Okay, so apparently 1/998,001=0.000001002003004…996997999000001002…
But isn’t it the case that, by definition, every non-whole rational number (like 1/998,001) can be expressed as the result of dividing some pair of non-zero integers? I mean, isn’t that just what a rational number is? And therefore the only reason 1/998,001 seems special is that we don’t expect something that looks like an ordered sequence to result from math performed on two innocent-looking numbers (viz., 1 and 998,001)?
In other words, doesn’t it have to be the case that, for any repeating string of numbers you can imagine, there is a non-zero number x and a non-zero number y such that x/y equals 0.that repeating string you imagined? So why is anyone startled by 1/998,001?
For greater “amazement” factor, why not find the x and y where the series 000-999 is complete for x/y, and express amazement at that? Which brings up the question, for a given rational 0.???.., how hard is it to determine x and y for x/y=0.???..?
It’s really easy to craft a repeating decimal. If you want to repeat a specific number, all you need to know is the number of digits in that number, and use the same number of digits, all 9, in the denominator. For a three digit number, say 0.579579579579…, the fraction would be 579/999. The 1/998001 is just cool because the naive implementation requires 2997 digits in the denominator, but the fraction reduces to something much more manageable.
Also, there’s no need to exclude whole numbers. The denominator in that case is just 1.
Note that the neatness of this is that it generates all the 3 digit numbers (except 998). The video explains how it works, and that you do 2, 3, 4, 5, etc repeating digit decimals.
Essentially yes. But note that whole rational numbers can also be expressed as a division of integers (8 = 8/1, for example). Also, only the bottom number can’t be 0 (0/9 is a perfectly rational number).
Sort of. The number in question can be expressed as:
This fraction reduces to 1 / 998001. So, if you were looking to create this weird decimal expansion from scratch as it were, it wouldn’t take that much work. It just turns out that this fraction reduces quite a bit.
Yes. Except that 0.000000000 … as a fraction is going to have a 0 in the numerator.
Because of the amount that the fraction reduces and by the unexpected pattern that the decimal expansion gives. If you’re not amazed by it, that’s okay, but I thought it was pretty interesting the first time I saw it.
Because that fraction is a mess. I don’t have a lot of time to calculate at the moment, but 0.012345679 repeating works out to 1/81, while 0.123456789 repeating works out to 13717421⁄1111111111.
It’s not hard at all. I showed how to to it above.
Well, not quite by “definition”, but yeah, you pretty much got it in one.
If you have a repeating decimal (no matter that there’s a million or a billion or a trillion digits before repeating), the number can be expressed as a ratio of two integers.
Human psychology is definitely involved. Humans are natural pattern matchers. It’s a great survival trait. So when we see patterns, we attach greater significance than they might otherwise warrant.
For example, if the lottery came out 1-2-3-4-5, we would see that as a special event, though 12-16-17-33-56 (or any combination of 5 numbers) is just as likely. The pattern causes us to attach significance to a random event.
When you were a kid, did you ever sit hunched over the back of the front seats to peer at the odometer turning to 30,000? A bunch of zeroes is completely expected. People still like it when particular sequences pop up. Humans are hardwired to see patterns. And we create them if they’re not to be found. It should be celebrated.
You don’t have to go out as far as 1/998,001. 1/49 is a doozy.
0.02040816326 (that would be 64 expect that the next doubling is 128 and that pushes the 4 to a 5.)
Perhaps not as impressive, but anything (other than a multiple of 7) divided by 7 gives you the repeating sequence …142857… starting somewhere in there.
Anyway, yeah, there’s stuff in Math. Some is interesting to one person but not to another. So I’m not the least bit amazed or anything with the fraction in the OP. Very, very ho-hum.
OTOH, I love the idea of finding rational approximations to real numbers using the Stern–Brocot_tree. Cooool.
Yes. Note that there’s two interesting things about this. One is just that it cycles through all the three digit values in order except for 998. That is, cycling through 999 different 3 digit values, for a cycle length of 2997.
But ANOTHER way of looking at is that it passes through ALL numbers from 0, 1, 2, 3, on up, without end, starting each successive value three digits further down and letting the carries cascade. For example, when we see 997; 999; 000; 001; …, it’s not that we skipped 998 and restarted the cycle, but rather, we reached the point in the sequence for 997; 998; 999; 1000; 1001; …, but squeezing these into three digits apiece made it into 997; 998; 99(9+1); 00(0+1); 00(1+1); …, which then comes out after carrying into 997; 999; 000; 001; 002; …
So there’s something interesting here already: Writing out values in order, without end, but with a fixed number of digits of separation, results in a cyclic pattern. That’s probably not obvious until you first see it.
Er, yes 1/998,001 is the result of dividing 1 by 998,001, quite clearly
Sure. A rational number is one integer divided by another.
Well, why would you expect that, if you hadn’t been aware of it previously? It is, for most, surprising.
This is also true, but different from the observation you just made (that rational numbers are defined as ratios of integers). This is true, but not at all obviously true, to the person who has never seen it before.
The reason this is true is that, if S is some finite string, and N is a string of the same length consisting of all 9s, then 0.[repeating S] = 0.[repeating N] / N * C = 0.999… * C/N, which, since 0.999… = 1 (cute controversy, but it’s the appropriate interpretation of decimal notation for this context and reasoning), comes out to simply C/N.
More generally, even 0.[some initial other string][now some repeating cyclic string] will also be rational, and indeed every rational will be of this eventually periodic decimal form, by similar reasoning.
But this isn’t obvious until you reason your way to it.
Well, again, there are multiple things going on:
A) It’s not obvious until you see it that this nice repeating pattern of 3-digit numbers in a row (sans 998) should come from a simple division like this
B) Even after you realize that SOME rational should produce this cyclic pattern, it’s not obvious that it should be such a nice rational as to have a numerator of 1 and a denominator which is very nearly a round number
In fact, the reason it comes out this way is that, again, this cyclic pattern can be seen as the unending pattern 0 * 0.001 + 1 * 0.001^2 + 2 * 0.001^3 + 3 * 0.001^4 + … . That is, the number which, when you multiply it by 1000, grows by 0.[001 repeating] = 1/999. Well, if you solve 1000X = X + 1/999, you’ll find that X = 1/999^2 = 1/998,001.
Sure. That’d be 000001002…997998999 / 999999999…999999999, with 3000 digits in the denominator.
It’s quite straightforward, as noted above (in this post, and by other posters before me).
Or in any base that’s a perfect square (except for the trivial example of 1/2 in some sense always working out).
The phenomenon we’re interested in is that, for each x not divisible by 7, the fractional component of x/7 matches that of 1/7 * a power of 10. In other words, in mod 7 land, the powers of 10 include all the nonzero values (i.e., all the values strictly between 0 and 7).
More generally, we may replace 7 by some m and our decimal base by base B, and ask whether this still works.
You can quickly convince yourself this would be impossible in any base if m weren’t prime (in mod m land, powers of values coprime to m remain coprime to m, while powers of values non-coprime to m remain non-coprime to m, so powers of B couldn’t include both; it’d have to be the case that every value not divisible by m were coprime to m, which is to say, that m is prime).
Now, m = 2 always works, but only trivially so (every multiple of 1/2 is either straight-up an integer or an integer + 1/2; there’s only one fractional component that ever comes up, so it’s not very interesting).
Every other prime m will be odd. And in that case, the m - 1 nonzero values mod m pair off into (m - 1)/2 values up to negation, producing (m - 1)/2 distinct square values, so not all of them will be squares mod m. Yet, if b is a square, then so must be every power of b; thus, square b will be unsuitable.
This explains why there’s no “cyclic fraction” like 1/7 for base 16 or any other square base (except in the trivial sense of 1/2).
[A nice addendum to the above is that EVERY prime m comes with at least 1 suitable b (indeed, precisely as many suitable b as there are values coprime to m - 1). This is one of the great hidden (to laypeople) treasures of mathematics, I think. I’ll phrase the argument for this a bit more jargonly: arithmetic modulo a prime forms a field, and thus, there are at most n values satisfying x^n = 1, for each n. Thus, its multiplicative group has at most n elements of order dividing n, for each n.
Any finite such group is cyclic by a simple counting argument: Any value of order precisely n induces, via its powers, all n elements of order dividing n, of which precisely phi(n) will have order precisely n; thus, there are either 0 or phi(n) elements of order precisely n. Summing this up over all the proper divisors of the size |G| of the group, it follows that there are left over at least (and thus exactly) phi(|G|) generators of the group.]
Well, we can do better. Consider the number X = 0.[000001002…997998999 repeating] repeating. This has the property that 1000X = X + 0.[001 repeating] - 0.[000…001000 repeating, with 3000 digits, almost all zero except that 1 towards the end].
Thus, X = (1/(10^3 - 1) - 10^3/(10^3000 - 1))/(10^3 - 1). That’s a nice simple description. If you want the lowest terms form, the denominator will be (10^3000 - 1)/(10^3 - 1) = 1001001001001… with 2998 digits, and accordingly the numerator will be 1003006010015021… (running through triangular numbers at three digit separation, except for some cutoff phenomenon towards the end) with 2992 digits.
You can see why, compared to 1/998001, that’s a less snappy fraction!
Which brings us back to what’s “amazing” about 1/998001 = 0.[000001002…997999 repeating]. That it has such a nice form, with 1 in the numerator, and such a comparatively small denominator!
But, again, what’s most interesting isn’t the phenomenon in isolation, but the background understanding and generalization: that in ANY base B, in exactly the same way, we have that 0.[all digits other than B - 2, in order, repeating] = 0.[0123… through ALL values, at one digit separation, with carries] = 1/(B - 1)^2 (which has a simple-as-dirt numerator and a much smaller denominator then would be expected for a random cyclic fraction of same cycle length).
The phenomenon in the OP is the base 1000 example of this.
On the other hand, in ANY base B, we have that 0.[all digits in order repeating] comes out to, well, [all digits in order]/[B many (B - 1)s in a row]. If B is even, you can divide out a (B - 1) from numerator and denominator here, if B is odd you can divide out a (B - 1)/2, but in general, there’s not much reduction available. So that’s rather less nice.
Well its decimal representation has to be terminating or repeating.
In this case its repeating. Why so many digits repeating ?
Since the repeating decimal has to be representable as n/9 or n/99 or n/999 or n over some number of 9s… the number of repeating digits is determined this way… find a new numerator that multiples top (trivially, 1 times x is x ) and bottom of that fraction so that the new denominator is all 9’s. eg 9999999 ?
Well it turns out the smallest all 9’s denominator which has 998001 as a factor is thousands of digits long.
look at another example, For 1/7, the smallest all 9’s denominator is 999999 , with the numerator 142857 … For 1/3, its 9 , giving 3/9 . For 1/6, its not possible to find the integer numerator because its even, which correlates with the decimal 0.1666666 … ie the first digit not being in the repeating bit, But it is 0.1 + (6/9 )/10 of course, so the pattern is that the repeating decimal part can go over 9’s.
What’s going on with 1/49, incidentally, is that this is 2/(100 - 2), and in any base B, we have 2/(B - 2) = (2/B) + (2/B)^2 + (2/B)^3 + … . So, in particular, in base 100, we get 2/100 + 4/100^2 + 8/100^3 + …
Actually, for any polynomial Q(x), we get that 1/Q(x) (or, indeed, P(x)/Q(x) for any polynomial P) has an expansion following a simple pattern closely related to the coefficients of Q(x); specifically, it follows a linear recurrence described by Q.
For example, letting Q(x) = x^2 - x - 1, if we look at 1/Q(x), we get 1/x^2 + 1/x^3 + 2/x^4 + 3/x^5 + 5/x^6 - …, whose coefficients are the Fibonacci series, each term being equal to the sum of the previous two. [I’ve expanded in terms of 1/x instead of x for convenience in what’s to come, though it makes no essential difference]. It’s not hard to see why it should follow this recurrence: x^2 - x - 1 has to multiply by this to a sequence of coefficients which is (after the initial 1) all zeros, so “Two terms ahead” - “One term ahead” - “This term” has to be (after the initial zeros) always zero.
And just plugging in 10 for x, we therefore get the same phenomenon visible in the decimal expansion’s digits: 1/Q(10) = 1/(10^2 - 10 - 1) = 1/89 = 0.0112359550, which is essentially the Fibonacci series, spaced out at one digit intervals, with the carries eventually making it less familiarly recognizable. We could space it out further by using 1/Q(100) = 1/9899 instead (this comes out to 0.00010102030508132135…).
Of course, every rational number A/B is P(10)/Q(10) for some polynomials P and Q straightforwardly given by the digits of A and B, but if you can find cleaner polynomials to express the same number, you get a cleaner understanding of its digit-expansion. (E.g., 1/49 is 1/(4x + 9) with x = 10, but this is not as pleasing for understanding it as recognizing it as 2/(x^2 - 2) with x = 10, the polynomial x^2 - 2 describing a more pleasing pattern than the polynomial 4x + 9 or the polynomial 5x - 1 or what have you, though all yield accurate descriptions of what is going on here)