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Again, I don't think this is necessarily that fruitful a discussion, since the whole point of the guy's argument, as I see it, is that you can slightly adjust t

by drcode 17y ago

Again, I don't think this is necessarily that fruitful a discussion, since the whole point of the guy's argument, as I see it, is that you can slightly adjust the meaning of "equality" as it relates to infinite fractions and maintain a completely self-consistent mathematical system. > You can not have an infinite series of something, > followed by something else. Why is that? How about the number 19...91? why are you denying me that number? :) > If you need convincing, start the long division > on paper and keep going until you're convinced. When I do that, I constantly am left with a remainder. If I accept that 1/3=0.3333..., as you're arguing, then the remainder disappears, for unexplained reasons. An alternative argument is that the remainder persists in the form 0.0...1 and then you don't have to "hand wave away" the remainder. > Recall that introducing one contradictory premise > into a logical system allows you to prove any statement I agree with that completely, but I haven't seen the "contradictory premise" yet... all I see (at this point) is some alternative premises that seem fully internally consistent. > In order to show the two are distinct entities, you > need to show a situation in which they behave > differently. I think that's actually the best argument you've put forward so far. Basically, an argument in favor of choosing the more pragmatic way of handling these situations.

jerf 17y ago
"When I do that, I constantly am left with a remainder. If I accept that 1/3=0.3333..., as you're arguing, then the remainder disappears, for unexplained reasons." No, it doesn't. You stopped finitely soon into an infinite process. You apparently didn't take it far enough. No disrespect intended, but a HN comment is not a place to post a full description of infinity. Hie thee hence to a textbook. And read it like a textbook, not an internet post. You know, where you go over it with a fine-tooth comb looking for opportunities to leap up and call your opponent a Nazi. Yeah, I know that style. Without that, counterarguments are getting mangled on their way into your brain, because you don't actually understand them in Math. That can't be corrected by people batting down various ill-posed English-based objections. (If it could, I would have seen it happen by now.)
- drcode 17y ago
  I agree this discussion can't be taken much further on an HN thread for the reasons you give :)
ErrantX 17y ago
ok some key points about infinity ot address here :) >then the remainder disappears, for unexplained reasons. Actually no - it doesnt disappear. That's why it works. Dont forget the definition of infinity is "forever". The remainder is always there - and it always reduces further (edit: if you continue completing the sum. As you can see below the sum is already completed for infinite length anyway - your just manually bound checking it at the beginning) In your counter argument where the 0.0....1 acts as a remainder. The set of 0's before the 1 can only be finite. One reason people fall for this is because they imagine the infinite set growing by adding a 0 on the end each time. Or perhaps pushing on one the front (which is what would have to happen for your requirement). This is *wrong( entirely - the infinite set exists, in entirety for the whole of it's infinity.
jibiki 17y ago
0.0...1 is a very interesting number. Normally, one indexes digits by integers (there's a first digit, a second digit,...) But here you have digits seemingly indexed by ordinals. The omega-th digit is 1. I don't think you can make a consistent number system this way. Here's the problem: what is 10*0.00...1? It can't have omega-th digit "10" because 10 isn't a digit. It can't have omega-th digit 1 and omega+1-th digit 0, because that should be the same as 0.00...1, under any logic I can think of. So that number system doesn't work. The real numbers, by contrast, have the advantage of working (so far as we know.)
- drcode 17y ago
  OK, that's the most convincing argument I've heard so far. The omega+1-th digit would need to be 1, for your example. I'm not entirely sure that it has to be the same as 0.00...1, as you argue. However, you're definitely showing that this way of handling fraction would probably get to be pretty ridiculous at some point.

by drcode 17y ago

jerf 17y ago

"When I do that, I constantly am left with a remainder. If I accept that 1/3=0.3333..., as you're arguing, then the remainder disappears, for unexplained reasons." No, it doesn't. You stopped finitely soon into an infinite process. You apparently didn't take it far enough. No disrespect intended, but a HN comment is not a place to post a full description of infinity. Hie thee hence to a textbook. And read it like a textbook, not an internet post. You know, where you go over it with a fine-tooth comb looking for opportunities to leap up and call your opponent a Nazi. Yeah, I know that style. Without that, counterarguments are getting mangled on their way into your brain, because you don't actually understand them in Math. That can't be corrected by people batting down various ill-posed English-based objections. (If it could, I would have seen it happen by now.)

drcode 17y ago
I agree this discussion can't be taken much further on an HN thread for the reasons you give :)

ErrantX 17y ago

ok some key points about infinity ot address here :) >then the remainder disappears, for unexplained reasons. Actually no - it doesnt disappear. That's why it works. Dont forget the definition of infinity is "forever". The remainder is always there - and it always reduces further (edit: if you continue completing the sum. As you can see below the sum is already completed for infinite length anyway - your just manually bound checking it at the beginning) In your counter argument where the 0.0....1 acts as a remainder. The set of 0's before the 1 can only be finite. One reason people fall for this is because they imagine the infinite set growing by adding a 0 on the end each time. Or perhaps pushing on one the front (which is what would have to happen for your requirement). This is *wrong( entirely - the infinite set exists, in entirety for the whole of it's infinity.

jibiki 17y ago

0.0...1 is a very interesting number. Normally, one indexes digits by integers (there's a first digit, a second digit,...) But here you have digits seemingly indexed by ordinals. The omega-th digit is 1. I don't think you can make a consistent number system this way. Here's the problem: what is 10*0.00...1? It can't have omega-th digit "10" because 10 isn't a digit. It can't have omega-th digit 1 and omega+1-th digit 0, because that should be the same as 0.00...1, under any logic I can think of. So that number system doesn't work. The real numbers, by contrast, have the advantage of working (so far as we know.)

drcode 17y ago
OK, that's the most convincing argument I've heard so far. The omega+1-th digit would need to be 1, for your example. I'm not entirely sure that it has to be the same as 0.00...1, as you argue. However, you're definitely showing that this way of handling fraction would probably get to be pretty ridiculous at some point.