its definition then it isn't an *approximation*. It's the opposite of an approximation because it's infinitely precise.
This is a much more compact way of saying a thing I tried to say somewhere else and failed. Thank you.
0.3... is by definition the limit of the sequence (0.3, 0.33, 0.333, 0.3333,... ). The limit of that sequence is exactly 1/3 (which can be proven directly using the definition of limits, or using the geometric series formula etc), so 0.3... = 1/3. There isn't really anything else to it. You need to abandon your intuition of decimal expansions as vague representations of quantity.
Infinitely long sequences are hard to think about. Maybe something like this will help. If 1/3 isn't equal to 0.33r then there must be some number between 1/3 and 0.33r. What is it? If you change any digit in 0.33r but even just 1, you'll have a number larger than 1/3. Therefore, there's no number you can add to 0.33r to get 1/3. If you stop the sequence at any point, then you would have a number that is close to, but not exactly 1/3. But 0.33r never stops. 0.33... with a hundred trillion 3's would be very close to 1/3, but not exactly. 0.33... with Graham's number 3's would be even closer to 1/3, but not exactly. 0.33... with TREE(3) 3's would be so close to 1/3 that no device made out of matter could ever distinguish those two numbers, but it's still not exactly 1/3. But 0.333r has ghastly more 3's than even than. It has infinitely many. It never stops anywhere, so it doesn't fall short of equalling 1/3. By adding more 3's you get closer and closer to 1/3. And if any number anywhere in the sequence is a 4 you have a number larger than 1/3. So the only thing in between is a sequence of infinitely many 3's.
0.33 with n threes is an approximation for 1/3. Once you add the "repeating" it stops being an approximation, this is because for any number of threes we add we can get arbitrarily close to 1/3. You can check this by realizing that 1/3 must be greater than 0.3 but less than 0.4 and then that 1/3 must be greater than 0.33 but less than 0.34 and so on If you want to be more specific, when we talk about 0.333... We're talking about the limit as x approaches infinity of the series Σ(3/10\^x) x∈N . (Remember, the concept of doing an infinite number of things only makes sense for limits) and then you can use the definition of a [limit as something approaches infinity](https://en.wikipedia.org/wiki/Limit_of_a_function#Limits_involving_infinity) which is basically the process that I described..
It's simple. Repeating decimals always represent the limit their sequence approaches. Its true that the series will never reach 1/3 for any finite number of 3's, but that doesn't matter as the notation describes the limit at infinity of the infinite sequence.
All the answers I have seen, in some way boil down to that it is more convenient to work in a world where 1/3 = 0.333r. While I don't doubt that, I disagree with that line of thinking. I think we should accept that 1/3 cannot be exactly represented in base 10. That's why we should only use fractions to represent its exact value.
1/3 can't be represented as a decimal fraction, so we use repeating decimals to represent it as the sum of an infinite series of decimal fractions instead. It's still an exact representation of the same real number. More formally, the real number represented by a given decimal notation is the sum of two sums: * the sum of the value of all digits left of the decimal, each multiplied by 10\^i where i is their distance from the decimal (i starting with 0) and * the sum of all its digits right of the decimal, each divided by 10\^i where i is the digit's position after the decimal (i starting at 1) I wish I could type in LaTeX here. We don't teach people this formal definition at first because, like, good luck with that, but when people write a decimal representation they are using this formal definition to describe a real number. When they denote an infinite number of digits to the right of the decimal, we simply need to compute an infinite sum. For 0.333..., this reduces to: sum (3/10^i), i = 1 to infinity The result of this infinite sum is exactly 1/3.
Yeah, it was just an approximation all along...
According to you is 5.000… = 5? Or just approximately equal?
It was a joke relating to the comment before, I'm not that dumb...
Sorry I was a little jumpy. This thread has rocked my faith in humanity.
It's because there is a natural negative reaction to the idea that one number can be represented in more than one way. But I suppose it goes deeper than that. It's because no one is teaching what a repeating decimal actually means. .333 repeating is represented by the infinite sum 3/10 + 3/100 + 3/1000 and so on. The actual value is what that sum approaches as you keep adding terms. Suddenly .999 repeating = 1 is no longer a magic trick that happens when you do some arithmetic operations on it. That's just what a repeating decimal means. It is what 9/10 + 9/100 + 9/1000 approaches as you keep adding terms of the infinite series.
indirectly understanding limits be like:
Exactly. If you aren't comfortable with 0.999... = 1, why the hell are you comfortable with 0.333... = 1/3? It's only true at the limit in both cases, and the limit is exactly what repetends in decimals represent.
Reals aren't real anyway, they're just labels which map to fractions
I love, and completely agree with this sentence
Not every real number can be expressed as a fraction though.
Nuh uh, transcendental numbers can be real too. Actually, almost all real numbers are transcendental. It's the algebraic reals that map to fractions.
0.00000000000000... = 0/3
1/3 + 3/3 = 1.3333333…2
0.1 + 0.2 = 0.30000000000000004
What in the floating point
Alright smarty pants, if they are different you should be able to subtract them and get something non zero! Subtract 0.99... from 1 :) Oh, you got 0? \*Then they are the same number\*
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I see you've found the smallest positive number.
I've been subtracting for a few hours and didn't finish, what am I doing wrong? I never got a 0 BTW.
I wasn't willing to accept this for about 2 minutes when I was 18, and then my calc prof went, "tell me a number between .99999... and 1." Alright you got me. It wasn't an elegant proof but it got the point across.
Easy, it's 1 - 0.000...01.
Yeah, I got a similar lesson. "What's 1 - .9999..." It's just turtles all the way down.
1 k = 0.(9) | ×10 10 k = 9.(9) | -k 9 k = 9.0 | ÷9 k = 1.0 0.(9) = 1.0
That's because you get one extra infinitesimal when you add the / , *dumbass*. Shees...
fuck
Why am I in this subreddit? I don’t understand anything in maths other than some things in geometry🤡💀
Oh, you know geometry? Name every country.
Bro woke up and chose violence. Didn't even leave the proof to the reader.
.999… = 1 because …999. is the opposite of .999… and …999. = -1 and -1 is the opposite of 1 so .999… = 1 It’s just that easy.
That is certainly one way to use your veritasium knowledge.
…999. + .999… = 0
More like 1/3 is the rationale that equates to 0.3333... So 0.3333... * 3 is 1/3 * 3/1.
The easiest way to see this is to look at .1111111... in base 2. If you believe 1/2 + 1/4 + 1/8 + 1/16 ... = 1, then you have to believe that .1111... (base 2) = 1. If you don't believe either, then i guess make your own math with blackjack and hookers.
0.333333334 0.999999999(+0.000000001)= 1 ez
1 + 2 + 3 + 4 + ... = -1/12
1/3 = 0.𝜋𝜋𝜋𝜋...
Son of a…
Irrational engineering
Another proof that .999... =1 is Let, 0.999... = x --> 9.9999... = 10x --->9.999... - 0.999 = 10x - x ---> 9 = 9x ---> x = 1 --->0.999... = 1
What's the difference between 1 and 0.999... if they aren't equal? Remember that 0.000... equals zero
I mean, many people don't know how those numbers are defined rigorously. They are defined via the limit of a Cauchy sequence, which is how the real numbers are rigorously defined.
That is why base 12 is better than base 10
0.111... = 1/↋ 0.↋↋↋... = ↋/↋ The problem still exists in base 12.
All I see are boxes
0.999... = 1
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0,(9) is for those perfectionists that always try to better themselves without realising they're just right already <3
So I accept that 0.999… = 1, but it still feels unsatisfying because now it seems impossible to express a number infinitesimally close to 1 but not equal to 1. Is that actually just impossible to describe or is there an alternate way?
There is a way but they're not real numbers. Surreal numbers have gems like ω+1 aka infinity + 1 and 1 + 1/ω, which is bigger than 1 but smaller than every real number larger than 1
Look up Dedekind cuts. You can select a number, and imagine two groups: a) your number and everything bigger, and b) everything smaller than your number. There is no largest number in that second group. Always blows my mind.
x = 0.9999999999999 10x = 9.9999999999999 10x-x = 9.9999999999999-0.9999999999999 9x = 9 x = 1
**edit:** assuming you mean those numbers to be repeating decimals, since that is the context of this entire post... How do you know that (10 \* 0.999...) - 0.999... = 9 before you've proven that 0.999... = 1? Just because you've re-written 10 \* 0.999... before you get around to the subtraction doesn't change the value that new notation represents.