6 ms·

Whoa! What? Not a mathematician in any way (in case that isn't obvious), but I'd have totally thought 0.999... asymptomatically _approaches_ 1, but never reache

by eszed 2y ago

Whoa! What? Not a mathematician in any way (in case that isn't obvious), but I'd have totally thought 0.999... asymptomatically _approaches_ 1, but never reaches it, and so is <1. Is there a short-form explanation (that I might have a chance of understanding, lol) of why that's incorrect? I'd love to have my mind blown.

techno_tsar 2y ago
There's a rigorous proof on Wikipedia, but there's simpler ways to show it. For example, we know that 1/3 = 0.333... 3 * 1/3 = 3 * 0.333... 1 = 0.999... You can also do it with subtraction. For example, 1 - 0.999... = x. Assuming x is greater than 0, then it should evaluate to 0.000...1. But we can't have the digit 1 after an infinite number of zeros. If there truly were a "1" after infinite zeros, it implies reaching the end of infinity, which is a logical contradiction. So x can't be greater than 0.
justsomeshmuck 2y ago
1/3 + 1/3 + 1/3 = 3/3 = 1 .3… + .3… + .3… = .9… = 1 https://en.m.wikipedia.org/wiki/0.999 https://en.m.wikipedia.org/wiki/0.999...
xypage 2y ago
I think the simple one is x=0.999..., 10x=9.9999..., 10x-x=9.999...-0.999..., 9x=9, x=1
evanb 2y ago
Sequences can approach things. The sequence 0.9, 0.99, 0.999, 0.9999 and so on asymptotically approaches 1. The difference between 1 and the Nth term in the sequence is 1e-N, which goes to 0 with N. 0.999...[forever] is not a sequence, it is a number. Numbers have values, they don't approach things. The misleading part is that 'forever' is not something about evolution or the passage of time. It's not 'happening' or 'sequential' like the sequence. There is no 'and then another 9'. All the 9s are really there, at once. And it is closer to 1 than any term in the sequence. Since the sequence gets closer and closer to 1, converging to it asymptotically, 0.999...[forever] cannot differ from 1; if it did the sequence wouldn't converge.
- eszed 2y ago
  Thank you, and everyone else who answered (I hope they see this reply). Your distinction between "sequence" and "number", along with the mathematics of 0.333... = 1/3, convinced me - and my mind is successfully blown. Follow-up: is it the same for other repeating sequence-looking numbers? As in, would 0.9333... = 0.94?
  - eigenket 2y ago
    Its true of numbers whose with a decimal representation which ends in an infinite string of 9s, so 0.939999... = 0.94. This is because we write numbers in base 10. If you write numbers in base 2 its equal to numbers whose binary representation ends with an infinte number of 1s e.g. 0.11111... (base 2) = 1.
  - evanb 2y ago
    0.9333... is equal to 9/10 + 1/30. To get 9/10 + 4/100 I think what you're aiming at is 0.93999...
atq2119 2y ago
An alternative to the sibling comments: In this context, the notation 0.999... does not represent a process. It represents a fixed number. Which number? Well, if you reason through it you'll find that it has to be the same number as that represented by the notation 1. An insight that is crucial (and pretty obvious in hindsight, though many people seem not to be exposed to it) is to distinguish between a number and its representations. "1" is not a number, it is merely the representation of a number. And numbers have many different representations. As a member of this forum, you can probably appreciate that "12" and "0xC" are two different representations of the same number. In the same way, 0.999... and 1 are different representations of the same number.
tsimionescu 2y ago
0.xxx... is just a notation for certain fractions (specifically, the fraction x/9). If we set x = 9, then the notation 0.9999... is just a notation for 9/9 = 1. So it's just a silly notation for 1.
- nkrisc 2y ago
  This actually makes the most sense to me. It’s an artifact of our chosen numerical notation system.
rachofsunshine 2y ago
There are a lot of proofs of this, but they all rely on a certain level of rigor about what a real number is - and that, it turns out, is a much more difficult question than it sounds like. You don't typically get a rigorous definition of the real numbers until well into a college-level math education. ----- First, you're making a category error. "0.9999..." is a single value, not the sequence of values 0.9, 0.99, 0.999, 0.9999... Single values cannot "asymptotically approach" anything, any more than the value 2 or the value 7 can asymptotically approach anything. It's just a number like any other. To show what value 0.9999... takes on, we need to do two things. First, we need to show that this notation makes sense as a description of a real number in the first place, and second, we need to show what that real number is (and it will happen to be 1). ----- So, why is it a real number? Well, remember what we mean by place value. 0.9 means "0 ones, 9 tenths[, and zero hundredths, thousandths, and so on]". 0.99 means "0 ones, 9 tenths, 9 hundredths, [and zero of everything else]". Another way to say this is that 0.9 is the value 0 * 1 + 9 * 0.1 [plus 0 times 0.01, 0.001, and so on], and that 0.99 is the value 0 * 1 + 9 * 0.1 + 9 * 0.01 + [0 of everything else]. What that means is that if 0.9999... means anything, it means 9 tenths, plus 9 hundredths, plus 9 thousandths, plus 9 ten-thousandths, plus 9 hundred-thousandths, and so on and so forth forever. In other words, 0.9999... is the value of an infinite sum: .9 + .09 + .009 + .0009 + ... Infinite sums, in turn, are by definition the limit of a sequence. This is where that "asymptotic" thing comes back, but notice the distinction. 0.9999... is not the sequence, it is the LIMIT OF the sequence, which has a single value. To show that it's a real number, then, we need to show that the limit of the sequence 0.9, 0.99, 0.999, 0.9999... does in fact exist. But this sequence is clearly increasing, and it is clearly not greater than 1, so we can (among other things) invoke the Monotone Convergence Theorem [1] to show that it must converge (i.e., the limit exists). Alternately, you can think back to your algebra 2 or calculus classes, and notice that this is the geometric series [2] given by sum 9 * 10^-n, and this series converges. ----- Now, why is it equal to 1? Well, there's a few ways to prove that, too. But the simplest, in my book, is this: given any two different real numbers x and y, I'm sure you would agree that there is a value z in between them (this is not a difficult thing to prove rigorously, provided you've done the hard work of defining the real numbers in the first place). The average of x and y will do. But we can flip that statement around: if there is NOT a value between two real numbers, those two real numbers MUST be equal. In more symbolic terms, we claim that for all real numbers x and y such that x < y, there exists z such that x < z < y. So if there ISN'T such a z, then we must not have x < y in the first place. (This is the contrapositive [3], if you're not up on your formal logic.) So consider the values of 0.9999... and 1. What value lies between them? Can you find one? As it turns out, no such value exists. If you pick any real number less than 1, your 0.99[some finite number of nines]9 sequence will eventually be bigger than it - and therefore, since the sequence is increasing, its limit must be bigger than that value too. Since there are no numbers between 0.9999... and 1, they must be equal. ----- [1] https://en.wikipedia.org/wiki/Monotone_convergence_theorem https://en.wikipedia.org/wiki/Monotone_convergence_theorem [2] https://en.wikipedia.org/wiki/Geometric_series https://en.wikipedia.org/wiki/Geometric_series [3] https://en.wikipedia.org/wiki/Contraposition https://en.wikipedia.org/wiki/Contraposition
aguaviva 2y ago
The simplest argument I can think of is to ask yourself: "Are there any numbers between 0.999... and 1?" If not then it's logical to conclude they wind up at the same "place", that is, the same number. Or equivalently: it can't have any other value besides 1. If you care about such things, then you're a mathematician.
- ForOldHack 2y ago
  It is a limit. A very powerful tool. You can choose any arbitrary finite number in the sequence, and I can find a number greater. So the value of 1 exists, and its a continuous function, so therefore the limit exists. I do care about such things, and yes, a degree in math for me was just slogging through every math course the college had, with a long string of high grades, and honor roll achievements.
- floxy 2y ago
  1/6 = 0.166666...6 1/6 = 0.166666...6 1/6 = 0.166666...6 1/6 = 0.166666...6 1/6 = 0.166666...6 + 1/6 = 0.166666...6 ---------------------- = 0.999999...6 ...so that 6 times 1/6 is zero point 9 repeating, with a 6 on the end ;-)
deleted 2y ago
[deleted]

by eszed 2y ago

techno_tsar 2y ago

There's a rigorous proof on Wikipedia, but there's simpler ways to show it. For example, we know that 1/3 = 0.333... 3 * 1/3 = 3 * 0.333... 1 = 0.999... You can also do it with subtraction. For example, 1 - 0.999... = x. Assuming x is greater than 0, then it should evaluate to 0.000...1. But we can't have the digit 1 after an infinite number of zeros. If there truly were a "1" after infinite zeros, it implies reaching the end of infinity, which is a logical contradiction. So x can't be greater than 0.

justsomeshmuck 2y ago

1/3 + 1/3 + 1/3 = 3/3 = 1 .3… + .3… + .3… = .9… = 1 https://en.m.wikipedia.org/wiki/0.999 https://en.m.wikipedia.org/wiki/0.999...

xypage 2y ago

I think the simple one is x=0.999..., 10x=9.9999..., 10x-x=9.999...-0.999..., 9x=9, x=1

evanb 2y ago

Sequences can approach things. The sequence 0.9, 0.99, 0.999, 0.9999 and so on asymptotically approaches 1. The difference between 1 and the Nth term in the sequence is 1e-N, which goes to 0 with N. 0.999...[forever] is not a sequence, it is a number. Numbers have values, they don't approach things. The misleading part is that 'forever' is not something about evolution or the passage of time. It's not 'happening' or 'sequential' like the sequence. There is no 'and then another 9'. All the 9s are really there, at once. And it is closer to 1 than any term in the sequence. Since the sequence gets closer and closer to 1, converging to it asymptotically, 0.999...[forever] cannot differ from 1; if it did the sequence wouldn't converge.

eszed 2y ago
Thank you, and everyone else who answered (I hope they see this reply). Your distinction between "sequence" and "number", along with the mathematics of 0.333... = 1/3, convinced me - and my mind is successfully blown. Follow-up: is it the same for other repeating sequence-looking numbers? As in, would 0.9333... = 0.94?
- eigenket 2y ago
  Its true of numbers whose with a decimal representation which ends in an infinite string of 9s, so 0.939999... = 0.94. This is because we write numbers in base 10. If you write numbers in base 2 its equal to numbers whose binary representation ends with an infinte number of 1s e.g. 0.11111... (base 2) = 1.
- evanb 2y ago
  0.9333... is equal to 9/10 + 1/30. To get 9/10 + 4/100 I think what you're aiming at is 0.93999...

atq2119 2y ago

An alternative to the sibling comments: In this context, the notation 0.999... does not represent a process. It represents a fixed number. Which number? Well, if you reason through it you'll find that it has to be the same number as that represented by the notation 1. An insight that is crucial (and pretty obvious in hindsight, though many people seem not to be exposed to it) is to distinguish between a number and its representations. "1" is not a number, it is merely the representation of a number. And numbers have many different representations. As a member of this forum, you can probably appreciate that "12" and "0xC" are two different representations of the same number. In the same way, 0.999... and 1 are different representations of the same number.

tsimionescu 2y ago

0.xxx... is just a notation for certain fractions (specifically, the fraction x/9). If we set x = 9, then the notation 0.9999... is just a notation for 9/9 = 1. So it's just a silly notation for 1.

nkrisc 2y ago
This actually makes the most sense to me. It’s an artifact of our chosen numerical notation system.

rachofsunshine 2y ago

There are a lot of proofs of this, but they all rely on a certain level of rigor about what a real number is - and that, it turns out, is a much more difficult question than it sounds like. You don't typically get a rigorous definition of the real numbers until well into a college-level math education. ----- First, you're making a category error. "0.9999..." is a single value, not the sequence of values 0.9, 0.99, 0.999, 0.9999... Single values cannot "asymptotically approach" anything, any more than the value 2 or the value 7 can asymptotically approach anything. It's just a number like any other. To show what value 0.9999... takes on, we need to do two things. First, we need to show that this notation makes sense as a description of a real number in the first place, and second, we need to show what that real number is (and it will happen to be 1). ----- So, why is it a real number? Well, remember what we mean by place value. 0.9 means "0 ones, 9 tenths[, and zero hundredths, thousandths, and so on]". 0.99 means "0 ones, 9 tenths, 9 hundredths, [and zero of everything else]". Another way to say this is that 0.9 is the value 0 * 1 + 9 * 0.1 [plus 0 times 0.01, 0.001, and so on], and that 0.99 is the value 0 * 1 + 9 * 0.1 + 9 * 0.01 + [0 of everything else]. What that means is that if 0.9999... means anything, it means 9 tenths, plus 9 hundredths, plus 9 thousandths, plus 9 ten-thousandths, plus 9 hundred-thousandths, and so on and so forth forever. In other words, 0.9999... is the value of an infinite sum: .9 + .09 + .009 + .0009 + ... Infinite sums, in turn, are by definition the limit of a sequence. This is where that "asymptotic" thing comes back, but notice the distinction. 0.9999... is not the sequence, it is the LIMIT OF the sequence, which has a single value. To show that it's a real number, then, we need to show that the limit of the sequence 0.9, 0.99, 0.999, 0.9999... does in fact exist. But this sequence is clearly increasing, and it is clearly not greater than 1, so we can (among other things) invoke the Monotone Convergence Theorem [1] to show that it must converge (i.e., the limit exists). Alternately, you can think back to your algebra 2 or calculus classes, and notice that this is the geometric series [2] given by sum 9 * 10^-n, and this series converges. ----- Now, why is it equal to 1? Well, there's a few ways to prove that, too. But the simplest, in my book, is this: given any two different real numbers x and y, I'm sure you would agree that there is a value z in between them (this is not a difficult thing to prove rigorously, provided you've done the hard work of defining the real numbers in the first place). The average of x and y will do. But we can flip that statement around: if there is NOT a value between two real numbers, those two real numbers MUST be equal. In more symbolic terms, we claim that for all real numbers x and y such that x < y, there exists z such that x < z < y. So if there ISN'T such a z, then we must not have x < y in the first place. (This is the contrapositive [3], if you're not up on your formal logic.) So consider the values of 0.9999... and 1. What value lies between them? Can you find one? As it turns out, no such value exists. If you pick any real number less than 1, your 0.99[some finite number of nines]9 sequence will eventually be bigger than it - and therefore, since the sequence is increasing, its limit must be bigger than that value too. Since there are no numbers between 0.9999... and 1, they must be equal. ----- [1] https://en.wikipedia.org/wiki/Monotone_convergence_theorem https://en.wikipedia.org/wiki/Monotone_convergence_theorem [2] https://en.wikipedia.org/wiki/Geometric_series https://en.wikipedia.org/wiki/Geometric_series [3] https://en.wikipedia.org/wiki/Contraposition https://en.wikipedia.org/wiki/Contraposition

aguaviva 2y ago

The simplest argument I can think of is to ask yourself: "Are there any numbers between 0.999... and 1?" If not then it's logical to conclude they wind up at the same "place", that is, the same number. Or equivalently: it can't have any other value besides 1. If you care about such things, then you're a mathematician.

ForOldHack 2y ago
It is a limit. A very powerful tool. You can choose any arbitrary finite number in the sequence, and I can find a number greater. So the value of 1 exists, and its a continuous function, so therefore the limit exists. I do care about such things, and yes, a degree in math for me was just slogging through every math course the college had, with a long string of high grades, and honor roll achievements.
floxy 2y ago
1/6 = 0.166666...6 1/6 = 0.166666...6 1/6 = 0.166666...6 1/6 = 0.166666...6 1/6 = 0.166666...6 + 1/6 = 0.166666...6 ---------------------- = 0.999999...6 ...so that 6 times 1/6 is zero point 9 repeating, with a 6 on the end ;-)

deleted 2y ago

[deleted]