The original question was: 0.999… = 1 does not make sense with respect to my conception of the number line, I do not know much about number classes but for the number line from lets say 0 to 1 there is an infinite number of points, so there is a number right next to 1 we cant write it in entirety because the decimal expansion is infinite but the difference between that 1 and that number is 1*10^(-infinity) (sorry if I am abusing notation). so that number to me should be 0.999… but it is not where am I missing the point?
Physicist: In the language of mathematics there are “dialects” (sets of axioms), and in the most standard, commonly-used dialect you can prove that 0.999… = 1. The system that’s generally taught now is used because it’s useful (in a lot of profound ways), and in it we can prove that 0.99999… = 1. If you want to do math where 1/infinity is a definable and non-zero value, you can, but it makes math unnecessarily complicated (for most tasks). The way the number system is generally taught (at the math-major level, where the differences become important) is that the real numbers are defined such that (very long story short) 1/infinity = 0 and there isn’t a “next number” for any number. That is, if you think you’ve found a number, x, that’s closer to 1 than any other number, then I can find a number half way between it and 1, (1+x)/2, that’s even closer. That’s not a trivial statement. In the system of integer numbers there is a next number; for 3 it’s 4, for 26 it’s 27, etc.. In the system of real numbers every number can be added, subtracted, multiplied, and divided without “leaving” the real numbers. That leads to the fact that we can squeeze a new number between any two different numbers. In particular, there’s no greatest number less than one. If there were, then you couldn’t fit another number between it and one, and that would make it a big weird exception. Point is: it’s tempting to say that 0.999… is the “first number below 1”, but that’s not a thing.
The term “real numbers” is just a name for a “sand box” of mathematical tools that have become standard because they’re useful. However! There are other systems where “very very very slightly less than 1” , or more precisely “less than one, but greater than every number that’s less than one”, makes mathematical sense. These systems aren’t invalid or wrong, they’re just… not as pretty and fluid as the simple (as it reasonably can be), solid, dull as dishwater, real number system.
It is a fact, immutable and True, that every rook is safe for the next move according to the most widely accepted rules for chess. In other games that may not be the case. 0.999…=1 in the most widely accepted rules for real numbers.
In the set of “real numbers” (as used today) a number can be defined as the limit of the decimal expansion taken one digit at a time. For example, the number “2” is {2, 2.0, 2.00, 2.000, …}. The “square root of 2” is {1, 1.4, 1.41, 1.414, 1.4142, …}. The number, and everything you might ever want to do with it (as a real number), can be done with this sequence of ever-longer decimals (although, in practice, there are usually more sophisticated methods).
These sequences are “equivalent” and describe the same number if they get (arbitrarily) closer and closer to that same number forever. Two sequences don’t need to be identical to be equivalent. The sequences {1, 1.0, 1.00, 1.000, …} and {0, 0.9, 0.99, 0.999, …} both get closer and closer to each other and to the value “1” forever, so they’re equivalent. In absolutely every way that counts (in terms of the real numbers), the number “0.99999…” and the number “1” or “1.0000…” are exactly the same.
It does seem very bizarre that two numbers that look different can be the same, but there it is. This is basically the only exception; you can write things like “0.5 = 0.49999…”, but the same thing is going on.
A sequence is equal to its limit. For an increasing sequence its LUB is its limit. So this sequence is 1.
I have read all your posts and multiple times you have refused to use this.
The limit of the sequence (0.9, 0.99, 0.999, …) is 1. I agree.
Yet it doesn’t mean that 0.9 + 0.09 + 0.009 + … adds up to a point on the number line. Neither to 1 nor to any other point.
Do you agree that
[0, 0.9]∪[0.9, 0.99]∪[0.99, 0.999]∪… = [0, 1) and not [0, 1]?
Yes, but it doesn’t matter. The definition of an infinite decimal is its LUB. Nothing you’ve said changes that, and that LUB is 1.
@ netzweltler – I think I see the problem here. The LUB of a sequence doesn’t have to be a member of it. You’re assuming it does. This seems to be the problem. Maybe.
OK, again, for the nth time:
The value of an infinite decimal is defined to be its LUB.
The value of 0.999 . . . is defined to be its LUB, which is clearly 1.
You wrote: “The limit of the sequence (0.9, 0.99, 0.999, …) is 1. I agree.
Yet it doesn’t mean that 0.9 + 0.09 + 0.009 + … adds up to a point on the number line. Neither to 1 nor to any other point.”
You need to be more precise about what you mean by “adds up to a point on the number line”.
@Alan Feldman
This is like saying “We _define_ 0.999… to be 1. Therefore, 0.999… equals 1.”
To me this is like comparing 1 – which is a point on the number line – to something which is not.
@Alan Feldman
“You need to be more precise about what you mean by “adds up to a point on the number line”.”
0.9 + 0.09 adds up to point 0.99 on the number line.
0.9 + 0.09 + 0.009 adds up to point 0.999 on the number line.
0.9 + 0.09 + 0.009 + … doesn’t add up to point 1 on the number line.
I tried to make that clear in my very first post in this thread.
Do you agree that
[0, 0.9]∪[0.9, 0.99]∪[0.99, 0.999]∪… = [0, 1) and not [0, 1]?
Sir, just how are _you_ defining 0.999…? Obviously in a different manner. So there is no contradiction.
We are not defining 0.999… itself to be anything; we are defining what a repeating decimal is. And it is defined to be the LUB of the sequence of adding more and more digits.
What is 0.9? It is defined to 9/10.
What should we do if we require the decimal form of say, 1/3? What would you write? Most of us would write 0.333…. I suppose you would say you can’t write it down because of your points on a number line bit. The rest of us would like to write it down, so we have chosen to write a sequence whose LUB is 1/3.
_We_ have specified what we mean by a repeating decimal. You, OTOH, have talked about points being or not being on a number line. _You_ are using a _different definition_, whatever it is (you haven’t said, other than the vague idea of adding up to a point on the number line), so there is no surprise that you come up with a different conclusion.
I don’t think I can make it any clearer without spending an entire day or two on it.
“Do you agree that
[0, 0.9]∪[0.9, 0.99]∪[0.99, 0.999]∪… = [0, 1) and not [0, 1]?”
I’ve already replied to this. Yes. But it doesn’t matter. The LUB ******DOES NOT HAVE TO BE A MEMBER OF THE SEQUENCE OF PARTIAL SUMS******. For example, the LUB of the negative real numbers is zero, which is not a negative number. And neither is 1 a member of (0,1).
Again, a infinite decimal is defined to be its LUB, which is in general _not_ a member of the set of the partial sums. All you’re saying, ABAICT (or AFAICT), is that none of the partial sums is 1. BUT THAT DOESN’T MATTER. An infinite decimal is defined to be its LUB, which in general is not a member of the set of the partial sums of the series.
By “defining” we are saying that, for example, the way to write 1/3 as a decimal is 0.333…. Multiply that by 3 and you get 0.999…, which is one.
We are defining a way to write a number in decimal form that cannot be represented by finite decimals.
I’m sorry about the “screaming”, but you keep missing this crucial point.
@Alan Feldman
I still think it is not clear what I am saying. To serve as an LUB, the LUB needs to be a number, i.e. a point on the number line. I cannot see how 0.999… can be assigned to a point on the number line – for the reasons given.
OTOH I fully agree, that 1 IS the LUB of the sequence (0.9, 0.99, 0.999, …). I just want to make that clear, because you are still trying to find arguments for that.
Let me try a more general approach…
Infinities are hard. They are simply impossible to deal with through brute force. It is easy to write 0.9 + 0.09 + 0.009 + … but you cannot actually add up all the terms manually. It is easy to write 1/2 + 1/4 + 1/8 + … but you cannot actually add up all of the terms manually. However there are lots of these infinite sequences, and it would be really helpful to have an agreed standard way of working with them. So how do we deal with them?
We could say: bah humbug, we do not accept infinite sequences at all and we refuse to ever deal with them. This is not useful.
We could say: every individual can make their own guesses about sequences. Then we might get:
Alice: I guess that 0.999… is equal to 1
Beth: I guess that 0.999… is equal to some number less than 1
Carrie: I guess that 0.999… is equal to 35
netzweltler: I guess that 0.999… is not equal to any number
This is also not useful.
So what happened is that the concept of “limit” was created, and for a sequence the numeric value of the sequence was defined to be the limit of the sequence. This concept of a limit of a sequence is helpful. It is pretty well behaved, you can add and subtract and multiply by a constant. It gives a well defined answer that Alice, Beth, and Carrie can all agree upon. They will even agree if they take different routes to get to the answer:
Alice: I use the definition of limit directly, do the calculation of the limit, and I see that 0.999… = 1.
Beth: I use the definition of limit to prove that for an increasing sequence the limit is the LUB, do the calculation of the LUB, and I see that 0.999… = 1
Carrie: I use the definition of limit to prove it is okay to do addition and multiplication by a constant, then I use that to do the x = 0.999…, 10x = 9 + x proof, and I see that 0.999… = 1.
@The Cool Dude:
“What about the floor function?
Certainly floor(.9999…)=0.”
No. floor(0.9 + 0.09 + 0.009 + … ) = 1, because 0.999…=1.
@Anthony Rose:
“Interesting. It has always ‘offended’ 🙂 my mathematical sensibilities that two numbers, intuitively different by definition of what writing means…”
I think you don’t understand how writing numbers works. Any number can be written in infinitely ways. 2 = 4/2 = 8/4 = … The writing doesn’t make two numbers different, so the writing isn’t a definition.
“(since 0.999… is always less than 1.000… if you stop writing 9s and 0s at any limit of decimal digits)…”
If you stop writing the 9s at some point, then the resulting number you’ve written is NOT 0.999…, but rather some other number. However, IF the number of 9s is infinite, as in 0.999…, THEN this number is NOT less than 1. By definition, it cannot be, because such a number represents an infinite sum, and we know how to evaluate infinite sums, and the sum evaluates to 1.
“But after reading this today it has just dawned on me that at some point the difference becomes vanishingly small in *reality* – literally vanishingly.”
Not just vanishingly small, but exactly 0. There is NO difference.
“For example, if you cut a cake in half, and have 0.4999… of it on one side, then (assuming no crumbs adhere to the knife) you have actually cut it perfectly in half to the very last molecule, which is way bigger than the difference between 0.4999… and 0.5.”
No, because 0.4999… = 0.5, so 0.5 – 0.4999… = 0. Not just in practice, but also in theory. They are the same number.
“So in practice 0.4999… is really the same thing as 0.5 because you can’t, in practice, cut real things small enough to see the difference.”
The difference is 0, so it cannot be seen in practice or in theory.
“In fact, because the number is recurring, it can always stay ahead of you, even if you went sub-atomic.”
No. Numbers aren’t recurring or non-recurring. This is nonsense. Numbers don’t “move’, they aren’t living entities or machines that do processes. No, numbers just are. 4 is just that: a number with a value of 4. It isn’t recurring or non-recurring. 0.999… is a number which is defined to be an infinite sum, and the reason it is defined this way is because of how the base-10 system is created. When we evaluate the sum using calculus, we obtain that this sum is equal to 1. It always is. So the two numbers are equal. There is no recurring or non-recurring anything here.
“Maybe this is a dumb layman’s way of viewing things – probably is.”
It is, but I’m glad you said it so you could be corrected.
“I just thought I’d share my light-bulb moment anyway because if it helps me stay sane, maybe it’ll help someone else.”
Mathematics isn’t about sanity. Mathematics and sanity don’t go well together.
@Brendan Hayden:
“If I find a number x = 0.99999999….. so that it is one planck length away from 1 on the number line can you find a number that is closer?”
Yes. Let us represent the Planck length by L. If x is L away from one, then by definition of what distance is in the real number line, x = 1 – L. We know L equals in the decimal representation, and it would not be equal to 0.999….
“…remembering that half a Planck length is not something that makes sense.”
It makes sense in mathematics, because the Planck length is a restrain on the physical world, not on numbers.
“Sure you can do it mathematically (the above mentioned (1+x)/2 ) but that is not a valid length in this scenario.) So in any real world scenario a number closest to 1 does exist…”
But not in math.
“…and if maths doesn’t have to relate to real word scenarios we may as well study unicorn poetry…”
No. Math is the reason physics exist, unicorn poetry on the other hand is not a useful tool to acquire knowledge about the real world.
@Waytfm:
“Now, if you were fiddling around in Non-Standard Analysis, where you have number systems that can deal with non-zero infinitesimals. You might could define a function from that number system to the integers such that floor[0.999…] = 0.”
Even in the hyper-real number system, the base field for non-standard analysis, and in its extension, the surreal numbers, 0.999… = 1. This is because the infinite sum does not involve infinitesimals whatsoever, since it is a sum which exists in the real numbers. 0.999… is NOT the same as 1 – epsilon, because 1 – epsilon cannot be represented as an infinite sum of real numbers. epsilon by definition cannot be represented as a combination of elementary operations performed on any number of real numbers.
@Stan:
“From the point of view of the pure theory o.99999… is not 1.”
There is no such a thing as pure theory. You’re talking nonsense.
“But there is a problem. In reality infinitesimals do not exist. In reality the nines in 0.99999… have to end.”
No, not necessarily. We can add an infinite amount of numbers. The infinite sequence (0.9, 0.09, 0.009, …) is a sequence in which we can add every single element together, even if the sequence is infinite. We can add them and we will obtain 1 every time. If we only add finitely many of these numbers in the sequence, then the number we obtain is less than 1, but if we add ALL the elements, the sum is exactly 1.
“Things in reality could be ever changing but they are defined at every single moment.”
This is irrelevant. A number isn’t a changing thing which evolves over time.
“Rational periodic numbers are periodic only because of the number system we use.”
This is true, but their values do not change. 0.999… would still be equal to 0.999… in base 12.
“We can always find another number system where rational numbers will not be periodic. 1/3 is not 0.3333333… It is more. There will always be residual of 1 which is the cause for the ever repeating threes.”
There only is a residue of 1 IF the process of division is finite, but because the number 0.333… stands for a decimal number for which the division has been completed, there is no residue.
“I am not sure if distinction between 1 and 0.999999… is needed in math. For now – seems not. But in theory there is a difference – purely abstract, because all manifested things are finite.”
No, there is no difference in any given theory, not even in non-standard analysis. It is also false that all manifested things are finite. The size of the universe is maybe infinite. The density inside a black hole is presumably infinite. The range of the gravitational effect on objects is infinite.
“As I remember the real numbers include sq., cube, forth … n-th roots of numbers. N-th root (n->infinity) of any positive number is 1, if we accept that 1/inf=0 (or that 0.99999…. and 1.0000….1 = 1). Now let’s just try to invert the calculations. It’s impossible. Example: inf.-th root of 0.5 = 1. But 1 to the power of infinity is 1 and not 0.5.”
1 to the power of infinity is NOT 1. We can prove this in multiple ways, using calculus or algebra. 1 to the infinite power is actually indeterminate, this is to say, equal to every number simultaneously. Part of the reason for this is because any number to the null power is 1, so inverting this process must yield every number at once.
“So there are infinitesimals not equal to zero even in the real numbers.”
No, there are not. You just failed to understand a simple limit of calculus I.
“For me the only reason o.99999… to be equal to one is because it is more convenient practically.”
No, the reason is because it is true.
@The Cool Dude:
“In the most simple terms possible, the abstraction we’re speaking of is equal to
lim(n→∞) of the sum(9*10^-i) from i=1→n.
Basically, .9+.09+.009+.0009, forever.
Using calculus and geometric series, you can easily show that the limit is 1, the limit being the thing it never reaches but gets infinitely close to.”
Yes, because the sum of finite numbers by definition is never infinite. However, if we consider the full infinite sum, then by definition it equals the limit to infinity.
“The difference can also be shown by using the floor function, such that, if you take the floor of the limit of the sum, it’s 1, but if you take the limit of the floor of the sum, it’s always zero.”
Actually, the floor of the sum is only 0 in the situation that N is finite. Also, your conclusion is obvious. The floor function is NOT continuous, so you cannot commute the limit and the floor function. This is to say, whenever the limit as x approaches any integer m of the floor function is undefined, since the limit on both sides is different. However, you can take the floor of a limit. There is no difference between the two numbers, you’re simply performing and invalid manipulation by wrongly assuming you can switch the order of the limit and the floor function. You can only switch the order is the limit exists & if the function is continuous. Neither of the above is true in this case.
“Hence why it isn’t a matter of definition, it’s a matter of thinking you can reverse the order of the limit and some other function.”
Yes and no. It is still a matter of definition. For any infinite sequence (a(1), a(2), a(3), …), the sum of every element in the sequence is the limit of the finite sum of N terms as N becomes infinite. This is because the finite sum of N terms is closer to being infinite, but it never reaches the limit. However, the infinite sum is has to be limit by inspection.
“@Stan
The conundrum you’re considering is a result of the fact that while lim(n→∞) of c^(1/n) is always 1, lim(n→∞) of 1^n is also always 1,…”
Not true. 1^infinite number is always indeterminate, and this is something that every high school student learns in calculus I.
“…and it comes from the order of the limits. If you operate one limit after another, those two statements will always be true, but if you do them at the same time, the result could be anything.”
Exactly.
“Since 1=n^(1/∞), if you raise 1^∞, it’s the same as n^(∞/∞). Since ∞/∞ is indeterminate, it produces a universal set, thus 1^∞ is technically anything and everything, so long as you do both limits at the same time.”
Yes, although your explanation is sloppy.
“That is, lim(n→∞) of c^(n/f(n)), such that lim(n→∞) of f(n) is infinite as well. The result will depend on how much faster or slower f(n) raises to infinity in comparison to ‘n’.
In Calculus, there are various cases in which something like this happens, and it’s part of solving limits. It usually happens whenever you can show that the solution to a function at a certain point must, in some variation, contain the entity 0/0. Other common variations of this entity that show that a limit can’t be directly evaluated include 0^0, 1^∞, ln(0)*0, ∞*ln(1), (1/0)^(1/0), ln(∞)*0.”
Correct.
@Stan:
“Even if infinity is irrational there still in theory has to exist a base unit, if the existence of the irrational number (composed of base units) is a fact.”
You’re not making sense. Infinite numbers are not expressible in terms of bases. Whether or not a number is an integer is not a matter of bases either. You’re using straw-man arguments here.
“But even if infinity is not an integer there are at maximum 2 options: 1)1^inf = 1 or 2) 1^inf =+/-1. If we work with complex numbers we will have n outcomes for 1^(1/n), but for real numbers they are one or two, depending on is n odd or even. If n is irrational we’ll not know if -1 is possible outcome, but still we’ll know that 1 is for sure. But main thing is that result other of +/-1 is impossible, and -1 is possible only if infinity is not an integer, irrational in the part after the decimal point and under the condition it is not compound of any kind of base units. Sad thing is we can’t have all that information and will never know is -1 is a possible outcome. But this is not important. The important thing is 1^inf can never be let’s say 0.4. Which mean that 0.4^(1/inf) can never be 1. And that means there is a difference between one and infinitely close to one.”
Stan, this has nothing to do whether a number is infinitely close to 1 or not. 1^ Inf is indeterminate. Point blank. End of that conversation. 0.4 is a decimal much less than 1 here, but in the complex numbers, it is impossible to talk about a number being greater than another number since the complex numbers are not an ordered set. This has nothing to do with infinitesimal values. There are no infinitesimals in the real numbers, and the reason why 1^Inf is indeterminate is completely unrelated to all of this. Also, 0.4^(1/Inf) = 1 because 1/Inf = 0 in limits.
“Actually the problem is not mathematical, but here is not the place for other analyses.”
Yes, the problem IS mathematical. Stan, numbers are inherently and nothing but mathematical. Infinity is nothing but mathematical.
“Whether we’ll have infinitely repeating nines or infinetely repeating 1*1 – where is the difference?”
The difference is that in the first case, the sequence of numbers (0.9, 0.09, 0.009, …) is infinite, and you perform an addition over all these numbers to obtain 1. In the second case, you have an infinite sequence (1, 1, 1, …), and you perform a multiplication over all of them.
“It is truth that we have to be very careful when operating with infinity. For example:
(1-1/inf)*10 – (1-1/inf) = 10-1/inf – 1 + 1/inf = 9
but we can have
(1-1/inf)*10 – (1-1/inf) = (1-1/inf)(10-1)=9-1/inf”.
No, Stan. (1 – 1/N)*10 – (1 -1/N) = (1 – 1/N)*9 = 9 – 9/N = 9. 9/N is equal to 0 when N is infinite.
“Yes infinity is not a real number. But what’s above is the well known proof:
x=0.999999…
10*x=9.99999….
10*x-x=9.999…. – 0.9999…
9*x=9
->0.9999… = 1”
No, it isn’t, because you performed the first proof wrong.
“I just represented 0.9999… as 1-1/inf.”
Yes, and 1/inf = 0. Stan, the limit as x –> inf of c/x = 0, where c is any real number. Congratulations, you just proved 0.999… = 1.
“And my calculations above are not less correct than 9.9999…-0.9999…=9. (I’m not saying they are correct)”
Yet your calculations proved that 0.999… = 1.
Pingback: Q: In base ten 1=0.999…, but what about in other bases? What about in base 1? | Ask a Mathematician / Ask a Physicist
if 0.9999…=x, then 10x would be 9.9999… subtract 1x from 10x and you get 9, 9x=9, so 1x =1
Anthony.R.Brown says:
December 3, 2017 at 4:18 am
The Original author of the Proof below…used in this Thread 🙂
http://www.mathisfunforum.com/viewtopic.php?id=6069
It’s impossible for Infinite/Recurring 0.9 (0.999…) to ever equal 1
Because of the Infinite/Recurring 0.1 (0.001…) Difference! 🙂
Only by the using the + Calculation can it be possible (0.999…) + (0.001…) = 1
Anthony.R.Brown
Alan Feldman says:
December 3, 2017 at 8:55 am
@Anthony.R.Brown
Nope. The definition of the value of an infinite decimal is the lowest upper bound of the sequence. So if you have the sequence 0.9, 0.99, 0.999, . . ., the lowest upper bound is clearly 1.
Anthony.R.Brown : Reply…
0.9, 0.99, 0.999, . . ., the lowest upper bound is clearly 1. ???
Infinite/Recurring 0.9 (0.999…) has No! lowest upper bound of 1 ???
Because the 0.9 (0.999…)’s are Continuous otherwise there will be an Obvious Contradiction of the Term Infinite/Recurring
I don’t understand what you’re saying. The LUB is the smallest number that is greater than or equal to the numbers in the set.
Consider the series 0.999 . . . . It is the LUB of the series 0.9, 0.99, 0.999, . . . .
None of them are equal to or greater than 1. So the LUB clearly cannot be greater than 1.
The LUB cannot be less than 1, because there will always be some number in the series that is greater than it. If you choose 0.99, a greater number will be 0.991. So 0.99 cannot be the upper bound. This is true for all numbers less than 1.
Therefore, the LUB is 1. No number in the series is larger than 1, and for any number x that is less than 1 there is a number in the series that is greater x and less than 1.
The main argument is “by definition”. 0.999… = 1 by definition.
What does the definition say? That the infinite decimal is equal to its limit.
That’s the issue.
Does 0.9(infinite times) even exist as a number?
Counter arguments are “Find me a number between 0.999… and 1” assumes that “0.999…” is a number in the first place. IMO, 0.999…. is 9/10 + 9/100 + … which is not a number, but rather an endless summation of numbers.
Why is there a requirement to accept 0.999… as a representation of a number in the first place?
Who said that we must represent 9/10 + 9/100 + 9/1000 … as 0.999… and this is a representation of a number ? To me, it’s a representation of an infinite string (infinite summation), but there is no final sum, hence ‘infinite’. It’s not a representation of any number. If we use the limit, then L = 1. But then maths says 0.999… = L.
Does 9/10 + 9/100 + … = LIM (9/10 + 9/100 + … ) ?? That is what maths is telling us.
By using limits and 0.999… (and the partial sums), we know that there is an epsilon for every single partial sum, yes? L – every partial sum = an epsilon > 0.
Point-9ers will argue that 0.999… is not a partial sum and that the 9s are “all there”.
I counter argue that the epsilons are “all there” for every 9. And that they are all > 0.
In a real scenario, you have to travel 1 unit, by traveling in 90% increments of the remaining distance. You never get to the end. There IS no end. But you will know where you WON’T get, and that is 1.
LIM [SUM (n=1 to N) 9/10ⁿ ] as N→∞ = 1
SUM (n=1 to ∞) 9/10ⁿ = 0.999…
And if we read the above, we are literally saying: “n = 1 to n = infinity”, however, n cannot “equal” infinity.
Which is exactly the same issue here:
0.9(n-times) and then we jump to: 0.9(infinite times) and we are happy calling it a number!
SUM (n=1 to ∞) 9/10ⁿ = 9/10 + 9/100 + … = 0.999…
and
LIM [SUM (n=1 to N) 9/10ⁿ ] as N→∞ = 1
They are not the same thing, even though maths define them to be.
It “tends to” the limit.
As for the “algebra trick”, someone else already pointed out the problems with it. You pull a 9 from infinity-land, and still have the “same amount” of 9’s left in the bag, and then do finite math operations on it.
A more rigorous way is this:
x = 0.999… = SUM (n=1 to ∞) 9/10ⁿ
10x = 9.999… = 10 times SUM (n=1 to ∞) 9/10ⁿ = SUM (n=1 to ∞) 90/10ⁿ
10x – x = ?
Who said we can just take infinite decimals and subtract off infinite digits?
Subtraction for finite decimals starts from the right, which can never get to.
For infinite series subtractions, we do this (index by index subtraction):
10x = SUM (n=1 to ∞) 90/10ⁿ
x = SUM (n=1 to ∞) 9/10ⁿ
10x – x = [SUM (n=1 to ∞) 90/10ⁿ] – [SUM (n=1 to ∞) 9/10ⁿ]
10x – x = SUM (n=1 to ∞) [90/10ⁿ – 9/10ⁿ] = SUM (n=1 to ∞) (90 – 9)/10ⁿ
10x – x = SUM (n=1 to ∞) 81/10ⁿ
9x = 8.1 + 0.81 + … = ?
Partial sums are {8.1, 8.91, 8.991, 8.9991, … , 8.9(n-1 times)1, …}
can we even write that exactly? 8.999…991 ??
Some will say “the 1 never comes” but actually it is there for every partial sum.
So it’s 8.999… ?
Then we have 9x = 8.999…
9x / 9 = 8.999… / 9
x = [ SUM (n=1 to ∞) 81/10ⁿ ] / 9
x = SUM (n=1 to ∞) [81/10ⁿ x 1/9]
x = SUM (n=1 to ∞) 9/10ⁿ
x = 0.999….
So when the ‘algebra’ or ‘digit manipulation’ is actually done correctly, you start with x = 0.999… and end up with the same. As you should! Of course, this assumes “0.999…” is a series, and we treat is as such, and do SERIES operations on it.
The absolute best way to understand that Infinite/Recurring 0.9 (0.999…) is 1
Is by simply understanding that (1) is an Integer value…
And (0.999…) is Not!
A.R.B 🙂
Mr. Brown,
Do you consider 0.333… to be 1/3?
The … is defined to be the LUB. The lowest upper bound of 0.999… is 1. That’s what the three dots mean. You are using a meaning that is in contradiction with that. I don’t know what more to say?
Well, I can say this:
0.9, 0.99, 0.999, . . ., the lowest upper bound is clearly 1. ???
Yes. Obviously! What is the limit of that sequence. It’s 1. What is 0.999 . . . defined to be? The limit of that sequence, which is also 1, being that it’s the same thing. You are implicitly assuming some different definition of “. . .” without acknowledging that, and then prove my, and mathematicians, definition, wrong. You can’t prove a definition wrong! You can say it’s inconsistent with yours, but mathematicians use mine. Or more accurately, I use theirs!
Since there are no reply links, I will now address Mathman, who wrote:
“The main argument is “by definition”. 0.999… = 1 by definition.
What does the definition say? That the infinite decimal is equal to its limit.
That’s the issue.
Does 0.9(infinite times) even exist as a number?”
You just proved my point. The definition of “. . .” is the LUB of the sequence. Then you assume a different definition, and argue that mine, which is the accepted one in math circles, is wrong. IOW, you’re trying to prove my definition wrong by implicitly using your own different definition. Do you believe 0.333… = 1/3? If you say yes, the logic forces you to accept 0.999 . . . = 1. Question: What is the limit of 0.9, 0.99, 0.999, . . . . It’s 1. Now, “. . .” is DEFINED by mathematicians to be the LUB of the sequence it implies, which in this case is 1. YES, it is DEFINED that way.
Ask a M and P: Can you please fix this site so that it has reply links? The only reason I can reply is that an email directed me here. Or is this how it’s supposed to work?
Mr. Brown, 0.999 is DEFINED to be an integer, mainly because the integer 1 _is_ the limit of the sequence.
Mr. Brown wrote:
“It’s impossible for Infinite/Recurring 0.9 (0.999…) to ever equal 1”
It is true that no element of the infinite set {0.9, 0.99, 0.999, . . .} equals 1, but 0.999 . . . is defined to be the limit of the sequence 0.9. 0.99, 0.999, . . . . If you argue that the definition is wrong, it means that you are using your own different definition. You insist that it must be a member of that set, but it is not. Do you think 0.333 . . . is 1/3? If so, than by the same token, 0.999 . . . equals 1.
Mathman writes:
“The main argument is “by definition”. 0.999… = 1 by definition.
What does the definition say? That the infinite decimal is equal to its limit.
That’s the issue.
Does 0.9(infinite times) even exist as a number?”
Yes, by definition. That’s what definitions are!
You’re implicitly assuming a different definition for 0.999 . . . than what mathematicians use. “0.9(infinite times)” is not how 0.999 . . . is defined. It’s defined as the limit of the sequence (or LUB, as it says in my analysis textbook). Check your analysis textbook.
Looks like my post from earlier today didn’t get posted.
Why do you say it’s not when it is defined to be exactly that? You are simply repeating your mistake of making up your own definition and claiming it disproves the conventional definition. Are you bothering to read what I’m writing? You’re simply repeating the same incorrect thing without explaining what’s wrong with the conventional definition.
BTW, do you believe 0.333… = 1/3 ?
Something Strange happened with my First post above the very important is missing ??? So I post again!
The absolute best way to understand that Infinite/Recurring 0.9 (0.999…) is 1
Is by simply understanding that (1) is an Integer value…
And (0.999…) is Not!
A.R.B 🙂
@Mathman
“To ‘solve’ the infinite 9 issue in 0.999…, we take the LIMIT, and we end of up with 1. It’s the same reason something like 1/10^∞ is poor symbology. However, the limit 1/10ⁿ as n→∞ = 0. This does not mean 1/10^∞ = 0.”
You are correct: it does not mean 1/10^ ∞ =
0 because the expression is complete nonsense, because ∞ is not a number and it is not a quantity you perform arithmetic with.
“Then how in the hell was I doing engineering calculus in high school? sheeesh.”
The same way most students in high school were.
“You and your condescending attitude again.”
Me pointing out you don’t understand something isn’t condescending when I’m able to just prove it.
“I understand the definitions fine.”
No, you do not. You fail to even understand the basic construction of numbers and what constitutes a number. You misuse terminology and conflate it, such as interchanging “infinite sum” and “sequence”. And the fact that you try to argue against DEFINITIONS and not mere premises only further strengthens my point.
“The only failure of understanding here is you seeing my point.”
I see your point clearly, and it is absolutely unfounded and incorrect.
“A few of you all have that same attitude. “If you don’t accept everything our way, you are wrong, and you don’t understand anything!!”. It’s pretty sad.”
Indeed, if you don’t accept facts and definitions, then you are indeed wrong. I’m even tempted to say that is almost equivalent to the definition of being wrong. This is not about “our way” of doing things. Maths are objective: they are not based on personal ways of doing things. There is one way to do things in math, and the way is laid out by the rules and definitions. You can reject these rules and definitions, which constitutes rejecting math altogether, which is what you are doing, or you can not do that. But there is no “your way” or “my way” to do it. I hate it to break it to you, but math could care less about your feelings and your way of wanting to do things. Math isn’t an art.
“Once again, I understand what YOU are saying about the definitions and concepts just fine.”
Then prove it by not butchering and misusing the terminology. I’m really not asking for much here.
“I understand what you want to me understand just fine. You just want me to “shut up and just do it how we tell you to”.”
Feeling oppressed there? What a scapegoat.
”So according to you, being incorrect that 0.999… not equaling anything, and the limit 0f 0.999… equaling 1, has what consequence?”
If you want to declare 0.999… not a number, then we must also declare that 9 is not a number, and I already outlined the inference rule that allows me to conclude this in my previous comment. 9 is a sequence, according to your theory, which I will call pseudo-mathematics.
“I asked if I was attacking you, and you didn’t answer. You are STILL attacking me.”
Telling you are wrong is not attacking you. You can continue to believe otherwise, but you would continue to be wrong in that regard then too. Also, red herring.
“You changed what I said. Because I didn’t say that.
The “infinite sum” (or string of digits), 0.999… forms the partials sums sequence {0.9, 0.99, 0.999, …}”
A string of digits cannot form a sequence, let alone some partial sum sequence. I already discussed this. Do you understand what strings of digits are?
“The LIMIT of the sequence is equal to the LIMIT of the infinite sum, both being 1. 1 = 1.”
No, the phrase “limit of the infinite sum” is an oxymoron and demonstrates a misusage if the word “limit” (once again). The infinite sum is the limit of the sequence of partial sums. That is the definition. Saying it is anything else is wrong by definition, and that is objective, because math is objective, because does not depend on what I believe or you believe, and it does not care about personal feelings.
“You haven’t. If you had, I wouldn’t be responding with all this right now.”
No, the real reason you are responding right now is because you think your high school engineering trumps a full math degree and you refuse to accept basic mathematics or any kind of definition that exists in the field. I proved my statement, and I can prove that I did. If you insert my claims into a proof-evaluating software, then I can bet you big money it will confirm what I stated. It is not the first time I have this conversation with someone, after all.
“The LIMIT of the infinite sum 0.999… = 1.”
Once again demonstrating you don’t know what a limit is. Try again next time using the actual definition.
“1 – SUM 9/10ⁿ (n→∞)”
Yes, the very expression you just wrote there is what mathematicians and engineers and physicists and economists and high school teachers call 0. And by 0, I mean zero, the number that was discovered by Indians a very long time ago to represent the additive identity. Yeah, that one.
“Ok, if 1 and 0.999… are both digital representations of the same number ‘one’, then what are the 2 digital representations of the result 1 – 0.999… ?”
There are more than 2. There are infinitely many. This is true for all rational numbers. One has 1.0, 1.00000….., 01.0, 0.9…, etc. And for 1 – 0.9…, one has 0, 0.0, 00.0, 0.000……, etc. They are all the same number. They all represent 1 – 1.
“There is no possible summation which gives 0.000…1
1 – SUM 9/10ⁿ (n→∞)”
There is no possible summation which gives 0.0…1 because you cannot have something after an infinite amount of something else while not belonging to the original infinity. Because lim n —> ∞ n = lim n —> ∞ (n + 1). Or, if you want it in cardinal numbers, Aleph-0 + 1 = Aleph-0. The expression you wrote is just 0.
“There are ‘infinite’ 9’s, thus, there is never a sum in the first place. There is only a limit. There are partial sums, because you have a sum; it’s done. You can’t have infinite and sum for every case.”
I already responded to this.
“Maybe zero. And what the hell is “infinitely many” ?? How many is that exactly?”
Pretending to be okay with “infinite” operands and not “infinitely many” operands is quite petty. But in any case, the obvious answer is Aleph-0, since we are dealing with sequences: you cannot have such a thing as an uncountable sequence for the same reason you cannot have a dark photon.
”Wrong. There are many ways to calculate the LIMIT of the infinite sum, or the convergence of the sequence of partial sums. I have no issue with either of those. I have an issue with saying “zero decimal infinite 9’s is equal to the limit of the infinite sum of them” The “having infinite 9’s” is just nonsense. You can’t have that to even start with.”
I asked for a proof of this. I’m still waiting on it. You’ve done a lot of talk, but a whole “nada” of proving your claims.
“We can have arbitrarily large amount of 9’s if you like. But this argument about “having all infinite 9’s there at once” is just assertion.”
Okay, but you have yet to prove it. Can you find any axiom that shows you cannot have this? Can I not have Aleph-0 of something in an object? Because if such is the case, then this implies “You cannot have Aleph-0 of the natural numbers in a set” which means you actually cannot have such a thing as the set of natural numbers. And with that, you’re rejecting all notions of infinity altogether, since you cannot have the concept of infinity and simultaneously reject the set of natural numbers. So tell me, why can I not have Aleph-0 digits in a number, why can I not have Aleph-0 of anything in a particular something? I’ve been waiting for your glorious proof, so please give me one. In fact, just ignore everything else that has been said in the conversation so far and instead just give me this proof. That is literally the only thing you need to do to end this conversation.
“What does the definition say? That the infinite decimal is equal to its limit.”
No, the definition says that infinite decimal is the limit [of some sequence].
“That’s the issue.
Does 0.9(infinite times) even exist as a number?”
Can you prove it is not a number?
“Counter arguments are “Find me a number between 0.999… and 1” assumes that “0.999…” is a number in the first place. IMO, 0.999…. is 9/10 + 9/100 + … which is not a number, but rather an endless summation of numbers.
Why is there a requirement to accept 0.999… as a representation of a number in the first place?”
I already explained in a previous comment. Unless you can explain why there cannot be infinitely many operands of addition (Hint: there can be. There must be. That’s pretty much the foundation of set theory. And rejecting it means rejecting any concept of the infinite, which also means rejecting the existence of the set of all natural numbers as we know it), 0.9… is as much of a number as any other number. I’m assuming you also reject the existence of π and e? For the same reasons.
“Who said that we must represent 9/10 + 9/100 + 9/1000 … as 0.999… and this is a representation of a number ?”
Do you understand how decimal digit strings work? Because if you do, then you already have your answer.
“To me, it’s a representation of an infinite string (infinite summation), but there is no final sum, hence ‘infinite’.”
Infinite only makes a reference to the number of summands in the expression, it tells you nothing about the ability to complete or evaluate the sum.
“It’s not a representation of any number. If we use the limit, then L = 1. But then maths says 0.999… = L”
Here are your choices:
1. Infinity does not exist [Corollary: the set of all natural numbers does not exist] [Corollary 2: irrational numbers are not real either. Poor circles: they now have no ratio between the circumference and the diameter. Oh boy]
2. We cannot with complete generality represent numbers as digital strings, and digital strings are not numbers [even though decimal notation is literally defined and constructed to just be an addition abbreviated]
3. 9 is not a number
4. Prove that somehow all the above are false AND YET 0.(9) still cannot be a number because of some weird unknown reason that apparently no mathematician in the universe is aware of (and if you believe in God, this would have to include God too).
“By using limits and 0.999… (and the partial sums), we know that there is an epsilon for every single partial sum, yes? L – every partial sum = an epsilon > 0.”
You are not quoting the definition of the limit correctly here, so no, there is not “an ε for every partial sum”. The definition is far more complicated than that. You are missing a δ, several quantifiers, and some inequalities with absolute values. And you have to match each quantity with the others correctly.
“Point-9ers will argue that 0.999… is not a partial sum and that the 9s are “all there”.
I counter argue that the epsilons are “all there” for every 9. And that they are all > 0.”
Well, no, because that is not what a limit is. But also, it is completely pointless on your part to try to use the limit to justify your argument when you presumably don’t even believe in infinities.
“In a real scenario, you have to travel 1 unit, by traveling in 90% increments of the remaining distance. You never get to the end. There IS no end. But you will know where you WON’T get, and that is 1.”
Yeah, math doesn’t care about “real scenarios”. This is not physics. I love physics too, but math is not physics. They are very different things. In a real scenario, perfect polygons don’t exist. You also cannot have actual derivatives or continuous functions in real life. Why? Because atoms and quanta. They take away continuity. You could never draw a perfect circle. So do we say circles don’t exist then? Well if you want to do that, you go ahead. But we’re talking math here in this article, so I’m going to go with my math education, not my physics research.
“LIM [SUM (n=1 to N) 9/10ⁿ ] as N→∞ = 1
SUM (n=1 to ∞) 9/10ⁿ = 0.999…”
The two things are equal. Again, that is how it is defined. If you want to reject the definition, then once again, you can do that, but at that point we wouldn’t even be talking math anymore. And rejecting the definition would mean your usage of the word infinity is completely different from how everyone else uses it, or just rejecting infinity altogether. Or perhaps you are defining addition differently than everyone else.
“And if we read the above, we are literally saying: “n = 1 to n = infinity”, however, n cannot “equal” infinity.”
It need not be. SUM n = 0 to n = Infinity is a notational shortcut for lim N —> Infinity SUM n = 0 to n = N.
“Which is exactly the same issue here:
0.9(n-times) and then we jump to: 0.9(infinite times) and we are happy calling it a number!”
Okay, so you reject infinity. Well then. You could have said this from the very beginning and saved me what probably has been over three hours worth of total writing since our original conversation started. But no. Of course you made it more difficult than that.
“They are not the same thing, even though maths define them to be.”
Okay then, so you literally are confessing to the fact [which I had known all along] that you are rejecting the current state of affairs of mathematics; abandoning math as it exists, in short. Alright then. I don’t need to respond to anything else because you literally just confirmed my point. You don’t agree with mathematics. You’re not trying to make a mathematically correct statement, you actually just think math is wrong. I wish you could have admitted this from the beginning, but you eventually did. This conversation need not continue.
“As for the “algebra trick”, someone else already pointed out the problems with it.”
And I already took the time to debunk it.
“A more rigorous way is this:
x = 0.999… = SUM (n=1 to ∞) 9/10ⁿ
10x = 9.999… = 10 times SUM (n=1 to ∞) 9/10ⁿ = SUM (n=1 to ∞) 90/10ⁿ
10x – x = ?”
Doing it rigorously does not change the result than if you perform the subtraction algorithm digit by digit.
“Who said we can just take infinite decimals and subtract off infinite digits?”
Well, does it matter? You reject infinity and many axioms of most theories of math anyway.
“Subtraction for finite decimals starts from the right, which can never get to.”
Okay, no, that is just a helpful algorithm for doing the calculation by hand, but the operation itself does not depend on the algorithm chosen. As long as the algorithm is valid, it does not matter how you subtract the digits or in what order.
“For infinite series subtractions, we do this (index by index subtraction):
10x = SUM (n=1 to ∞) 90/10ⁿ
x = SUM (n=1 to ∞) 9/10ⁿ
10x – x = [SUM (n=1 to ∞) 90/10ⁿ] – [SUM (n=1 to ∞) 9/10ⁿ]
10x – x = SUM (n=1 to ∞) [90/10ⁿ – 9/10ⁿ] = SUM (n=1 to ∞) (90 – 9)/10ⁿ
10x – x = SUM (n=1 to ∞) 81/10ⁿ
9x = 8.1 + 0.81 + … = ?
Partial sums are {8.1, 8.91, 8.991, 8.9991, … , 8.9(n-1 times)1, …}
can we even write that exactly? 8.999…991 ??
Some will say “the 1 never comes” but actually it is there for every partial sum.”
We care about the limit of the partial sums, not the what is in the partial sums.
“So it’s 8.999… ?
Then we have 9x = 8.999…
9x / 9 = 8.999… / 9
x = [ SUM (n=1 to ∞) 81/10ⁿ ] / 9
x = SUM (n=1 to ∞) [81/10ⁿ x 1/9]
x = SUM (n=1 to ∞) 9/10ⁿ
x = 0.999….”
All yo did was reverse the operations. This hand-waviness can be easily debunked in several ways. 1) consider x = 0.333… = 1/3. 2) Use the fact that 9/10 + 9/100 + 9/1000 +
••• is a geometric sum to evaluate it.
“So when the ‘algebra’ or ‘digit manipulation’ is actually done correctly, you start with x = 0.999… and end up with the same. As you should! Of course, this assumes “0.999…” is a series, and we treat is as such, and do SERIES operations on it”
Series are numbers. Period.
Try again because for some Strange reason this site does not accept Math symbols ?
The absolute best way to understand that Infinite/Recurring 0.9 (0.999…) is Not! equal to 1
Is by simply understanding that (1) is an Integer value…
And (0.999…) is Not!
Alan E. Feldman says:
December 21, 2018 at 11:31 pm
Mr. Brown,
Do you consider 0.333… to be 1/3?
No! Because 0.333… is a Decimal Number it is not 100% a Third!
Whereas 1/3 is a Described Third! without Actually showing the Number!
Basically it’s Impossible to show exactly a Third of something in a Number format,
Just the same as 0.999… can Never be shown to be exactly equal to 1
A.R.B 🙂
@Mr. Brown, who writes:
“Something Strange happened with my First post above the very important is missing ??? So I post again!
The absolute best way to understand that Infinite/Recurring 0.9 (0.999…) is 1
Is by simply understanding that (1) is an Integer value…
And (0.999…) is Not!
A.R.B 🙂”
0.999… is DEFINED to be 1. The … is defined to mean the limit of the sequence. You can’t prove a definition wrong. For the zillionth time, you are using a different definition, and are trying to use it disprove the definition that mathematicians actually use. But you can’t disprove a definition!
“Is by simply understanding that (1) is an Integer value…
And (0.999…) is Not!”
Not much of a “proof” there; if it proves anything — and pardon me for being blunt here — it’s that you don’t understand what a definition is.
Wait a minute. Didn’t you just contradict yourself? You say it’s 1, and then you say it’s not 1. Well, which is it?
Alan E. Feldman says:
December 22, 2018 at 2:32 pm
Wait a minute. Didn’t I just contradict Myself? I say it’s 1, and then I say it’s not 1. Well, which is it?
A.R.B 🙂
The above shows exactly how a Math type person ? can contradict Him/Her/ Self!
Whereas I have no such contradiction…
Infinite/Recurring 0.9 (0.999…) is Not! equal to 1
1 is Not! equal to Infinite/Recurring 0.9 (0.999…)
And to Prove it…
try the Calculation Integer (1) = 1
and the Calculation Integer (0.999…) = 1
Wait a moment A.R.B must be wrong because they both equal 1 ? 🙂
OK! Now we just proved that the only way Infinite/Recurring 0.9 (0.999…) can ever = 1 is by rounding up the never ending .9’s by using the Integer Function! 🙂
And we also just proved how easy it is to catch out a Math type person by letting them do the Calculation themselves! and falling into the {Famous Infinite/Recurring 0.9 (0.999…) Integer Trick Trap} 🙂
A.R.B 🙂
@ARB,
I’ve explained the problem clearly several times, that you’re trying to use your own definition (which you never give) to disprove the real one, and you have yet to address that. You simply keep repeating the same nonsense.
You clearly don’t know what you’re talking about, or you’re a troll, or both.
Bye! (^_^)
The Original author of the Proof below…used in this Thread
http://www.mathisfunforum.com/viewtopic.php?id=6069
It’s impossible for Infinite/Recurring 0.9 (0.999…) to ever equal 1
Because of the Infinite/Recurring 0.1 (0.001…) Difference!
Only by the using the + Calculation can it be possible (0.999…) + (0.001…) = 1
Anthony.R.Brown
@Angel Mendez Rivera
>It is also false that all manifested things are finite. The size of the universe is maybe infinite. The density inside a black hole is presumably infinite. The range of the gravitational effect on objects is infinite.
None of those things are manifested.
This has nothing to do with the tautology of 0.9…/1.0…, of course, but it’s also a tautology that nothing infinite can be manifest.
@Anthony.R.Brown
>Only by the using the + Calculation can it be possible (0.999…) + (0.001…) = 1
This doesn’t work because the decimal extension isn’t 0.001…, it’s 0.000…[1].
If you never get to the last 9, you also never get to the [1]. So the difference is always 0.000…
@Anthony.R.Brown
>Only by the using the + Calculation can it be possible (0.999…) + (0.001…) = 1
This doesn’t work because the decimal extension isn’t 0.001…, it’s 0.000…[1].
If you never get to the last 9, you also never get to the [1]. So the difference is always 0.000…
Your Biggest mistake is that you are trying to Isolate the 1 by putting it in brackets [1] ?
When in fact the Infinite Recurring .1 runs along side the Infinite Recurring .9
Where the .1 is the Infinite Recurring Difference! as shown below…
0.999…
0.001…
No matter how many Decimal places are Calculated the Calculation will always only equal 1 by adding the two values together as shown below…
0.999… + 0.001… = 1
0.999… Cannot equal or become 1 in some mysterious magical way on it’s own without the above Calculation! 🙂
Alan Feldman says:
December 3, 2017 at 8:55 am
@Anthony.R.Brown
Nope. The definition of the value of an infinite decimal is the lowest upper bound of the sequence. So if you have the sequence 0.9, 0.99, 0.999, . . ., the lowest upper bound is clearly 1.
You clearly have it wrong again! 🙁
Your Quote: “So if you have the sequence 0.9, 0.99, 0.999, . . ., the lowest upper bound is clearly 1.”
It’s impossible for 0.999… to have an upper bound because it’s not bound by anything!
The .9’s are not Restricted in anyway…they NEVER! END!!
Only by the using the + Calculation can it be possible (0.999…) + (0.001…) = 1
Anthony.R.Brown
I do not even see what that means. The sequence 0.9, 0.99, 0.999, . . . Will! Always be < 1 🙂
If Not ? 0.999… is a Contradiction! Because if 0.999… ever reaches ? or equals 1 then it will not be or equal to Infinite/Recurring 🙂