[eszed]

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]

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.

[atq2119]

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.

[evanb]

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]

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]

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]

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...

[justsomeshmuck]

1/3 + 1/3 + 1/3 = 3/3 = 1

.3… + .3… + .3… = .9… = 1

https://en.m.wikipedia.org/wiki/0.999...

[tsimionescu]

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]

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

[xypage]

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

[aguaviva]

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]

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]

     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 ;-)

[rachofsunshine]

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 [2] https://en.wikipedia.org/wiki/Geometric_series [3] https://en.wikipedia.org/wiki/Contraposition

[eszed]

This is my (new) favorite answer. You've made something counter-intuitive seem simple and obvious. Fantastic!

(I've never done this before, but: a pox on those down-voting my original question. Learning is the very essence of "hacker"-dom. Thank you to all who have seen this and taken their time to teach me something.)

[ForOldHack]

This needs to be repeated time and time again for people who deny the basic tools of Calculus, and will suffer the misuse of them. ( specifically the sum of the sequence of the reciprocals of the natural numbers is equal to 1/12. I get that it is a useful tool for quantum chromodynamics, but it makes my skin crawl. )

[eszed]

It's one of the regrets of my life that I didn't take start calculus my first term in college. I'd done pre-calc in high school, but I'm a Humanities guy, and maths - even when I "get" it (and I'd done fine in high school) - drops out of my head pretty quickly. By the time I had an opportunity to continue, and signed up for a "refresher" pre-calc course, I realized I'd have had to go right the way back and re-learn enough algebra and geometry stuff that it was too heavy a lift. Maybe when I'm retired and have time on my hands I'll sign up for some courses at a community college. I still remember how satisfying it was to grok a problem or a concept that had seemed impossible.

So, ahem: is there any possibility that I'll be able to understand "the sum of the sequence of the reciprocals of the natural numbers is equal to 1/12"? I understand (I think!) each of the words in that sentence, but I can't make them make sense! I've got as far as

1 + 1/2 + 1/3, etc.

So how does something that starts off with 1 + [something] end up < 1?

[rachofsunshine]

It doesn't, and in fact I believe the person you're replying to has it confused with another result where the answer would also be "it doesn't".

The sum 1 + 1/2 + 1/3 + 1/4 ... - what we call the harmonic series - runs off to infinity, although it does so very slowly (the nth partial sum is approximately equal to ln(n) plus a constant whose value is about 0.58). Showing that this series diverges is standard calc-textbook stuff (it's the textbook example of the integral test for convergence, although there are plenty of other ways to show it).

However, in math well beyond basic calculus, there are methods for assigning meaningful values to series that don't converge in the conventional sense. Those methods assign the same value to convergent series as the regular old calculus arguments would, but they can also assign values to some divergent series in a way that is consistent and useful in some contexts.

For example, the series 1 + 1/2 + 1/3 + 1/4 + ... is a specific example of a more general series 1/1^z + 1/2^z + 1/3^z + 1/4^z + ..., for some arbitrary number z. This series only converges in the standard calculus sense when the real part of z is > 1, but it turns out that that's enough to define a function called the Riemann zeta function, whose input is the value z in the series and whose output is the sum of the series.

The zeta function, as it turns out, can be extended to values where the original series didn't converge. And doing so gives you a method for assigning values to "sums" that aren't really sums at all.

-----

It turns out that even THAT won't get you a completely nice value for 1 + 1/2 + 1/3 + 1/4 + ..., because that sum corresponds to the zeta function's value at z = 1. But the zeta function isn't well-behaved at z = 1. If you do even more massaging to beat a number out of it, you don't get 1/12, you get the about-0.58 constant mentioned in the previous section.

Which brings me to the way the poster you were replying to is probably confused. I think the sum they meant to refer to was the even-more-obviously-divergent sum 1 + 2 + 3 + 4 + ... This sum happens to correspond loosely to the z = -1 value of the zeta function, since 1/2^-1 is just 2, 1/3^-1 is just 3, and so on.

Again, the sum 1 + 2 + 3 + 4 + ... does not converge, but if we choose to identify it with a zeta function value, we'd identify the sum 1 + 2 + 3 + 4 + ... with zeta(-1). And the value of zeta(-1) happens to be -1/12 (yes, minus 1/12).

----

So the answer here is just that no, 1 + a bunch of positive numbers is never < 1. It's actually a decent basic calculus exercise to prove that any series a_0 + a_1 + ... a_n, where all the terms are non-negative, either fails to converge or converges to a value >= a_0. But because we're taking some extra steps here to leave the domain of convergent series in the first place, it turns out we can get results that (for conventionally convergent series) would be impossible.

[rachofsunshine]

I suspect the downvotes are just because this is a well-known result whose proofs are rather easy to google. But I like running into one of today's 10,000, I guess. https://xkcd.com/1053/

[eszed]

Fair enough. I'm enough not-a-mathematician that I didn't know it was well-known, and wouldn't even have known how google the proof! (Nor have much confidence I'd understand the real thing once I found it, lol.)

Anyway, I love that concept, and enjoy being on either side of the exchange. Thanks again.

[autoexec]

> So consider the values of 0.9999... and 1. What value lies between them? Can you find one?

As long as it's true that you can always add two real numbers, there's always infinity + 1, or ultimately 0.9999... + 0. ...9