My math teacher showed me 0.999...=1. I was surprised & confused. Reddit r/theydidthemath
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- เผยแพร่เมื่อ 14 ต.ค. 2024
- We will discuss the classic math question if 0.999...=1 i.e. 0.9(repeating) is equal to 1 or not. We will use "an algebraic trick" to explain why 0.999...=1.
We first let x=0.999.. then multiply both sides by 10 so we get 10x=9.999...
Next, we will do 10x-x, which is 9.999...-0.999... and get 9x=9.
After dividing both sides by 9, we see that x=1.
But we set x=0.999... to begin with. So we can conclude that 0.999...=1.
Note that we can do the same trick to convert
0.999... as an infinite geometric series: 👉 • I had a grown adult sc...
This is from Reddit r/theydidthemath.
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We know sqrt(4)=2
Now what is sqrt(0.444...)=?
Answer here: th-cam.com/video/8XY3fprG-wg/w-d-xo.htmlsi=VnL5pgCe-fGU6vmX
Less than 2
0,666...
It's 4/9 so sqrt would be 2/3
there is an logical error in 2nd line, because 0.999...=-1-(1/10^n), n->inf, then if you multiply that by 10 you will get 10*(1-(1/10^n))=10x. Whatever you do you won't get rid of -(1/10^n), therefore there is only possible to say that 0.999... is close to 1
@@2EOGIY 10*(1-(1/10^n))=10x
10-1/10^(n-1)=10x
as n-> inf, we have:
10=10x
:.x=1
And because you defined x to be 0.(9), that means 0.(9) must equal 1.
For a proper proof, you can use inequalities to show there can't be a between 0.(9) and 1. And if there isn't a number between them, they must be the same number.
I also like the 1/3 + 1/3 + 1/3 = 1 but 1/3 is .3(repeating) and if you add 3 of them together you get .9(repeating)
Agree, I found this one is relative easy to understand and conceivable. If we agree that 1/3 is 0.33333... and 2/3 is 0.6666..., then 3/3 would be 0.99999... and since 3/3 is 1, 0.99999... also must be 1.
That is way simpler way to teach kids , thx a lot
While that one works, it isn't convincing enough for a lot of adults. I used to not believe this until I saw proofs like this one and the geometric series one. The reason this one didn't satisfy me is because, at the time in my head, you could argue that there is more than one way you can get to 0.999... so using 1/3 as 0.333... wasn't a perfect representation and that the only reason 1/3 is a repeating decimal is because of base 10. For people like how I think, it is better to prove it in ways that show it that it doesn't matter. It's going to equal 1 regardless.
Huh. This is a good point.
That's because 1/3 cannot be expressed as a decimal with 100% accuracy. Stop talking shit.
I entered 0.99999....-1 in my calculator and got 0 so I'm convinced.
How did u even enter an infinite series in a calculator??
it's the thing with the dot above it, I think pretty much all scientific calculators should have it
@@Kiwi-9381 some calculators can do finite sums but i dont think theres many that can do infinite sums
Most calculator algorithms use 12 point precision. Or maybe it’s 16? Regardless, it’s plenty more than most engineers need, yet far from what you would need to confidently use your calculator as evidence.
I use a philisophical calculator instead of a scientific one. It doesnt have that button.
My favorite argument is the "in-between" argument. If two numbers are not equal, you can always name a number "in between" the numbers. For 1 and 0.999...., there is no such number. Hence, they must be equal.
Now, something I learned recently from Mind Your Decisions is that if we are talking about the HYPERREAL numbers, then this equality no longer holds true.
After this video my favorite argument is the "show 0.999 in a/b form". A rational number is defined to be a number that can be represented as a/b where a and b are both integers and b is not equal to 0. We know numbers that repeat forever after the decimal point are rational, so what's the a/b form of 0.999...? It is 1/1.
I'm not really a fan of the in-between argument because if we say that 0.999...8 (0.9 repeating ending with 8 somewhere) then that's equal to 0.99999 repeating because you cant have a number between them right? So with this we can go to 0.999...7 and so on so forth till we consider that all numbers are equal to each other 💀although if im wrong about this please correct me lmao
@@akdb 0.9 repeating is infinitely many 9s you can’t just stop in the middle and replace a digit then carry on. That number is no longer 0.9 repeating. They are not equal because if you compare 0.9 repeating with 0.999n999… at any arbitrary number of digits tending to infinity the latter will always be smaller.
@akdb you cant say "repeating ending in an 8." Its either repeating, or it ends. Not both. If its repeating it literally does not have a last number. And if it is repeating it is rational.
i would argue that there 'is' a number between 1 and 0.9rep. which would be 1/(10^inf). and i feel like doing math with a repeating number as is inherently breaks algebra in some way, and repeating numbers should be treated as a distinct variable or at least replaced with a variable while manipulating it.
with a more true statement being 1 = 0.9rep only with an infinite number of precision. and with a nth level of precision it being 1 = 0.9rep + 1/(10^n)
It is also good to note that 1/2+1/4+1/8+1/16 ......... = 1 when written in binary (base 2) is just the repeating decimal: .111111 ...... which is indeed = 1 in base 2. Not to be confused with 1/9 in base 10.
Nice observation that I had not thought about. Thank you
Computer science/math major: .1bar is not 1 in base two, it is .1bar, and the series you listed is an infinite series, as soon as you truncate it you end up with a value less than one.
I've tried it. Very easy program to write.
@@TurboLoveTrain It is exactly 1 if you don't truncate it. That's the whole point. Infinite # of terms must be added to get equality - so no truncation in an infinite series. Yes it's true that a computer would need an infinite amount of time but you don't need that requirement to consider the infinite sum as a whole.
'Computer science/math major: .1bar is not 1 in base two'
You are a rather bad computer science/math major, then, because 0.111... (base 2) = 1/2+1/4+1/8+... = 1. Calculating the sum of that series is a very classic problem to give to students who are only starting to learn about series.
'as soon as you truncate it'
Nobody is truncating it.
You just seem to be very confused by the concept of non-terminating digital representations of numbers.
@ianfowler9340 that’s what people don’t understand about infinite, even if you had infinite time it would never reach the end. We are all assuming that there is an end because that’s how it is for our reality.
because for repreating values, it's x/9. and as we use 0.999 repreating over 9, you'll get 9/9=1.
We can write this as 0.9+0.09+0.009+.... which is a geometric series, so 0.99... =0.9/(1-1/10)=1
Your answer should use ≡, not =.
You're estimating the value of infinite series through truncation, you aren't solving it.
@@TurboLoveTrain he's not estimating by truncation, he's using the formula to evaluate a geometric series with a common ratio less than 1, so his result is correct.
@@luisaleman9512 he's evaluating an infinite series. You may find it useful, I see it as math voodoo that only works on paper. You're not going to convince me that 1 =.9bar because your interpretation ignores what is right in front of you as a result.
@@TurboLoveTrain I don't need to convince you of anything. If you don't want to learn that's your prerogative.
@@luisaleman9512 Common core is trash and you're the proof as to why. You don't understand how to handle infinity--common problem.
You should actually do the division *4/9* to show that you get *0.4 repeating.*
For anyone curious, you can do long division. 9 goes into 40, 4 times! Then you have a remainder of 4, and it just repeats, adding 4 after 4 after 4.
All repeating/terminating decimals are rational numbers. That's why we can think of the irrationals as non-terminating, non-repeated decimals. If you are prepared to stretch your thinking a bit, you could also think of a terminating decimal as a special case of a repeating decimal where the digit 0 is the repeating part.
1.24 = 1.240000..... Now it may seem a bit silly to consider the trailing zeros as an infinite geometric series, but in fact they are, as the formula: a/(1-r) still applies where a=0 and r=1/10 (correction). Thanks BigDBrian.
This allows us to say all rational numbers can be expressed as a repeating decimal. (and vice-versa)
shouldn't r = 1/10? if we're basing the series off decimal notation. the conclusion remains the same.
@@BigDBrian Well caught sir.
@@BigDBrian You are absolutely correct. My bad. Thanks for the correction.
That's not a stretch. All terminating decimals are actually repeating decimals. We just have a writing convention that we don't display infinitely repeating zeroes. But they are still there.
Same for whole numbers being rationals with the writing convention that we don't display the denominator "/1". And for reals that don't include the imaginary part "0i". And ...
The problem most people have with understanding maths is that its teaching logically starts with simple special cases and progressively open up to more general cases. Each generalization step means that you must review your comprehension of the simple numbers you have already learned and treat them as actual elements of the larger set, not as a separate unrelated set.
I took up to calculus 2 in college and I have never seen this before. Is this covered in a number theory or other course? This blows my head
Something like this should've been covered in calc 2, especially when doing series and partial sums.
I saw it in Introduction to Discrete Mathematics, a lower level math course, at the same time when I was taking Calculus II
@@carlosgomez701I also remember going over this in Discrete Math, but I don't remember which course I took first.
High School Calculus.
This should be calculus 2 in the same topic as deriving common Taylor/Maclauren series from an integral.
Many of the comments below have been a real eye-opener. When we say that lim[S(n)] as n---> inf = 1 we are indeed saying that S(inf) = 1. I wonder why it so difficult for some to get that. I guess they are having trouble with the fact that you actually need to consider the infinite sum as a whole before you get equality. And that does not conflict with the limit notation. Anyway, keep on truckin. Love your stuff. You help a lot of people.
obviously there is a flaw in mathematics
@@edwardmacnab354 I think it more likely that there is a flaw in some peoples ability to understand the concept of S (sub) infinity.
@@ianfowler9340 it does not matter what flaws i have if i can always supply an n whereby 0.9999....=1.0000 - 1/10^n forever if needed . I think your house of cards is on top of a table inside a glass house frankly . Sure , I can follow all your logic but Sorry, it does not add up in the end.
@@edwardmacnab354
'it does not matter what flaws i have if i can always supply an n whereby 0.9999....=1.0000 - 1/10^n forever if needed'
The problem is that you can't. Whichever n you pick, you will end up with 1-1/10^n = 0.999...9 < 0.999...
@@edwardmacnab354For infinitely repeating 9's your n would be ∞, and lim_{n -> ∞} 1/10^n = 0. So, 0.999... = 1 - 0 = 1.
Assume 1 - 0.9999... = x > 0 for some x > 0. Then there exists some natural number n s.t. 1/10^n < x. But .9999... + 1/10^n = 1.0(...)09999... with n zeroes after the decimal point.
Thus, 1 - 0.9999...
Just a question: When you say you take a decimal of infinitely many digits minus the same infinitely many digits, do we always get zero? It seems somehow like an infinity-infinity situation, but I may be wrong.
Don't confuse "infinity" with "infinitely many"
If their difference is not 0, they are not the same.
You know that there are the same number of whole numbers (0,1,2,3,...) as there are natural numbers (1,2,3,...). We know this by bijecting every number in the first set with one in the second set.
I'll label the first set's numbers with a w, and the second with an n.
0w matches to 1n, so in reverse 1n goes back to 0w.
1w matches to 2n, so in reverse 2n goes back to 1w.
Do this ad infinitum and you'll find each number matches to exactly one other, with no numbers left out. This means there's the same amount. So whilst there may appear to be 1 less, in truth that 1 less does *absolutely* nothing.
It's the same here. We have that many 9s, so if we "remove" one of the 9s nothing changes. We can still stack one on top of the other and draw a line from the first number's 9s to the second numbers 9s, leaving none left out, meaning there's the same amount.
@@dlevi67Explain the difference, please.
I mean, this is far from rigorous. But what you'll see if you consider writing out finitely many digits is that the "9 at the end" is the only term left over, worth 9/10^n, where n is however many digits we wrote out. Since you can make 9/10^n abritarily small by choosing arbitrarily large n, it's okay - in the limit, the difference is 0.
@@elio7610 Take a dictionary, and/or learn English. It's a lot faster and more accurate than asking strangers on the internet.
They're okay with .444444... but not .999999... because, if you divide 4 by 9 as a decimal you actually get an infinite number of 4's, but if you divide 9 by 9, you get a single 1, infinitely many 0's and no 9's at all.
Let’s take 0.333…, if you took each digit and added 1 to it, going down the number, removing the added 1 and moving to the next digit to get closer to 0.333…, when do you get the number exactly after 0.333…?
Never. You can’t get there. There is no next number. Hence, there must be an end to infinity according to the real numbers. Which means you can’t do 9.999… minus 0.999… and get 9. The 0.999… for the 9.999… (or 10x) is not the same as 0.999… (or x).
So, x is not 1.
What are your thoughts on this argument?
"end to infinity" 🤦@@encounteringjack5699
@@encounteringjack5699blackpenredpen made another video proving this one of his bprp basics channels, it involved rewriting 0.999... as an infinite geometric series, and then using the identity to figure out that it was indeed one (I forgot what the identity was called)
Also, how do you explain this: as any one digit number divided by nine is just that digit repeated forever(like 4/9 is 0.44444...), 0.9999999999... is just 9/9 which is 1.
Personally, I believe it's one
@@encounteringjack5699well, the number of decimal places in 9.99999 repeating Is the same is that in 0.999 repeating, so they can subtract to form 9.
let x = 1 - 0.9999
x = 0.0000...1
but you never reach the 1 because it is infinitely far away
This works in any positive whole number base greater than one. For example, 0.1 repeating in binary is one.That’s because zero plus one half,plus one fourth, plus one eighth, etc. is equal to one.
It is also true that any number that can be written in finitely many decimals has 2 decimal representations (apart from 0).
For example,
0.5=0.4999...
1.16=1.159999...
2=1.999...
@@methatis3013 True, but only the last example works out to an integer.
@@thomasschoeck9080 Similar things occur with, e.g., base golden ratio. (Look it up).
i would argue that the x=.444... proof wouldn't work, since you cannot perfectly represent 4/9 as a decimal, but can only represent it as an approximation.
I'd argue that you are wrong, as follows:
10 * 0.444... = 4.444...
=> 9 * 0.444... + 0.444... = 4 + 0.444...
=> 9 * 0.444... = 4
=> 0.444... = 4/9 exactly.
Actually, we do have a way to show 4/9 as a decimal. It's 0.4 with a dot or bar above the 4. Also note that 0.444... is a different number than 0.4 or 0.44, etc. In other words 0.4 with a dot is its own unique number.
Where can I buy the e poster to the right of you in the video?
The thing is that we’re always drilled into us the identity function x = x, with examples such as 1 = 1, 2 = 2, etc. Not seeing the same thing on both sides, and no mathematical operation (or variable) modifying a side, except for just a pesky overline, (which, if you had infinite writing space you COULD have written out fully if you had time, so by logic it’s just shorthand) feels wrong to them. No i’m not a psychologist or a mathematician
What about 1 = 2/5 + 3/5?
There's an interesting article by mathematician Fred Richman, where he takes this as an open question and develops a number system in which 0.999… ≠ 1. It can be done, but it has to violate some of the rules of arithmetic (it's not a field).
0.999… ≠ 1 is true in hyperreals, which do form a field.
@@BigDBrian : That's true (assuming that you take 0.999… to have 9 in all standard/finite places and 0 in all nonstandard/infinite places, rather than 9 in all places). In Richman's system, every number has a decimal expansion using only standard places, so he can make it work without bringing in the axiom of choice or anything like that, at the cost of not always being able to subtract.
I (maybe) just realized a really intuitive way of making peace with this: Imagine 0.0 (repeating). No matter how far into the decimals you look, you will never see any digit other than 0, so it should be safe to say that it equals 0.
So all of these should be true:
1 + 0 = 1
1 = 1
0.999... + 0.000... = 1
0.999... = 1
Am I correct?
I actually have another question that I wanted to ask when I happened on this way of thinking about it: In maths, infinitesimally small numbers are sometimes used. Could one add an infinitesimally small number to 0.999... and still get exactly 1? And is 0.000... an infinitesimally small number?
Both questions should have the same answer if my equations above are correct
... Is there even something like 0.0 repeating??
Edit: corrections
'Am I correct?'
Well, 0.999...+0 = 1, and 1-0.999... = 0 because 0.999... = 1, and not because the decimal '0.000...' looks a certain way.
'In maths, infinitesimally small numbers are sometimes used'
Not in the context of real numbers.
'Could one add an infinitesimally small number to 0.999... and still get exactly 1?'
Not in the space of real numbers - the context of the video - because infinitesimals do not exist there.
'And is 0.000... an infinitesimally small number?'
It's just another way to write '0'.
@@thetaomegatheta thank you! I found out that infinitesimals aren't part of the real numbers but hyperreal numbers soon after commenting, but I still wasn't sure about 0.0 repeating.
Anyway 0.000... just being 0 is quite intuitive and aligns well with 0.999... just being 1 I think. It helps me intuitively accept that 0.999... = 1, because adding these together understandably results in 1 and necessitates 0.999... being exactly 1.
It helps me to think about 0.9 + 0.1 = 1 and 0.99 + 0.01 = 1, and just imagining the last digit of each number moving farther and farther back, until each is "behind infinity", so it just never appears
@@frohnatur9806 Some people try to represent an infinitesimal with 0.000...1 and say that 0.999... + 0.000...1 = 1. They don't notice that it should be 0.999...9 + 0.000...1 = 1, or that perhaps it should be 0.999... + 0.000...1 = 0.999...1. If it means anything, then 0.000...1 would be 1/10^n where n is an unspecified natural number, and then 0.999... + 0.000...1 = 1.000...0999... where that last 0 is at position n. Also 0.999... + 0.000...1 would be 1.000...1
@@Chris_5318 yeah, I realized that the "final" digit (the only 1 for 0.0... and another 9 for 0.9...) needed to be at the same position in both numbers for this to work. I just think about both numbers terminating AFTER infinity, which just means "never" I guess, but it helps me to be intellectually at peace with the fact that 0.9... = 1, even though it's just an elaborate way of saying 0.9... + 0 = 1
@@frohnatur9806 I forgot to say that, in the current context, "infinite" means "endless", not "huge".
some simple explanation is to divide 9 by 9.
But instead of just saying it was 1 use long division and say it's 0.9 and repeat.
You'll end up having 0.99....
9|9 = 0.9*9 and remainder is 0.9
0.9|9 = 0.09*9 and remainder is 0.09
0.09|9 = 0.009*9 and remainder is 0.009 etc...
Anology between 0.(4) and 0.(9) fails if we take into account the way we represent real nubers in decimal format.
When you say x = 0. you automatically say that 0
'When you say x = 0. you automatically say that 0
"Yes, it is a valid representation of a real number." Provide a number that should written like that. Number 1 is written like 1. When you write the first digit of 1 as '0' that is a mistake.
Yes, the Uni has failed me because I do not confuse a number and a series. Of course that is math so you may introduce any symbol you want. You may choose to define "0.999... " as lim(n->inf) (10^n-1)/10^n then obviously 0.999...=1. Just keep in mind that symbol 0.999... is different from other infinite decimal numbers like 0.444... Because for 0.444 it is actually true that 0
'Provide a number that should written like that'
Decimals '0.999...' and '1' have the same referent.
'Number 1 is written like 1'
Yes, and, as covered early in calculus/real analysis, it has another decimal representation - '0.999...'.
The fact that real numbers can have more than one decimal representation is a common fact and a topic of exercise early in calculus/real analysis textbooks.
'Yes, the Uni has failed me because I do not confuse a number and a series'
No, it has failed you because you neither understand what a real number is, nor the fact that a real number can have multiple decimal representations.
"A real number has a single decimal representation." If this is true, it should hold in every numeral system. Counterpoint: in balanced ternary, 1/2 has two equally valid representations: 0.111... and 1.TTT.
"When you say x = 0. you automatically say that 0
@@areadenial2343 That's an interesting point. However "If this is true, it should hold in every numeral system." -- no, actually only for those that are in good relationship with the decimal system. Let's say there's f: R -> D so for any real there is only one decimal representation. And lets say there's S -- a numerical system such as g exists g: D -> S. Then there must be g(f(r)): R -> S. Then we could find r that has two representations in S and get a contradiction. However if S is such as there is no g: D -> S then this line of reasoning doesn't work. Since I'm not an expert on balanced ternary I want to ask you: are you sure there is no decimal repressentation that has two or more equivalent representations in balanced ternary?
If I had two continued fractions, how can I add, subtract, multiply, divide them?
Didn't you just upload a video about the same thing a few days ago?
That one he used a more calculus approach of the geometric series while here he’s using a “simpler” approach
I've had troubles understanding this forever but this explanation finally made sense
One should also note here that setting .9 repeat = x only works because .9 repeat is a convergent series (geometric in this case). You can now treat x as a real variable and perform all the operations that you would normally do with a real variable - mult by2, add 3, sutract 4x, etc..... Where people run into trouble is when they set a real variable, S, equal to the sum of a divergent series. Everything and anything that follows is nonsense.
So setting S = 1+2+3+4.... is a nonsensical statement since infinity is not a real number and therefore does not behave like a real number. You can't double S or add 3 to it etc... and expect to get meaningful.
yeah i do have a similar feeling about treating repeating numbers as a distinct variable.
0.(9) is a number, not a series.
0,9+0,09+0,009+0,0009... is a series that converges to a number 0,(9)
@@SURok695 0.(9) is defined to be the series Σ [n=0,inf] (9*10^)-n). And this series converges to 1.
@@ianfowler9340 You're indeed blunt to be wrong. Numbers aren't series, they can't converge or diverge. I don't think you're saying that 2 converges to 2. Numbers are just numbers and repeating decimals are the form of writing them.
Series are different thing in maths with different qualities and operations with them.
And 0,(9) is a number. 0,9+0,09... is a series. Series isn't a number. It's sum (if the series converges) is. So 0,(9) doesn't converge to 1, it just equals to 1. While 0,9+0,09+0,09... series converges to 1.
@@ianfowler9340 it's you who learnt nothing
The reason 0.99.... is equals to 1 is because there is no number to represent the value in between 0.99.... and 1
That answer makes more sense to me, but at the same time it also feels more like a shortcoming in notation than a real thing. Things are not the same just because you have no word for the difference.
0.9 has no 1 in the first digit, that makes it less than 1, and putting an infinite number of 9 behind the point makes it come infinitely close to 1, but it never gets there. Same for 4/9, it is not 0.4, not 0.4444 and not 0.4(repeating), but that's as close as we can get to describe it without fractions and I'm ok with that.
@@vinny142 1+3. 128/32. They are different ways of writing 4, that have no 4s involved. That doesn't stop them being 4.
@xinpingdonohoe3978 That’s why we have fractions, it’s because our shortcomings in decimals that deal with infinite.
For pretty much any base and the wrong divisor, we can get weird things like this.
Through long division we can get intuition on most repeating decimals, but x/9 is strange, and people can't really "derive" 0.999... from long division, so they don't believe that it has an equivalent fraction form.
So they internalize the 0.999 through "infinite series of nines" instead of "also equivalent to some fraction", and then either misunderstand infinity, or some other thing. But point is, there isn't a "fallback" for 0.999 so most people will treat it as a separate entity outside of the usual number system. Like you don't have to convince a normal person that 0.333... is 1/3, because they can do the long division algorithm they can see it.
the thing is .333... is not equal to 1/3 it is a futile attempt to reconcile the two
@@edwardmacnab354No, they equal. They are just different representations of the same number.
There is something behind the scenes wich causes the equation :0.9999...=1.
Let me show why the equation -1+∞=∞ is behind those scenes.
x=0.9999...
x=0.(∞ many nines)
x=0.([1+(-1)+∞)] many nines) | *10
10x=(1 many nines).([-1+∞] many nines)
Only if -1+∞=∞ :
10x=9.(∞ many nines)
10x=9.(∞ many nines)
10x=9+0.(∞ many nines)
10x=9+x
10x-x=9
9x=9
x=1
The specific ∞ in question is Aleph 0. It is well recorded that Aleph 0 + any integer = Aleph 0.
I always viewed these types of things as a rounding issue.
1 minus 0.9 repeating equals what? Well it’s 0.00000000000000000000000000000 at some point a 001, but it’s always a smaller decimal place than you can quantify. It’s effectively Zero.
'It’s effectively Zero'
It's literally 0, because there is only one non-negative real number less than 1/10^n for all natural n.
It's not a rounding issue, 0.999... is an unending sequence, not simply a very large sequence. There is no real number between 1 and 0.999..., so there is nothing to round as they are equivalent
This whole comment section is Dunning-Kruger in action
Do you say that because you agree or disagree with the video?
@@Chris-5318 I fully agree with the video, but there's so many people that are relying on limited intuition instead of more rigorous logical proofs which have been the gold standard of mathematics for hundreds of years for a reason
@@redjoker365 I agree with you. :)
We can write 0.'9' (9 is repeating) as 9/9 which is 1.
(0.'279' (279 is repeating) is 279/999 = 31/111, as another example).
However, to be exact, 9x = 8.'9'1...., and so, x is 0.'9', which is 1...
Let's do this with 10 repeats:
x=0.9999999999
10x=9.9999999990
-x=0.9999999999
=9x=8.9999999991
x=0.9999999999
This means that x=0.'9' where the 9 repeats infinity times, and
that 10x=9.'9' where the 9 repeats "infinity minus one" times...
I don't know how to prove this, but I'm pretty sure that infinity minus one is still infinity... :)
" 9x = 8.'9'1 "
This is not a valid representation of a number, it has no meaning
@@_Ytreza_ when there is no way to put a line over the repeating digit(s), you put quotes on either side.
There! You've learned something new today!
@@OrenLikes I understand this but you can't have a digit after a repeating sequence of digits (else it wouldn't be repeating anymore)
@@_Ytreza_
right. but... it repeats infinite number of times, right? and THEN there is a different digit... :) at position infinity+1... :)
'at position infinity+1'
No such position.
When i was a freshman in high school, proving 1=.9 repeating was one of our homework problems. I just used substitution method since 1/3=.3 repeating.
How do you know 1/3 is equal to .3 repeating though? It's the same question.
@@Hahahahaaahaahaa Let's prove 1/3 = 0.333...
Let's find the remainders using modular division until we run out of remainders to see if it's truly infinite
1 % 3 = 1
Interesting, let's find the next remainder
1 % 3 = 1
I don't think we're getting far, but let's find the next remainder
1 % 3 = 1
Hmm, looks like we have repeated remainders of 1 and it's not going to stop, so now we have
1/3 = lim(n->infinity) 3*(((1%3)/10) + ((1%3)/100) + ((1%3)/1000) + ... + ((1%3)/10^n))
1/3 = lim(n->infinity) 3*(1/10 + 1/100 + 1/000 + ... + 1/10^n)
1/3 = 3*(0.111...)
1/3 = 0.333...
I think this reflects more of a flaw in axioms than a proof that .9bar equals one.
If I convert 10x-x to 9.9bar - .9 bar the logical consistency holds the original equation.
If I wrote a program that told me 1 = .9 I would first look for binary conversion errors and then code errors--I wouldn't assume .9 = 1 because the program doesn't crash when it runs.
I think the rift in the math is .9 can not be expressed as a fraction... which clearly breaks the equation.
show me a p/q that equals .9bar... there isn't one QED.
'I think this reflects more of a flaw in axioms than a proof that .9bar equals one'
Well, you think wrong, then.
'Flaw in axioms' is just gibberish, and 0.999... does equal 1.
'I think the rift in the math is .9 can not be expressed as a fraction...'
It can be. A more general case is an exercise in calculus textbooks.
x = p/10^n+x/10^n
=> x = (p/10^n)/(1-1/10^n) = p/(10^n-1)
So,
0.999... = 9/10+0.999.../10
=> 0.999... = (9/10)/(1-1/10) = 9/(10-1) = 9/9 = 1/1 = 1.
'show me a p/q that equals .9bar... there isn't one QED. '
As I have shown, your baseless assertion is just silly. Math students are tasked with proving that repeating decimals in general all represent rational numbers in introductory calculus textbooks.
@@thetaomegatheta
0.999... does equal 1
It doesn't.
You're using an axiomatic conversion to get from an infinite series to an integer. The rest is you being unable to differentiate what your logic system is doing from what is happening right in front of you.
'It doesn't'
It does, and this material is covered in calculus textbooks as introductory material.
Understanding either construction of real numbers, or how series work makes the fact that 0.999... = 1 obvious.
You would know that if you were a computer science/math major and/or read at least one calculus textbook.
You can start by looking at Terence Tao's 'Analysis', volume 1, page 388, proposition B.2.3.
'You're using an axiomatic conversion to get from an infinite series to an integer'
'Axiomatic conversion' is not a term.
Not a relevant one, or one that I have heard of, at the very least.
In any case, the sum of a series the partial sums of which form a Cauchy sequence of rational numbers is a real number.
We also know that 0.999... is a rational number, and we can even calculate integer p and q for it, such that 0.999... = p/q. And, as I have shown, for 0.999... it's p = q, i.e. 0.999... = 1/1 = 1.
'The rest is you being unable to differentiate what your logic system is doing from what is happening right in front of you'
This is just more gibberish.
We are talking about real numbers and their decimal representations. We know that every terminating decimal representation of a real number has a corresponding non-terminating one that refers to the same number. For the decimal '1' we also have '0.999...'
Notably, you opted to not tell anybody what is supposedly 'happening right in front of you'.
It is pretty clear that you have never studied math, as you fail at the most basic calculus stuff.
@@thetaomegatheta I spent a few years correcting calculus books. I can't really imagine someone competent at maths defending public school math curriculum.
If you cant see the flip between a run to infinity and a whole number in the context of hand wavig. ... you fit right in with the public shoolers.
0.9 repeating isn't an infinite sum, it's the value of that sum.
There is no "rift". This is only a problem for students and teachers operating without proper definitions and thus getting bogged down in nonsense like yours.
In Serbia this is a 5th grade math question
i was shown it as 1/3=0.33333 recurring therefore 0.33333 recurringx3 = 0.99999 recurring = 1
You didn't prove that 0.333... = 1/3. (I know it is though).
What I don't get is if 0.9 repeating is 1, why aren't infinitesimals equal to 0? To my knowledge 1 minus an infinitesimal would give you 0.9 repeating. If 1 and 0.9 repeating are the same, the only number you can subtract from a number to get the same value is 0. If 1 minus an infinitesimal is not 0.9 repeating, what is it? I understand you can't add or subtract infinity, but infinitesimals are not infinite. Infinitely small, but not infinite in scale since there are boundaries on either side (0 is a hard boundary, and any non-infinitesimal number as a soft boundary).
'why aren't infinitesimals equal to 0?'
There are no infinitesimals in the space of real numbers.
A positive infinitesimal is a number that is greater than 0, but less than every positive real number.
'To my knowledge 1 minus an infinitesimal would give you 0.9 repeating'
What is that 'knowledge' based on? Because in the case of real numbers that is obviously not true.
'If 1 minus an infinitesimal is not 0.9 repeating, what is it?'
Just pick up a textbook on non-standard analysis if you want to get into in in any sort of depth.
@@thetaomegatheta You don't have to be condescending about it, I was just asking questions and stating my understanding. You've made it clear my understanding is wrong.
If you want to know what I meant by comment, here:
1 - 0.1 = 0.9
1 - 0.01 = 0.99
1 - 0.001 = 0.999
Following the pattern, if you had a decimal with an infinite amount of 0s followed by a 1 (the layman's interpretation of an infinitesimal), when you subtract that from 1, you would get a decimal with infinitely repeating 9s. Before you condescend to me further, yes, I know you can't have an infinite number of something followed by something else. That would make it inherently finite. I know that's not the true definition of an infinitesimal, it's just a simple way of explaining the concept for people like me who haven't built their lives around theoretical math.
I'm not going to pick up a college textbook in an esoteric field of mathematics just to answer 1 question I had in passing. If it's like my college calculus textbooks, that's like $300 and I got bills to pay. Why do you think I am on free website asking questions?
@@ThatGamerDude9000
'Following the pattern, if you had a decimal with an infinite amount of 0s followed by a 1 (the layman's interpretation of an infinitesimal), when you subtract that from 1, you would get a decimal with infinitely repeating 9s'
The problem is that in that sequence you never encounter either of the decimals '0.999...' and '0.000...01' with infinitely many '0's. In fact, the latter is not a valid decimal at all.
'Before you condescend to me further, yes, I know you can't have an infinite number of something followed by something else'
You can in general, but not in this case.
For example, consider infinite cardinals which 'follow' every natural number, which there are infinitely many of.
'If it's like my college calculus textbooks, that's like $300 and I got bills to pay'
Consider looking up 'libgen'.
@@ThatGamerDude9000 Re your
1 - 0.1 = 0.9
1 - 0.01 = 0.99
1 - 0.001 = 0.999
What you really are saying is that
1 > 0.9
1 > 0.99
1 > 0.999
...
ergo (by magic)
1 > 0.999...
How about
0.999... > 0.9
0.999... > 0.99
0.999... > 0.999
...
ergo
0.999... > 0.999...
Are we to infer that means that 0.999... is infinitesimally greater than 0.999...??? Nowhere did you mention 0.999..., yet you seem to think that you can make a conclusion about it. That makes no sense.
You: "if you had a decimal with an infinite amount of 0s followed by a 1"
You: "I know you can't have an infinite number of something followed by something else"
You want to have your cake and eat it too. The first case cannot exist, so you should have stopped there. Garbage in, garbage out.
Which of the infinitely many infinitesimals did you have in mind?
Why do you not think that 1 - d = 0.999... - d where d is an infinitesimal?
Why do you think that decimals can represent infinitesimals?
What would the decimal for 10 * 0.999... - 9 be?
@@Chris-5318 look I know I am uneducated in this subject. You do not have to rub my face in it. I have already admitted it. I am simply asking questions and stating my understanding of the situation. If you want people to learn, you don't mock them for their failures.
I was unaware that there were infinitely many infinitesimals. I thought it was positive infinitesimal, negative infinitesimal, and maybe at most a third infinitesimal in imaginary number space. Before my initial comment I did not know infinitesimals were not considered numbers in the real number space.
To answer your question on why I thought infinitesimals could be represented by decimals, I ask, other than fractions and decimals how else do you represent a number greater than 0 but less than 1? I don't mean this any condescending way. This is someone who took calc 1, 2, 3, and ODEs in college and that's the extent of my math knowledge. It's been years since college, I haven't used the knowledge, and I watch the occasional math video on TH-cam to at least prevent regression. The only other way I could think to represent an infinitesimal is a symbol, but usually symbols are stand ins for values we can't otherwise write out, not the actual value itself.
I forget if it was bprp or another youtuber, but I remember someone saying "there is no value in between 0.999... and 1, therefore it is 1." To my understanding, there is no value between a positive infinitesimal and 0, but it is not 0. It is the smallest non-zero value. To me that seems contradictory but my knowledge on the subject is limited. If the answer is "infinitesimals are not numbers in the real number space, therefore it's not 0," then why does 0.999... exist in the real number space? Or I guess, how do you write a number that is not 0.999... that is infinitesimally smaller than 1? Is that even a valid question? I genuinely do not know. Is it just 1-1/(infinity)?
I could be wrong, but I thought I was told in calculus, 1/(infinity) does not exist. Maybe when I was told this, it was meant "1/(infinity) does not exist in the real number space, but is a value." The way I interpretted it was that "1/(infinity)" is not a valid number, akin to thetaomegatheta saying 0.000...01 is not a valid number.
I have seen this many times, and as before, I have the same feeling that multiplication 10 * 0.99999... = 9.99999... is not correct. You are multiplying infinite continuation, as if you would get one 9 more. But if it infinite continuation, you can't get more 9s than that.
There's definitly some assumptions that everyone seems to make about this problem
But multiplying by 10 do work, there's still an infinite amount of numbers
0.44... * 10 is 4.44...., you can show that with fractions
Precisely because it is infinitely many 9, you _can_ do that. Infinite - 1 is still infinite.
@@dlevi67 Yes, and 10 times infinitely many 9s is infinitely many nines. But this decimal point is what makes it absurd.
@@mikahamari6420 To you, but not to mathematics.
@@dlevi67 These tricks are not mathematics, more like circus.
X = 0.5 repeating
10x = 5.5 repeating
Then 10x-1 = 5.5 repeating-0.5
0.999 = 1
9/9 = 1
Calculator (repeating numbers)
1÷9 = 0.11
2÷9 = 0.22
3÷9 = 0.33 basically 1÷3
4÷9 = 0.44
5÷9 = 0.55
6÷9 = 0.666 basically 2÷3
7÷9 = 0.777
8÷9 = 0.888
9÷9 = 1
0.33x3 = 0.99
0.66+0.33 = 0.99
10x-1 is 4.55555… not 5.5… - 0.5
@@deltalima6703he meant 10x - 1x
Exactly so .9 repeating logically has to be 1. Unless you want to somehow magically elevate whole numbers to behave differently to fractions so they exist slightly above the infinite expression.
Several ways to show, I'd show multiple ways to try and convert the doubters. Your way; the sun of a GP formula (which is of course proven using your way in the first place); taking 1, dividing by 3 (0.333...) ant multiplying by 3 again ("0.999..."), so 0.999... must be 1
It has clearly gone wrong, because on line two they have 10x=9x.
I don't know why they don't just explain limits as you go to infinity: two numbers are equal if the difference between them is less than E where E>0 is arbitrarily small...
'It has clearly gone wrong, because on line two they have 10x=9x'
What are you talking about?
The second line is '10*x = 9.999...', not '10*x = 9*x'.
'two numbers are equal if the difference between them is less than E where E>0 is arbitrarily small...'
Incorrect.
Two numbers are equal if the difference between them is 0, or, alternatively, less than all real epsilon > 0, i.e. if that difference is 0.
@@thetaomegatheta in the original post, it says 10x - x = 9x.
Their answer was still correct at the end, but on that one step they subtracted x from one side while multiplying the other by x, which is something you cannot do. It was just them forgetting to write "9.9 - x" and instead writing "9x"
@@dignelberrt
'in the original post, it says 10x - x = 9x'
Literally the second line on the blackboard is '10*x = 9.999...'.
'but on that one step they subtracted x from one side while multiplying the other by x, which is something you cannot do'
On what step?
@@thetaomegathetayou just repeated the statement that two numbers are equal if the difference is less than E>0 for any real E. It is not an incorrect statement, it wouldbe incorrect if the difference is E, but it clearly states it must be less than E. Which is just a complicated way to say it should be 0.
'you just repeated the statement that two numbers are equal if the difference is less than E>0 for any real E'
They did not say 'for any real E' or 'for all real E'. They said 'where E>0 is arbitrarily small'. 'Arbitrarily small' is not a term.
How many questions like this do we have to entertain?
Is 0 an integer?
42.
Brother he is a math teacher 😂
@@Robin-Dabank696 it would certainly seem so.
Converges != Equals
Disregarding the infiniteness of series, such as here, one should be ok with 1=2 through derivatives, or inf - inf = 1, and so on.
'Converges != Equals'
And 0.999... = 1. It is exactly 1.
'Disregarding the infiniteness of series, such as here, one should be ok with 1=2 through derivatives'
Lol no. And, of course, you do not back your claims with any basis.
LOL yes
Care to provide your supposed 'proof' that 1 = 2?
You've gone rather quiet. Seems like you never actually had any basis for your point.
Also, here's a quick but relevant exercise:
x = p/10^n+x/10^n - solve for x; you may assume that p and n are natural.
@@yuriiherbenko8381 Your "Converges != Equals" was exceptionally vague. You didn't say what was converging or what was equal to what.
The sequence 0.9, 0,.99, 0.999, ... converges to both 0.999... and 1. Hence 0.999... = 1.
either he is doing this pent trick since he was 2 or he has some untapped pick pocketing talent
Why multiply by 10 though? Seems arbitrary. 🤔
It is arbitrary. So what? It is extremely sensible because it just involves shifting the decimal point one place to the right.
Multiplying by 7, say, involves more work, as I'll now show:
7 * 0.999... = 7 * (0.9 + 0.09 + 0.009 + 0.0009 + ...)
= 7 * 0.9 + 7 * 0.09 + 7 * 0.009 + 7 * 0.0009 + ...
= 6.3 + 0.63 + 0.063 + 0.00063 + ...
= 6 + 0.3 + 0.6 + 0.03 + 0.06 + 0.003 + 0.006 + 0.0003 + ...
= 6 + 0.9 + 0.09 + 0.009 + 0.0009 + ...
=> 7 * 0.999... = 6.999...
I'll use that to prove that 0.999.. = 1 follows.
=> 6 * 0.999... + 0.999... = 6 + 0.999...
=> 6 * 0.999... = 6
=> 0.999... = 6/6 = 1.
That shows that the math is self-consistent.
Could also represent it as an infinite geometric series with an r value of 1/10 and an initial term of 9/10
So its 9/10 + 9/100 + 9/1000….
and by the rules of infinite geometric series it must converge because it’s r is between 0 and 1.
And it’s sum is its initial term divided by 1 minus its r value.
So 9/10(1-(1/10)) = (9/10)/9/10) = 1
芝诺跑得很快，但是有一只乌龟说：你永远追不上我！你的速度是10m/s，我的速度只有1m/s，但是我在你前面9m处你追我。这样，第一次你到达我开始的位置的时候，用了0.9s，这期间我前进了0.9m；然后你下一次到达我刚才的地方，又用了0.09s，但是我又前进了0.09m……无穷无尽，所以你永远追不上我！
芝诺说：那我追上你只需要0.999…s啊？
旁边的小学生：我们老师教过了，这是追击问题，时间等于距离除以速度差，9m/(10m/s-1m/s)=1s，芝诺追上乌龟就要1s！
等等，这是英文频道？😅
Huh, that's actually a rather nice approach to this that I hadn't considered.
Translation of OP:
Zeno ran very fast, but a turtle said: You will never catch up with me! Your speed is 10m/s, my speed is only 1m/s, but I am 9m in front of you and you chase me. In this way, the first time you arrived at my starting position, it took 0.9s, during which I moved forward 0.9m; then the next time you reached the place I just started, it took another 0.09s, but I moved forward 0.09m. ...Infinite, so you can never catch up with me!
Zeno said: Then I only need 0.999...s to catch up with you?
The primary school student next to him: Our teacher has taught us that this is a pursuit problem. Time is equal to the distance divided by the speed difference, 9m/(10m/s-1m/s)=1s. It takes 1s for Zeno to catch up with the turtle!
Wait, this is an English channel?
this proof is not mathematically rigorous, because the rules of adding and multiplying finite decimals dont necessarily apply to infinite decimals.
The relevant rules are applied just fine.
You should learn how series work.
It is easy to prove that adding and multiplying decimals works just fine.
They actually do, if you're adding a constant finite number to infinity, you get the same infinity
@@redjoker365 Your comment makes no sense because nowhere was infinity being used as a number. The OP was referring to infinite length decimals.
why does the mind rebel when it sees the notion that 0.9999 = 1.0 ? .....does it mean something is wrong with the mind or something is wrong with the idea presented to it?
Something is wrong with the thinking inside the mind, as it tends to think in finite quantities, where a finite number of "9s" would indicate a difference.
It's just that "the biggest number less than 1" is easy to conceive of despite having no meaning in the rational/reals, whereas you can't even claim .9 repeating is a real number without a bunch of subtle definitions for things like limits and completeness (which are certainly not taught before this, if ever).
@@jkid1134 Well, for that matter you can't claim that 1 is a real number without those subtle definitions...
What BPRP is doing is not presenting a rigorous proof, but one that should be acceptable based on the intuitive definition of a real number as "any finite number with a terminating or not terminating decimal expansion, whether repeating or not." If you want rigorous proof, then I agree an understanding of set theory is required, as well as an understanding of limits.
Because people on the whole are not very smart.
@@dlevi67 I mean to speak more to intuition that rigor here.
Can you elaborate Geometry way ?
Ugh....that example from the post is so full of WTF.... All it takes on line 3 is to multiple 9 times the original x to see how this doesn't work. This is more like all the 1=2 manipulations.
Well, this thing works only if 0.(9) is a real number, so you can do operations like that with it.
And I don't see the proof of that.
It is easy to get to the point of showing that 0.(9) isn't just a real number, but a representation of a rational number.
The simplest way of getting there is starting with 0.(1) = 1/9, or 0.(3) = 1/3. Long division quickly shows that in base 10, the number can never get an even multiple and each individual subtraction always leaves a 1 left over. If at this point you don't want to accept that 0.(1) = 1/9 or 0.(3) = 1/3, then there isn't much that can be said. Since the fraction's denominator is relatively prime to the number base then the decimal representation will never terminate. If you can look past other definitions and accept that 0.(1) represents 1/9 or 0.(3) represents 1/3 in base 10, and it is an exact representation, then you can build up to 0.(9) from there.
However, I will agree that an infinite amount of values after the . is rather difficult to reason with.
@@darranrowe174and now you assuming that 0.(1) and 0.(3) are real numbers that are equal to 1/9 and 1/3.
And where is the proof of that?)
Obviously you can write 1 as 0.(9) if you want, or you can write 1/3 as 0.(3), but question there is - can the number with actual infinity of digits in it be real? Or is it just a representaion, like if we say x = 1, but in that case 0.(9) is used instead of an x?
@@F1r1at I don't get what you're saying at all. "Representations"? What?
You want a proof of 1/3 being equal to 0.(3)? Do the long division. But that's apparently not enough for you. You can do operations on 0.(3) because it is equal to 1/3. Do you reject fractions too?
(1/3)+(1/3)+(1/3)=(3/3)=1
0.(3)+0.(3)+0.(3)=0.(9)=1
If your only issue is infinitely repeating digits, let me introduce you to the base 3 number system. Where the only digits are 0,1 and 2. Subsequent digits denote subsequent powers of three.
So, 1 is 0*(3)¹+1*(3)⁰ which is 1(base3)
3 would be 1*(3)¹+0*(3)⁰ which is 10(base3).
Dividing, (1/10)(base3). Oh would you look at that, that's 0.1(base3). No infinite digits. Now we can multiply by 10(base3). 0.1(base3)*10(base3)=1(base3) which is, unsurprisingly, 1(base10).
Definitely not a rigorous proof but the logic still holds. You seemed to have a issue with infinite digits so I denoted (1/3) in base 3. Or are you just going to call it a "representation" and reject the entire notion of different number systems? Remember digital electronics use the base 2 number system which is generally known as the binary system. If that's true, then this is true too. Or you could just nitpick mathematics and decide to not believe what you don't like.
@@dhardhar7540 I do not say that 1/3 is not a real number. I'm saying 0.(3) is not.
We can say that 0.(3) is a representation of 1/3 if we need to write it in a decimal fraction form, but I don't see any proof of 0.(3) actually being a real number and being exactly equal to 1/3.
@@F1r1atthe formal definition I learned for a real number is that it's the sum of a natural number and the limit of the sum of x(i) * 10^-i , i going from 1 to n , where n is going to infinity
There's a few more stuff that's needed, for any index i x needs to be between 0 and 9 , and if for any n along the way, the difference between this and the real number needs to be strictly smaller then 10^-n
Under these rules 0.99 repeating is NOT a real number that actually exist since it violates the last rule, but 0.33 repeating is and is equal to 1/3
I like the way to describe the 0.999.. = 1 “paradox” in little less mathematical way to people with less mathematical skills. Divide 1 by 3, what you get? The answer is usually 0.333… and now multiply it back by 3 …
The problem is the decimals in our system does not perfectly dictate infinity. It’s why we use fractions as such. It’s not true 1 when using .33…*3 it does however converge to 1.
Lim x-> 0 |x| = 0
Lim x-> 0 |1/xl = infinity
Lim x-> 0 |x/x| = 1
The last two show that it’s not true 0 but 0.00…1
That is why .99… does not truly equal 1 however it does converge to 1.
.999.. = 1 assumes we are skipping past infinity. While theoretically it will never end.
This will never affect us since how infinitely small the difference is.
@@NewesSkiller You know, I am an engineer. For me, 3 x .333 is 1.
@@oldadajbych8123 I am a mechanical engineer, this is a theoretical issue not practical in our reality. You are rounding those decimals.
The argument comes from .333... not .333
Fractions are exact for rational numbers, while decimals are not.
@@NewesSkiller
'The problem is the decimals in our system does not perfectly dictate infinity'
That's just gibberish.
'It’s not true 1 when using .33…*3 it does however converge to 1'
It is true that 0.333...*3 = 1.
It does not converge to 1 or anything else, as it's a real number, and not a sequence or a function.
Learn some basic math.
'The last two show that it’s not true 0 but 0.00…1
What is 'not 0 but 0.000...01'?
'That is why .99… does not truly equal 1'
0.999... does exactly equal 1. No approximation involved.
'.999.. = 1 assumes we are skipping past infinity'
Again, this is gibberish.
The expression 'skipping past infinity' has no meaning.
Just stop spreading your BS and learn some basic math.
@@thetaomegatheta It's not gibberish using Fraction with rational numbers are vastly more accurate than decimals. Do a tiny bit of research.
4:32 “So if you're okay with this you should also be okay with that.”
I don't agree because 4/9 actually evaluates to 0·4 recurring so, to my mind, the two are not equivalent.
Nobody cares what you think
Then what *is* 0.444... and what *is* 4/9? What's the fractional form of the first number, which it has because it is rational, and what's the decimal expansion of the second number, which it has because it is real?
@@xinpingdonohoe3978But the decimal expansion of 1 is 1·000…
This is why I say they are not equivalent.
@@jarvisa12345 so far into this, uniqueness of representation as a fraction or a decimal hasn't been considered. There may be exactly one, there may be more than one, that is going to be settled further in. First I need to know the fractional representation of 0.444... and the decimal representation of 4/9, so that we may have a better chance of asserting whether each decimal expansion or fractional representation is mandated to have a single "look" about it in base 10.
If this isn't completely self-evident to you, then you don't understand what a "limit" is.
These "little bit of knowledge" math memes are doing so much damage to maths education. This is not deep. It's utterly unremarkable.
The difference is less than any number greater than zero, therefore the difference is zero.
At first, I thought this was more of a math trick than a fact, but then I thought about this: if .3333333333... is equal to 1/3, and .66666666666... is 2/3, then .9999999999... must be 3/3 (or 1). Likewise, .111111111... is 1/9, .4444444444... is 4/9, .7777777777777777... is 7/9, so .999999999999... must be equal to 9/9.
If 0,999... = 1 wasn't true, which real number would fit between these two?
That’s another way of proofing i.e. 1-0.99999999… =0.0000000000000….
none! there for it is true. Nice way to express it.
There isn't a real number between the two, that's why they're equivalent
More of an artifact of how poorly decimals are written. .
Nonsense. When you multiply by 10 you bring in an extra. As the series does not converge you do not know what is, or you can make in a 9. When you do he subtraction the result is 1............9, Not 1.
'Nonsense'
The video is correct.
'When you multiply by 10 you bring in an extra'
An 'extra' what?
'As the series does not converge'
The series does converge.
The series' sequence of partial sums is a Cauchy sequence, and the space of real numbers is a complete metric space. It's literally impossible for a Cauchy series to not converge in a complete metric space.
'When you do he subtraction the result is 1............9'
There is no real number that is denoted as '1............9' in any commonly-accepted system of notation.
What extra are you talking about? The series is convergent. What is 1............9 supposed to be? Don't bother answering, you'll only spout more nonsense.
Surprised you didn't just show us.
(9x)/9 = 9/9
It's just that 9/9 = 1
When you just cut to "x = 1" without showing us the divide by 9 step, thought you were setting us up for the big reveal later.
Wot???? 9/9 = 1 and 9x/9 = x. There's not really anything to show. It's basic schoolkid stuff.
If you prefer:
Starting at 9x/9 = 9/9, multiplying both sides by 9 => 9x = 9
Dividing both side by 9 => x = 1
Some time ago someone showed me that 1/3 = 0.3 repeat, 2/3 = 0.6 repeat, if thats the case, what is 3/3?
.3 repeating only approaches 1/3 it will never equal 1/3. .9 repeating approaches 3/3.
@@mantis1412 then what is 1/3? If we know that 0.333... approaches it, you must be aware of the thing it's approaching. What is it approaching? Every real number has a decimal expansion, most of which are infinitely long with no form of pattern. What's the one for 1/3?
@@mantis1412 so you're saying that 1/3 cannot be expressed in decimal form? you failed math
@@mantis1412 So in your opinion 0.5000000000... is only approaching 1/2?
@@LeNoLi. if fractions result in non-terminating decimals then they can never be correctly expressed as decimals it will always be an approximation approaching the correct value.
I thought 0.9̅ was equal to 1₋, like when finding limits 1₋ and 1₊. Does it mean 1₋ = 1 = 1₊?
'1₋ = 1 = 1₊'
'1₋' and '1₊' are just nonsense notation. There are no such real numbers.
1- is 1, approached from a specific direction. There are many ways to take limits. Look on the complex plane. Every possible path that can lead to a point is a valid limit, and all of them occur at the same number, just following a different route.
And besides, if 0.999...=1-, then what is 1+? If one is quantifiable, both are.
Presumably it will satisfy (1+)+(1-)=2. Given that 1- is rational (by the given definition), and 1 and 2 are rational, then 1+ would also be rational. This means the digits of its decimal expansion can be expressed through a finite sequence. For example, 1/4 has d_1=2, d_=5, d_n=0 for n≥3.
What's the sequence for 1+?
Yep. 0.999… equals 1.
This makes sense. It will only equal 4/9 when it reaches the infinite digit of .4 repeating, same as it will only equal 1 when it hits the infinite digit of .9 repeating. In reality neither can actually quite get there.
This is mathematics. There is no reality. If this was real, as in a physics problem, maybe it could be 0.4446 plus or minus .0004 or something like that. You could say that is suspiciously close to 4/9. If this result kept showing up, you could take that as a hint from nature to look for a formula with 4/9 in it somehow, and try to explain why it was there. But in mathematics, things CAN get all the way to infinity, so .4 repeating just IS 4/9.
@@geirmyrvagnes8718so .9 repeating is 1 in mathematics.
@@geirmyrvagnes8718 is .9 repeating = 1 correct or do whole numbers get elevated above the infinite expression and fractions do not?
@@mantis1412 That is the thing. If you have repeating digits forever, you can write it as a fraction of whole numbers with a calculation similar to this video, so they are "rational numbers". And in this case that fraction is 1/1, if you will. So 0.9 repeating just IS a whole number, namely 1.
So N.99999... = N+1 ???(N is a natural number)
Yes.
At step three, why not divide by 10?
Also, at the .99 cent store all .99 cent items ring up one dollar.
Why would we? What would we achieve? If we have 10x=9.999..., dividing by 10 will give x=0.999...
We already know that to be the case, from the first line.
When you did 4/9, I noticed that it is indeed 0.4 repeating... using 1 to 8 for X, and deviding by 9 (X/9) the number used for X is also the repeating number...
Thus 9/9 would be 0.9 repeating... no wait... it's 1. Ain't that funny?
The truth is, every decimal number like 0,999... only TENDS to it's fractional form. Technically, they're not exactly equal to each other, because there will be a so called infinitesimal difference.
The second thing is, that okay, we are subtracting 0,999... from 9,999..., but do we really know that the quantity of the infinite repeating 9's are exactly the same? If so, then we know, how big infinity is. But we don't know how big infinity is (at least, I don'r know), so we can't say for sure that 9,999...-0,999... is EXACTLY equal to 9.
For 4/9 and 0,444..., it's the same thing. 0,444... only TENDS to 4/9, b/c we don't know how much 4's do we have after the decimal point. And inifnity is not a number, it's just an expression for a number being so big that we cannot define it.
The sequence 0.9, 0.99, 0.999 etc. tends to 0.999...
The sequence 0.9, 0.99, 0.999 etc. also tends to 1, from geometric series.
Limits are unique. Go figure.
The horizontal bar means that this number is infinitly repeating
So .4 repeating is EXACTLY equal to 4/9
The truth is, you don't understand rational numbers or their representation as expansion-of-powers-of-a-base.
Okay.
Let's have the sequence an=(6n-7)/(2n+1)
If we take the limit as n goes to infinity of this sequence, itt will approach 3. But the question is, does an in infinity really equal to 3, or just goes to it?
Same idea: take the limit as x goes to negative infinity of e^x. We know that e^x never can be 0.
But if we "put" neg. infinity into the x, then is the answer exactly 0, or it just aprroeches 0, namely, it's 0+?
And about the geometric series: in fact, it's also just a limit, but with a fornula. It's like every day you eat half of the remaining chocolate bar. Then you say, you will never run out the chocolate bar. Then you apply the geometric formula, and you get exactly 1, and this means, you ate the whole chocolate. Then the question: did you eat the whole chocolate bar or not? It's a contradiction.
In fact, every number you get from any limit question is ALWAYS an asymptote that you can NEVER reach, just tend to it, also, in neg. inf.
I don't disagree with the proof. But this always confused me. If I divide 9/10, I get 0.9. If I divide 9999/10000, I get 0.9999. I can extend that indefinitely to get .9 repeating. However, in that case the denominator is always greater than the numerator. Thus, how can that equal 1? Infinity is a wacky animal.
Here are two more proofs:
1) Consider relation R between Cauchy sequences of rational numbers: for any two Cauchy sequences of rational numbers a=(a_1, a_2, a_3,...) and b=(b_1, b_2, b_3,...) the relation aRb holds iff lim(a_n-b_n)=0.
Any given real number is an equivalence class of such sequences with respect to R.
Any given digital representation corresponds to a Cauchy sequence of rational numbers, for example, 0.999... corresponds to (0.9, 0.99, 0.999,...), and 1 corresponds to (1, 1, 1,...).
Let's check if (0.9, 0.99, 0.999,...)R(1, 1, 1,...):
lim(1-sum(9/10^k) for k from 1 to n) as n->inf = lim(1/10^n) as n->inf = 0. That means that (0.9, 0.99, 0.999,...)R(1, 1, 1,...) and 0.999... = 1.
2) If x is some real number, |x|
0.999... is not of the form 999...9 (n 9s)/1000...0 (n 0s) = 1 - 1/10^n. It is the limit of that as n->oo, and that is 1.
You need to expand your consciousness to grasp infinity, that's where the leap in logic is failing you
@@redjoker365 Hence why I said infinity is a wacky animal.
@@Chris-5318 so why can't n go to infinity? 999... infinity 9's/ 1000 infinity 0's still will be a rational number where the numerator is less than the denominator. And written as a decimal, wouldn't it be 0.9999 ...
Or you could show .9999 repeating = 9/9
Nothing in infinity related constructs is exact, that's normal, and why everything is, and should be kept, an approximation. It's like saying 3.14 is equal to pi just because...idk...the next digit is 1 which is close to enough 0 to negate the rest or something as dumb as saying infinite thing times 3 is equal to this infinite thing which is "EQUAL" to 1, a constant. The entire concept of 1/3 is 0.333... and 0.999... is 1 brings the byproduct of claiming (as fact) that infinity is a constant, which is paradoxical (being a constant means having a value that cannot change + infinity is bigger than infinity + 1 because infinity is meant to be *the* biggest = infinity is not a constant), leveraging the entire theory as 'Disproven by contradiction'.
I wouldn't mind counter arguments, but I would like something more competent than "I feel like it" or "It's true because it looks like it". This argument is pointless to begin with, I am not here to outright discourage and tell people what to think. It's not about the notion that 0.999... is not related to 1, it's that saying something strictly NOT constant is EQUAL to a constant is so stupid it's making ME, some random programmer from a negligible country and origin that just so happens to be good at making sense (often enough to be coherent), has to correct some straight up tunnel vision (not realizing the existence of the byproduct). I'm no Einstein, Tesla nor Newton, but still, what the heck is that.
Potential Counter Argument: Pi has infinite digits, why is still a constant? (Counter-Counter Hint: It's the length that's infinite, not the value)
And if anyone wants to, prove to me what 0.444... is equal to. I *hope* it's a constant! (An Infinite Geometric series will not work because the limit will result in 0.4. (no repeating part, the same as 4/10) This is why I disagree with the last video, it's just wrong (No offence though ], love your calculus level vids).
Also also, saying that you can multiply an infinite number by a constant and get a coherent *value* is an undefined operation.
Also also also, saying that you can willy nilly subtract an infinite number from another and get a constant is just plain playing with fire.
Infinity means not-finite and it's strictly true, even you want it to stop, it won't. The logical answer would be that and infinite number subtracted or added or whatever with another will just result in yet another infinite number, which still ain't a constant btw.
If you want to confirm something, define the value of infinity, let's see how that turns out.
Thanks for coming to my Ted-Talk! Hope you have a comfortable day!
The entire concept that 1/3 is 0.(3), using the notation that () surrounds the repeated values, comes simply as a by product of using base 10 as our primary number system. The only way we can represent 1/3 as a decimal is by using repeated digits. The only way we can accurately represent it is by specifying the repeated digits.
You can understand where the requirement for specifying the repeated digits comes from if you do the division using long division. It becomes very apparent that it repeats. However, this is an issue only in number bases that don't share a factor, or in other words, the number base is relatively prime with the denominator of the fraction. 1/3 in base 9 is a simple 0.3. 1/3 in base 12 is 0.4. So instead of thinking of the pointless ways to try to define it, think of it in the simplest way. The number is a rational number, while abstract in definition, it corresponds to a natural idea. Think that 0.(3) is the exact representation of 1/3 in base 10, nothing more and nothing less.
Pi is a constant because it is always the ratio of the circumference and diameter of a circle. The cumulative effect of the infinite post decimal digits very quickly tends to 0 which is why we can we can do stuff with it like land on the moon instead of missing it by millions of kilometres without knowing the value to infinitely many decimal points. We don’t calculate the value to trillions of digits because it’s important.
The infinite series does not go to .4. It goes to 4/9, which is not 4/10 or 2/5. limit as a goes to infinity of Sum(i=0 to a, .4*10^a). Note, the first term of the summation is what you claim the entire sum is. That seems to be a small error as there are all non-zero and non-negative terms, which means your epsilon keeps increasing. Thankfully, your epsilon doesn't grow infinitely. It only grows to be 4/90.
No idea what geometric series youre looking at that makes 4/9 = 4/10. Whatever it is, its wrong or you interpreted it wrongly. The proof he shows in here isnt some controversial area of math, this has been estabilished as fact for centuries, its just how numbers work. No amount of wierd justification will change that.
@@skanderbeg152 He has the right to question, but when he is claiming the sum with infinite terms is the same as the first term, I question whether it would be possible to prove to him at all. Eventually the terms are so small as to be negligible, but he claimed all the terms are that way after the first. That is extremely bad analysis.
0.9999.... + 0.1= 1.09999...... but 1.0 + 0.1 = 1.10000.... Only if you round off the number do they become equal
1.0999... = 1.1
No rounding up or down is involved.
Thy shalt not muck about with infinity like that.
I think they're okay with 4 over 9 because that's still a fraction, or a decimal, less than 1. 0.9 repeating shouldn't logically be allowed to equal 1.
9/9 is a fraction
I always say: If you cannot find any number between two numbers a and b, the numbers a and b are equal.
I feel like this uses the same logic in stating that the sum of all positive integers is equal to -1/12
In a sense it does, in another it does not. A repeating decimal can be represented with a convergent series, so it's much "easier" to deal with, and it's legitimate to treat the series as a number. If a series does not converge, then a lot more attention is required, and operations such as "multiplying by a constant" or removing terms and adding them back in another place are not possible.
Not really. Using algebraic manipulation, you will only find that 0.999...=1, and this is because it exists.
1+2+3+4+... does not exist. You can make it "equal" -1/12, but you can also just as easily make it "equal" -1/8. This is a contradiction, showing 1+2+3+4+... does not exist.
It's not. And the -1/12 value is widely misunderstood. Mathematcians are not saying the sum of all positive integers is equal to -1/12. But that through a process, such as zeta function regularization or Ramanujan summation, they can get a value like -1/12 for example, that provides important information about a divergent series, such as the sum of all positive integers. They are NOT saying it equals -1/12 though, that's obviously nonsense. I know there was a click-baity video by another math youtube channel that really confused a lot of people and upset others, unfortunately they did not do a good job explaining that it's not an equality, just an important value that's useful in things like complex analysis and quantum field theory.
But when they say 0.999... = 1, they mean exactly that. And there's a bunch of rigorous proofs for it in higher levels of math. BPRP goes over some on his calculus channels.
@@dlevi67That it converges is what ought to have been shown and what he tacidly assumed, making the entire confusion worse.
@@MrCmon113 I very much doubt that the confusion in these comments (in general, not this particular thread) stems from the fact that the series was/was not shown to be convergent. We are one or two (dozen) levels of sophistication down, with people not understanding the concept of real or rational number.
We usually used some wavey = sign to emphasize that it’s a rounded number back in school
It has nothing to do with rounding, it is about equality.
It's not rounded though, it _is_ 1.
WRONG!! Oh boy, this is such a common mistake when dealing with infinite sequences/sets - the trouble is the notation is not sophisticated enough and loses information.
Try using notation of "..." to represent an infinite but consistent quantity (throughout the solution).
Consider n number defined as 1. 'infinite 0's followed by a 1 (1.00...01) now multiply by 10 (10.00..10) divide this new number by the first and you get 10.00...10); it's easier to comprehend if you start with "..." meaning one digit: 1.00001 multiplied by 10 equals 10.00010 not 10.00001
In addition the math of the second example is (with ... being infinite but consistent quantity of 4s or 9s):
x = 0.44...44
10x = 4.4...40
10x-x = 3.9...96
x = 3.9...96 / 9
x = 0.44...44
So 4/9 = 0.44...44 is actually inaccurate but by an infinitesimally small amount (specifically 0.00...04/9)
'WRONG!!'
The video is correct.
'Oh boy, this is such a common mistake when dealing with infinite sequences/sets'
Oh, I bet you have published refutations of basic theorems/statements about properties of sequences, then?
'the trouble is the notation is not sophisticated enough and looses information'
Point out to the 'lost information' (I assume you misspoke and did not mean 'loose information'), then.
'Try using notation of "..." to represent an infinite but consistent quantity (throughout the solution)'
This is incredibly vague. I.e. this is gibberish.
'Consider n number defined as 1. 'infinite 0's followed by a 1'
A number is not a sequence of digits. That's a decimal, i.e. a representation of a number. And a number can have multiple decimal representations.
If by that you mean 'a number that can be represented as 1.000...01 with infinitely many "0"s', then that's just nonsense, as well, because 1.000...01 = 1+1/10^n for some natural n - there can only be finitely many '0's there before the second '1'. And if you are looking for some number x, such that 1
@@manticore5733
'Here is the solution worked with ... representing an infinite but consistent number of 9s:
x = 0.99...99'
You seem to be unable to distinguish between the decimals '0.999...' and '0.999...9'. The former is non-terminating, the latter is terminating, i.e. the latter has only finitely many '9's in it. 0.999... > 0.999...9. The video is about 0.999...
'In addition the math of the second example is (with ... being infinite but consistent quantity of 4s or 9s):
x = 0.44...44'
The second example involved 0.444..., and not 0.444...4.
Rather amusing how you fail at basic reading comprehension, but have the gall to pretend that it's every single mathematician alive who is wrong.
@@thetaomegatheta To address your points.
1. I do have a paper I wish to publish on the topic yes, but have no means to do so.
2. typo: 'loses information' (corrected)
3. infinite but consistent quantity - in terms of countable sets, two sets are of equal size if you can map elements of the sets in a one-to-one relationship: which holds true even if the sets are infinite. I.E. throughout the solution "..." represents an infinite quantity that can be mapped in a one-to-one relationship based on position. Directly related to the pigeonhole principle. Hope that clears up the 'gibberish'.
4. You are taking the limit of 1+1/10^n as n tends to infinity. Yes, that is 1. However I was referring to a specific number that is infinitesimally larger than 1. As infinity is a quality not a quantity it can't be substituted for n in your given notation. You say there can only be finite number of 0's in a natural number that is of the form 1.0...01? and yet accept infinitely recurring numbers? I am only defining the nature of the number to highlight the point and the indeed your argument also shows how concepts involving infinite sequences collapse when converted to notation used to depict finite quantities.
5. Simple arithmetic, consider the following sequence of arithmetic steps:
n -> 10n -> 10n-n -> 9n/9
apply to increasing examples:
0.1 -> 1.0 -> 0.9 -> 0.1
0.11 -> 1.10 -> 0.99 -> 0.11
0.111 -> 1.110 -> 0.999 -> 0.111
Extending this to 0.11 recurring does not cause it to collapse, only pushing the values into notation that fails to preserve the effect of manipulating infinite quantities does.
As an aside the best example of this is the infinite room paradox, an infinite number of occupied rooms - all guests move to the next room freeing up a room for a new guest: represent as set R and set G for rooms and guests, there is a one-to-one mapping between set R and G. Moving all guests means there is either a different mapping and all rooms are still occupied OR there is no longer a one-to-one mapping and you can no longer consider R and G to be of equal size even though they are both infinite.
This principal is being applied in my argument regarding manipulating the digits... in fact it's easier to visualize than the paradox:
There are 9 guests in each room of The Decimal Place Hotel - which has infinite rooms.
You build a new room to the left (outside the front door if you like) and then every guest moves one room to the left (multiply by 10).
Nine guests now checkout of each of the original rooms (subtract original).
Tell the remain guests that only one in 9 of them can stay (divide by 9).
And bingo there is only one guest left and they are in the room to the left of The Decimal Place...
...but here is the thing, when every guest moved one room to the left the room furthest to the right is left empty, so nobody could checkout of that room.
@@manticore5733
'You are taking the limit of 1+1/10^n as n tends to infinity'
I don't. If you want to argue that I do, quote the relevant portion of my reply.
The fact that lim(1+1/10^n) as n->inf is the infimum of the sequence (1+1/10, 1+1/100, 1+1/1000,...) - i.e. the thing that I actually evaluated there - is not particularly relevant.
'However I was referring to a specific number that is infinitesimally larger than 1'
There are no infinitesimals in the space of real numbers, so there are no such numbers.
However, there is also another problem - if you wanted to work in a space of hyperreal numbers, there would be many numbers that satisfy your description, and not just one.
'As infinity is a quality not a quantity it can't be substituted for n in your given notation'
That's not why it can't be substituted for n there. n is natural. Infinity is not an element of the set of natural numbers. That's it.
'You say there can only be finite number of 0's in a natural number that is of the form 1.0...01?'
Numbers of the form 1.000...01 are not natural. And they all have finitely-many '0's before the '1's in their decimals.
'and yet accept infinitely recurring numbers?'
Terminating decimals are repeating decimals with trailing '0's. There is no issue there.
0.999... = 9/10+9/100+9/1000+...
Meanwhile, 1.000...01 = 1.000...01000... = 1+0+0+0+...+0+1/10^n+0+0+0+... - it has a '1' at the nth position after the decimal dot, where n is natural. That's literally what the notation means. If you want a real number x, such that 1
Welp, guess I can no longer be okay with 4/9 = 0.444…
Thanks, bprp! 😊
Then what is 4/9?
Its a fraction. Glad I could help.
@@deltalima6703 indeed, and so we can conclude that it is a real number. Every real number has a decimal expansion. It's essentially mandated by the completeness axiom. Given that 4/9 is a number, what is a decimal expansion of it?
@@xinpingdonohoe3978 Every **rational** number has a decimal expansion, there are real numbers like pi which have no decimal expansion
I think it's just an interesting play with the numbers
But it's still a contradiction
Let's think it through
Is 0.9 equals to 1?
No, it's 0.1 off
Is 0.09 equals to 1?
Still no, it's 0.01 off
We can follow this path as far as we want, 0.0....09 will always be 0.0....01 off
You can't have 0.0...01.
You also can't have 0.00...09
The three dots means there's nothing after it. Never a 1, just zeroes endlessly.
@@Daniel31216 what prevents me from having infinitely many zeros and either a 1, or a 9 at the end?
@@Vanhaomena the dots mean infinitely many zeros
And at the end either a 1 or a 9
I’m not arguing with the two proofs you provided, but I’m still not convinced based on the meaning of the word repeating
Does this not imply the infitesimal is exactly equal to 0? Thats bizzare. Or then that the infitesimal doesnt exist as it would violate its own definition?
'Does this not imply the infitesimal is exactly equal to 0?'
No, because there are no infinitesimals in the space of real numbers.
@@thetaomegatheta Right so it implies it doesnt exist in real numbers which is pretty profound.
@@OsirusHandle
It's not profound, considering what infinitesimals are - a positive infinitesimal is a number that is less than any positive real number, but greater than 0. There is obviously no such real number, as it would have to be less than itself.
@@thetaomegatheta but its not that there is no real infitesimal, but that any maths that would use infitesimals couldnt apply to real numbers, because they would be paradoxical. so a number system would surely be incompatible with reals, even if reals are compatible with it
@@thetaomegathetaor is it that the infitesimal is larger than the gap between 1 and 0.9_? i suppose that couldnt be represented by reals
.999... is not math, it is philosophy
.02
What you wrote is gibberish. Likely your attempt to cope with your bad intuition.
@@thetaomegatheta If you say so...
Precisely what disqualifies it from being math?
I always have the issue with this proof that you're essentially manipulating an infinite series but you can only do that for convergent series so aren't you already assuming the answer to be 1 when you multiply by 10?
The series 9/10+9/100+9/1000+... is absolutely convergent. Understanding that does not require the assumption that 0.999... = 1.
the common ratio is r = 1/10, which satisfies -1 < r < 1, so it easily converges.
Yeah this is kind of wild but I get it. You show at the very end that 0.4444444… is the same as 4/9, so additionally you can say 0.999999… is the same as 9/9, thus 1
All of this mathematical tom-foolery arises because some fractions cannot be expressed as an exact number in a particular number base. 0.1 base 10 has no exact representation in binary for instance. However, 1/3 is 0.4 base 12 exactly so it is all comes down to be nothing more than an isuue of number representation. Incidentially, 1 - 0.999.... could also be regarded as the smallest number not equal to 0 which might prove to be useful to someone. However, as a retired engineer, there are practical limits as to how many decimal digits numbers need to have - infinite digits are just not practical in real life
'0.1 base 10 has no exact representation in binary for instance'
Yes, it does. Every rational number has a corresponding digital representation in such a base.
'Incidentially, 1 - 0.999.... could also be regarded as the smallest number not equal to 0'
No, it couldn't, and if you knew anything about real numbers, you would have known that there is no such thing as the smallest positive real number (which is what you meant, because there is even more obviously no such thing as the least negative real number, which would be less than any positive real number).
1-0.999... = 0 exactly.
You're conflating an infinite series with a very large yet finite series, your logic is flawed and you should be ashamed for putting your ignorance on proud display
I get the logic and proof behind why 0.999...9 is equal to 1. It just intuitively feels like 0.999...9 is asymptotic to 1 and not equal to 1. It's emotion overriding logic. 😢
0.999... is the limit of an infinite series (specifically 9/10+9/100+9/1000+...). It already is the asymptote.
It is not 0.999...9, it is 0.999...
As Tony pointed out before me, 0.999...9 is not 1. 0.999... is 1. The first one ends after an unspecified (not infinite) number of 9s, the second is one contains an infinite sequence of 9s.
What would it even mean to say one number is "asymptotic" to another number? They're both just numbers, and they're either equal or they're not. Asymptotic means "approaches" which makes sense for a function or a sequence which takes on many values, but a number has only one value and therefore it's nonsense to say it approaches anything.
@@OMGclueless it's a buzzword, usurped by people who can't handle the reality of the situation to try and get their points across using argument by authority.
Ahh once more the same question :)
I guess people dont accept that 0.999 = 1.
Can you prove that 0.999... is 1 without multiplying with 10 first ?
I am just wounder if there are other ways to get to the same answer (1).
'Can you prove that 0.999... is 1 without multiplying with 10 first ? '
Yes.
Two more proofs:
1) Consider relation R between Cauchy sequences of rational numbers: for any two Cauchy sequences of rational numbers a=(a_1, a_2, a_3,...) and b=(b_1, b_2, b_3,...) the relation aRb holds iff lim(a_n-b_n)=0.
Any given real number is an equivalence class of such sequences with respect to R.
Any given digital representation corresponds to a Cauchy sequence of rational numbers, for example, 0.999... corresponds to (0.9, 0.99, 0.999,...), and 1 corresponds to (1, 1, 1,...).
Let's check if (0.9, 0.99, 0.999,...)R(1, 1, 1,...):
lim(1-sum(9/10^k) for k from 1 to n) as n->inf = lim(1/10^n) as n->inf = 0. That means that (0.9, 0.99, 0.999,...)R(1, 1, 1,...) and 0.999... = 1.
2) If x is some real number, |x|
Certainly. 1-0.9999999999…=0.000000000000000000…
2 * 0.999... = 1.999...
=> 0.999... + 0.999... = 1 + 0.999...
=> 0.999... = 1
7 * 0.999... = 6.999...
=> 6 * 0.999... + 0.999... = 6 + 0.999...
=> 6 * 0.999... = 6
=> 0.999... = 6/6 = 1
I'll prove that 2*0.999... = 1.999.... (It'd just more of the same for 7*0.999...)
2 * 0.999... = 2 * (0.9 + 0.09 + 0.009 + ...)
= 2* 0.9 + 2 * 0.09 + 2 * 0.009 + ...
= 1.8 + 0.18 + 0.018 + ...
= 1 + 0.8 + 0.1 + 0.08 + 0.01 + 0.008 + ...
= 1 + 0..9 + 0.09 + 0.009 + ...
= 1.999...
So by the same logic, 1.999999 reoccurring must be 2 then? Ad infinatum. What utter crap.
is that not true
@@tzbq obviously not, any decimal cannot equal an integer
1.9999... = 2.0 IS true though.
1.999... (no need to write out 6 '9's before saying 'recurring' or 'reoccurring') does equal 2.
@@thetaomegatheta 0.9..... x 0.9..... is that 1, it's obviously not. An infinite number isn't a real number, you're all fucking thick
why are we working under the assumption that the algebra for “repeating numbers” works the same as for real numbers? does it form a vector space over the real number and are we performing scalar multiplication? what is the operation and why is it well defined
Because "repeating" numbers are real numbers, they can also be represented as rational numbers
On the number line what number fits between infinitely many post decimal 9s and 1?
@@rogerszmodis None. There are no points between them. They are both represented by the same unique point on the real number line. They are equal after all.
Normally people just refer to "repeating numbers" by their actual name: Rational Numbers. And yes, scalar multiplication and other algebra operations are well defined for them.
Because if we work under the assumption that they don't work, then we can't make any progress. That's like assuming ε≤-1. It's just going to get us nowhere.
1 - .9999... = .000...01.
Small problem for you, 1 - 0.000...01 = 0.999...99, not 0.999...
Your logic is flawed, 0.999... is an infinitely long sequence summation, not a finitely large one. There is no "last" digit to be had, you just can't comprehend infinitesimals
You can't put something at the end of an infinite sequence. That last "01" has no where to go. It does not exist
Limits
Yeah, limits and infinity contradict each other.
@@elio7610 in what way?
0.999... is the limit as N approaches ∞ of the sum from n=1 to N of 9×10^-n. It's infinity and limits working together.
0.999...9 = 1 - 0.000...1
... and 0.999...9 < 0.999...
@@Chris-5318 i didn't find an overlaying overscore symbol, hence the dots
@@kuchesezik I use ... most of the time. Whatever, your 0.999...9 is terminating decimal, and so it is less then 0.999...
0.999... - 0.999...9 = 0.000....0999...
Another way to indicate recurring/repeating is with ( ) . e.g. for 0.123123123... you can write 0.(123), and that's much more useful than ... and much easier to use than the vinculum (overline).
@@Chris-5318 ok, 0.(9) = 1 - 0.(0)1
@@kuchesezikexcept that 0.000…1 doesn’t exist, as it’s impossible to have any number after an infinite number of 0s
So 0.999… - 1 = 0
I'm not a math expert by any means, but there's a pretty popular PhD Math professor from I believe Bangladesh that explained why this is incorrect on Reddit. Again, the conversations that took place on that thread was waaaay above my level of understanding, but that professor convinced every other advanced level math commentor that he was right. It was an amazing read, although most of it was foreign to me. I'll edit my comment and post the thread if I can find it.
The result is a math fact. Any mathematician that says otherwise is a crank (e.g. Norman J. Wildberger).
I just saw your, "but that professor convinced every other advanced level math commentor that he was right."
Either you made that up or the advanced commentor's were clueless muppets.
@@Chris-5318 I didn't. Go look for yourself and get back to me 👌🏾.
@@E_Cleazy I have better things to do than spend hours searching Reddit for an unknown crank professor schooling a bunch of ignoramuses. It is a math fact that 0.999... = 1. I have around 10 proofs, one of which is in the video above.
@@E_Cleazy I also note that you haven't attempted to pass on the argument that the professor made. Was the prof. N J Wildberger?
Profe Juan si fuera chino
Am I stupid ? Shoudlnt be the first exmaple be 1/9 because to get from 9x=9 to x=9 I would have to divide bothsides by 9. I see it is true for the other example with 4 but not for the first.
Where di you see him go from 9x = 9 to x = 9? I saw him go from 9 = 9x to 9/9 = 9x/9 to 1 = x
三三不尽，六六无穷。谢谢老师。
wow
Якщо Х=0,999...(N разів), то 10*Х=9,99....((N-1) разів), тому 10*Х - Х =8, 999...1 і НЕ ДОРІВНЮЄ 9, а значить Х не дорівнює 1 !!! Тому не потрібно вішати лапшу на вуха з єхідним виразом обличчя махінатора-інсинуатора&&&&&
So how many 9s are there in the first sequence, and how many are there in the second sequence?
If there are finitely many, so that we can talk about there being 1 less, then you've selected the wrong number.
If there are infinitely many, they can't differ by 1. The fact is that we can biject each 9 in the first to a 9 in the second, making the amount the same.
There are the same number of even numbers as there are integers. That alone should make you realise that "removing" a single 9 will not change the infinity at all.
@@xinpingdonohoe3978Не треба розказувати свої дебільні байки, бо зменшене на одну цифру 9 нескінченне число і нескінченне число з цифр 9 - це НЕОДИНАКОВІ ЧИСЛА!!! Тому видалення однієї цифри 9 змінить нескінченність &&&&&&&&&
@@xinpingdonohoe3978 Сінь Пінь Хуе, ти взагалі не тямиш, про що говориш, бо нескінченність і на 1 менше від нескінченності розуміється лише в контексті якогось визначеного достатньо великого, але все ж таки конкретного числа. Тому твоє твердження що видалення однієї цифри 9 не змінить нескінченність абсолютно неправильне, а тому НІКЧЕМНЕ&&&&&&&&&&
@@xinpingdonohoe3978 Висловлювання СіньПіньДонаХ...я про те, що видалення однієї цифри 9 не змінить нескінченність абсолютно необгрунтоване, а значить безпідставне!!!!
@@xinpingdonohoe3978 Так може стверджувати лише дилетант з ясельної групи, який абсолютно не розуміє, що нескінченність - це віртуальне поняття, і тут потрібно оперувати конкретними дуже великими числами, тому число з К цифрами 9 та (К-1) цифрами 9 - це абсолютно різні числа&&&&&&