Why can't mathematicians, after 2000 years of mathing, find out if 0.999... = 1?
Because they're everungiving.
>>16871180Or otherwise*
[math]0.333..._7=0.5[/math]
Start at 0 on the number line and move 9/10 to the right, then 9/100 to the right, then 9/1000 to the right, and so on. Some numbers will eventually be to your left if you take enough steps. Some numbers will never be to your left no matter how many steps you take. 0.999... is by definition the number at the boundary, and that number is 1.
For those who want to know more:Consider the set of numbers {0.9, 0.99, 0.999, ...} that you can move to with enough steps.>Some numbers will never be to your left no matter how many steps you take.We call one of these numbers an upper bound on the set.>Some numbers will eventually be to your left if you take enough steps.These numbers fail to be upper bounds on the set.With some thought, you can see that the boundary between the upper bounds and the numbers which are not upper bounds must belong to the upper bounds. So this number is called the least upper bound, or supremum for short. We define 0.999... as the supremum of {0.9, 0.99, 0.999, ...}.
>>16871170Why after 2000 years are people still trolling with this boring shit?
Proof that 1 is the least upper bound of {0.9, 0.99, 0.999, ...}:At each step in the sequence, the number falls short of 1 by [math]\frac{1}{10}[/math] of the distance by which the previous number fell short of 1, so the numbers in the sequence will always be less than 1. Thus 1 is an upper bound. Now consider any number x < 1. Let n be the number of digits in [math]\left\lfloor \frac{1}{1-x} \right\rfloor[/math]. Then [math]\frac{1}{1-x} < 10^n[/math] and [math]x < 1 - \frac{1}{10^n} = 0.\underbrace{999 \ldots 9}_{n}[/math]. Since any number less than 1 is not an upper bound, 1 is the least upper bound.
Even if we restrict ourselves to the rational numbers, for any repeating decimal, there is a least upper bound on its truncations, which we can define to be the value of that decimal.First we show a rational number is an upper bound on all the truncations of a repeating decimal if and only if it is an upper bound on the subset of truncations which consist of the nonrepeating part followed by a whole number of repetends. The forward implication is trivial; for the backward case, for any truncation of the repeating decimal, append digits until you reach a decimal of the desired form, and since the extended decimal is less than or equal to the upper bound on the subset, so is the original decimal. If we can find a least upper bound on this subset, it is a least upper bound on the whole set.The truncations in the subset can be written in the form [math]A + \sum_{k=0}^{n-1} B \cdot 10^{-mk} = A + B \frac{1 - 10^{-mn}}{1 - 10^{-m}}[/math] where A and B are non-negative rational numbers and m is the number of digits in the repetend. The number [math]X = A + \frac{B}{1 - 10^{-m}}[/math] is an upper bound. Let y be a rational number smaller than X; to show y is not an upper bound, we need to find an n such that [math]A + B \frac{1 - 10^{-mn}}{1 - 10^{-m}} < y[/math]. This holds iff [math]B \frac{10^{-mn}}{1 - 10^{-m}} < X - y[/math]; if B = 0, it is true for every n. If B > 0, then it holds iff [math]10^{mn} > \frac{B}{(1 - 10^{-m}) (X - y)}[/math], so a sufficient n can be obtained by counting the digits in [math]\left\lfloor \frac{B}{(1 - 10^{-m}) (X - y)} \right\rfloor[/math], dividing by m, and rounding up. Thus X is the desired least upper bound.
>>16871170
>>16871170Why are you incapable of understanding that a number can have many different representations?One, 1, 2/2, 3 * 1/3, 0.999..., etc
>>16871242[math]0.999..._7=1.5[/math]
>>16871325>Muh upper bound
1 is the least upper bound of 0.999, it isn't equal to 0.999. That only happens if summation takes place. If you don't summarize the decimal expansion, then it does indeed go on forever and never equals 1. However, you can't prove something "goes on forever", that just means you've encountered a limit. You can prove the limit exists, so you can prove 0.999 = 1 if and only if you "take the sum" of the sequence in order to discover the limit exists. If you just have 0.999 and no additional steps like summation of the expansion, then it is just what it is, 0.999, not 0.9, not 0.99, nlt 0.999. Just 0.999. Anything else requires additional "steps" which involve a transformation of the notation to reveal the limit of the sequence.
>>16871378if I'm reading this brainrot right, then apparently 0.999 is not 0.999, just 0.999
>>16871967>then apparently 0.999 is not 0.999, just 0.999damn, that's heavy
>>16871353>the factorial of 0.999...=1yeah
>>16871967Deep
x = 0.999...10x = 9.999...10x-x = 99x = 9x = 1what am I missing here?
>>16871170Because 2000 is a bit less than an infinity.
>>16872372but how much less?, a grain of mustard?, 2?
>>16872393Like I said... a bit, you just toggle the bit that indicates whether the value is finite or infinite.
>>16872211>10x-x = 9but it's not...
>>16872455why
>>16871312Impossible. It's missing the last little ...01 at the end.
>>16872510impossible, there is no end
>>16872511No, it's infinitely many nines.So it looks like [math]0.\underbrace{999 \ldots 9}_{\infty}[/math]You need the last little bit to make it equal to 1.
>>16872515Just because you can write a trailing 9 like that does not mean that there is a last 9 to that number. You are fundamentally misunderstanding infinity.
>>16872510>>16872515Are you arguing that after [math]\infty[/math] steps, you should be [math]\frac{1}{10^\infty}[/math] short of 1? Because "where are you after [math]\infty[/math] steps?" is not the question. The question that defines 0.999... is "where is the boundary between numbers that you can put to the left of you by taking a sufficiently large but finite number of steps and numbers that will never be to the left of you no matter no matter how large a finite number of steps you take?"
>>16871170It's a part of the definitions to what a number isIt isn't actually about numbers
isn't it something like this?
>>168728810.999... means the smallest upper bound on the terminating decimals {0.9, 0.99, 0.999, ...}, which your picture correctly identifies as 1.
>>16872901its not 1. 0.999 starts with a 0 1 starts with a 1 so they cannot be the same number retard
>>16872910Numbers aren't just meaningless strings of digits. That string of digits, called a numeral, is a name for a point on the number line. There can be multiple ways to name the same point.
Let Sn = 1 + q + q2 + q3 + ... + q^{N-1}Multiply Sn by q: q×Sn = q + q2 + q3 + q4 + ... + q^NSubtract q×Sn from Sn: Sn - q×Sn = 1 + (q - q) + (q2 - q2) + (q3 - q3) + (q4 - q4) + ... (q^{N-1}-q^{N-1}) - q^NSn×(1 - q) = 1 - q^NSn = (1 - q^N)/(1 - q)When taking the limit as N -> infinity and if |q| < 1, we get S = 1/(1 - q)Subtract 1, S - 1 = q/(1 - q)Set q = 0.1 => S - 1 = 0.1/(1 - 0.1) = 1/9Multiply by 9, 9(S - 1) = 1Tada!
I tried to explain least upper bounds to a sixth grader today. I think I ended up tiring her out too much, especially when I tried to talk about how we know there isn't a smaller upper bound than 1/3, which was probably too ambitious. What I really needed in retrospect was a simple Desmos demonstration where you can adjust the candidate upper bound and see visually if it's working or not.One thing I've noticed with the middle schoolers is that it's not obvious to them that these sequences are bounded above. It makes sense; there are infinitely many steps, so the naive expectation is that you eventually pass any bound. If there's a super easy way to show these things are bounded above at a middle school level, it would be useful to know.
>>16871170if you mean = as in the algebraic sense of an equivalent class, then yes 1 = 0.999...If you mean are they exactly the same and literally the same thing literally muh ligma balls? no, fucking dumbass they're literally different strings, you stupid ass retard.
1/3 is 0.333...3/3 is 0.999...Therefore 0.999... = 1
