But don’t bits after the point in base ½ signify numbers greater than 1?
The first bit to the right of the point should signify 1/½, i.e. 2, the next one 1/(½)^2, i.e. 4, and so on. So how can an infinite string of bits after the point add up to less than 1?
Heck, in any base, call it X, the individual digits are:
X[sup]n[/sup]… X[sup]3[/sup] X[sup]2[/sup] X[sup]1[/sup] X[sup]0[/sup].X[sup]-1[/sup] X[sup]-2[/sup] X[sup]-3[/sup] … X[sup]m[/sup]
If X is a fraction, raising it to a negative exponent turns it into a number greater than 1, while raising it to a positive exponent yields a smaller result, hence the “backward” significance.
Like I said, we can use an unusual notion of convergence to make sense of this (one where higher and higher powers of 2 are considered to go closer to zero, rather than farther from it, in direct address of the issue that we want 0.0000…1 to be considered as small when there are lots of 0s before the 1).
Or we can just ignore worries about “convergence”, as such, and simply postulate that we still get to use all the same sort of familiar arithmetic and algebraic reasoning that establishes that 1.1010101010… = (110 - 1)/(100 - 1) = (b[sup]2[/sup] + b - 1)/(b[sup]2[/sup] - 1) in ordinary bases b, and see where that takes us even for extraordinary b.
Put another way, we can define addition and multiplication for digit-expansions in base 2 easily enough even for expansions which contain infinitely many nonzero digits to the left of the decimal point. And once we’ve done that, we can readily see that …0101010101011 + …0101010101011 + …0101010101011 = Well, there’s a 1 in the units digit, and a 1 that carries to the twos digit, producing a 0 in the twos digit and a 1 that carries to the eights digit. In the fours digit we get a 0, and in the eights digit (incorporating the carry from the twos digits), we get a 0 along with a 1 that carries to the thirty-twos digit. In the sixteens digit, we get a 0, and in the thirty-twos digit (incorporating the carry from the eights digits), we get a 0 along with … = …000001.
Accordingly, in some sense, …0101010101011 represents 1/3 in base 2, which, in mirror image, means that 1.101010… represents 1/3 in base 1/2.
[In some sense. Sure, in another, more familiar sense, this just adds up to ∞. But that sense is boring, so why not discuss the more interesting one?]
Huh, by my calculations, in base ½ :
…11111111.0 = 1
You can add more 1s to the left to get arbitrarily closer to the value of 1, but it’s like adding 9s to the right of 0.9999999… i.e. less significant.
And the value of one-third in base ½ is, as best I can figure:
…1111111000110101010101010101010101010101010101010101010101010.0
The digits to the left are not all 1s, of course, but that’s just the limit of precision I can get out of Excel without getting even more anal about it. I suppose the actual representation might be “10” repeated, and I just made a mistake or Excel isn’t precise enough. I don’t feel like doing it by hand to find out.
No, …111111110.0 would represent 1. Remember, the last digit before the point represents 1 by itself.
…111111111.0 would be 2. (The series …1/512 + 1/256 + 1/128 + 1/64 + 1/32 + 1/16 + 1/8 + 1/4 + 1/2 + 1 approaches 2)
It should be “10” repeated. As I said in post #54,
⅓ = 1/2[sup]2[/sup] + 1/2[sup]4[/sup] + 1/2[sup]6[/sup] + 1/2[sup]8[/sup] + 1/2[sup]10[/sup] + …
So you skip every alternate power of 2, i.e. every alternate digit in the base-½ expansion.
Of course. Funny thing is, I pointed this out myself earlier and still made the misake because when I made my Excel worksheet, I started my exponents with 1,2,3,4… and not (as would be correct) 0,1,2,3,4 …
My oversight, no doubt, but I was responding to Indistinguishable’s
If you allow yourself infinitely many nonzero digits to the left, then base 1/2 expansions are not unique, as you’ve already seen with …111110 = …000001. That’s fine, but it is interesting to note that if you instead make the traditional restriction to only finitely many nonzero bits before the “decimal” point and whatever you like after it, you get a system which can be understood as giving each rational (of any sign) a unique representation.
Hardly surprising, as if you allow yourself infinitely many nonzero digits to the right then base ten (or any base>1) expansions are not unique. As we all know here, if you allow infinite digits then 0.99999…(base 10) = 1.
Yes, exactly. But the interesting thing is that you can get uniqueness of representations in base 1/2 if you allow yourself infinitely many nonzero digits to the right but only finitely many to the left [which is to say, you can get uniqueness of representations in base 2 if you allow yourself infinitely many nonzero digits to the left but only finitely many to the right], and what’s more, you don’t need an extra sign bit.
Base 1/p expressions (or even just the eventually periodic ones), in this sense, for any prime p, directly comprise a field, which is not true of rightward expansions in any ordinary base. [One cannot subtract ordinary decimal expansions directly without having to keep track of negation signs, or even, in that context, reason from a + b = a + c to b = c [as 0.999… + 0.999… = 1.999… = 1.000… + 0.999…] without identifying distinct digit-sequences. But one can do all these things in base 1/p.]
Anyway, the reason I mentioned non-uniqueness of representations was because of Bryan Ekers apparently feeling that …101010.0 being a representation of 1/3 in base 1/2 was in conflict with 1.101010… being a representation of 1/3 in base 1/2. There’s no conflict; those are both representations of 1/3, because once we allow leftwards-infinite expansions in base 1/2, there’s no uniqueness of representation.
No, because that’s not a spoiler–you’re told that from the beginning. But, even if you weren’t, it’s a part of popular culture, so you wouldn’t be spoiling anything.
This is not so for this book. Just because it’s old (and it’s not that old, really), doesn’t mean it isn’t a spoiler. Ordinarily, the title of this thread would be sufficient to let people know to avoid it for spoilers’ sake, but, in this case, there’s both a book and a movie, and this is only a spoiler for the book.
I would definitely have preferred that I was having an important part of the book spoiled by coming in here. So, yes, I would request “[book spoilers]” be added to the title.
I have no objection. Mod, please add “open spoilers” to the subject.