tag:blogger.com,1999:blog-5212354357884270733.post7697114431292243408..comments2017-07-28T13:33:26.127+01:00Comments on Shiny Pebbles and other stuff: 0.999... it just keeps on going.Rob Lownoreply@blogger.comBlogger12125tag:blogger.com,1999:blog-5212354357884270733.post-74153595212064399302017-07-28T13:33:26.127+01:002017-07-28T13:33:26.127+01:00Infinite decimals have a long and troubled history...Infinite decimals have a long and troubled history. 1/3 could be represented in bases that had 3 as a factor, like base 12, but it could not be represented in base 10. Mathematicians wanted all quantities to have a base 10 representation, so in the 16th century Stevin created the basis for modern decimal notation in which he allowed an actual infinity of digits.<br /><br />For over 200 years mathematicians were troubled by infinite decimals. In order to determine that 9/10 + 9/100 + ... all add-up to 1 would require an infinite amount of work. Another problem is that if we add more terms then we have still only added a finite number of terms. It seemed that the addition of infinitely many terms was impossible.<br /><br />They used the trick of saying 0.9 + 0.09 + ... cannot add up to anything else, so it must add up to 1.<br /><br />There were still more problems. If you take any of the infinitely many terms in the series 0.9 + 0.09 + 0.009 + ... then the sum to your chosen nth term is given by the expression 1 - (0.1)^n which means that the sum is always a non-zero distance away from 1. This holds for ALL of the infinitely many digits meaning that no term CAN POSSIBLY EXIST where 1 is reached. This variation on Zeno's dichotomy paradoxes appeared to be solid proof that 0.999... cannot equal 1.<br /><br />The argument that there is no number between them fails. We begin by assuming that the series with the nth sum of 1 - (0.1)^n is a different number to 1. <br /><br />Now if we say 1 is the series that has the nth sum of 1 - 0^n then we can easily find a series halfway between 0.999... and 1, which is the series with the nth sum 1 - (0.5)(0.1)^n and so it is easy to find as many 'numbers' as we like between 0.999... and 1. We cannot presume that when we convert these series into decimal form they will all become equal to 1, because that would mean that our starting position is that 0.999... already equals 1.<br /><br />In the early 19th century Bolzano and Cauchy introduced the apparatus of limits and convergence. Now you should no longer think of 0.999... as the endless sum 9/10 + 9/100 + 9/1000 + ..., instead you should think of it as being the 'limit' that the increasing (partial) sum is approaching.<br /><br />The serious problems with this approach are still there, but are less obvious.<br /><br />With the limit approach, when you see the symbol 0.999... you should think of its value as being what is returned from the function: THE-LIMIT-OF[9/10 + 9/100 + 9/1000 + ...].<br /><br />This approach was generalised, so that all decimals were then said to contain endless digits. For example, 2.5 would now be 2.5000... (i.e. it contains 'infinitely many' trailing zeros).<br /><br />The first problem is that if this limit function returns a decimal value, then in order to assess the value of that decimal we again need to call the limit function. We end up in an endless loop of calling the limit function. To avoid this problem, we claim that when we call this limit function for 0.999... then it returns the rational 1.<br /><br />But the limit cannot always be described as a rational. For example, THE-LIMIT-OF[4/1 - 4/3 + 4/5 - 4/7 + ...] cannot return a rational and it cannot return a decimal, all it can return is the symbol pi. We have to imagine that this symbol can equal a constant value. Finitists do not accept this imagined existence.<br /><br />A second problem is that in order to convert 'infinitely many' terms into a constant like pi, the function would have to do an infinite amount of work.<br /><br />Thirdly, if it processes more and more terms it will still only have processed a finite number of terms. How can the function process the actual infinity of terms in pi to find its constant value?<br /><br />If we think of pi and the square root of 2 as functions that allow us to get as accurate a real-world measurement as we need, then we don’t have the infinity-related problems that we have if we try to think of them as constants. Sadly this approach does not fit with the mainstream Platonist position which is that perfect forms, like a perfect circle and a perfect diagonal of a unit square, MUST somehow exist. Karma Penyhttps://www.blogger.com/profile/14206806631191720441noreply@blogger.comtag:blogger.com,1999:blog-5212354357884270733.post-16808218355709884172017-06-07T13:53:01.798+01:002017-06-07T13:53:01.798+01:00It's mostly the (easily fixed) problem of mult...It's mostly the (easily fixed) problem of multiplying an infinite sum by a constant giving the same result as multiplying each term by a constant. The distributive law only tells you that for finite sums, so there's a proof of convergence needed. (And if you're very picky, subtracting infinite decimal expansions; the algorithm never terminates, so again you have to assume that something 'obvious' in this particular case, but a little technical, is true.) Not built on sand, just lacking a little bit of detail. <br /><br />The usual gaps in the argument for the infinite geometric series are this 'infinite distributive law' and showing that a^n really does give 0 in the limit.<br /><br />So whatever you do, it all ultimately rests on some properties of limits and convergence. With the 0.999... example, the properties are so 'obvious' it's hard to remember that you are using them.Rob Lowhttps://www.blogger.com/profile/01202748963636548483noreply@blogger.comtag:blogger.com,1999:blog-5212354357884270733.post-68825873674290123982017-06-07T06:48:17.417+01:002017-06-07T06:48:17.417+01:00I can see that the explanation with 1/3 has a big ...I can see that the explanation with 1/3 has a big assumption at the beginning...but I am very curious as to what is wrong with the 9S version.<br />Aside from seeming sound it also is v similar to the (now alleged!) proof I use for the sum of a geometric sequence...<br />Is my (limited) knowledge all built on sand and I am just regurgitating maths comforters to the masses?<br />Unknownhttps://www.blogger.com/profile/00711851124679291978noreply@blogger.comtag:blogger.com,1999:blog-5212354357884270733.post-3977069533742471192017-06-05T16:55:12.631+01:002017-06-05T16:55:12.631+01:00If 0.9999etc was a price and i paid with a pound t...If 0.9999etc was a price and i paid with a pound the till would calc my change for ever(boring and I'd died of old age) skip to eternity no change thou. If the price was a pound I'd be out the shop, can be more different george bullhttps://www.blogger.com/profile/01317389742729893360noreply@blogger.comtag:blogger.com,1999:blog-5212354357884270733.post-15360958191723883112017-06-05T16:51:06.679+01:002017-06-05T16:51:06.679+01:00If 0.9999etc was a price and i paid with a pound t...If 0.9999etc was a price and i paid with a pound the till would calc my change for ever(boring and I'd died of old age) skip to eternity no change thou. If the price was a pound I'd be out the shop, can be more different george bullhttps://www.blogger.com/profile/01317389742729893360noreply@blogger.comtag:blogger.com,1999:blog-5212354357884270733.post-68220520725848847132017-06-05T08:25:09.184+01:002017-06-05T08:25:09.184+01:00I don't know: of course, there are lots of non...I don't know: of course, there are lots of non-rigorous arguments, and different ones will convince different people. If you try it out, I'd certainly be interested in how successful you find it. Rob Lowhttps://www.blogger.com/profile/01202748963636548483noreply@blogger.comtag:blogger.com,1999:blog-5212354357884270733.post-18925018944043960222017-06-05T08:09:43.244+01:002017-06-05T08:09:43.244+01:00Has anyone tried this approach to convince skeptic...Has anyone tried this approach to convince skeptical students?<br />1<br />= 0.9+0.1<br />= 0.9+0.09+0.01<br />= 0.9+0.09+0.009+0.001<br />etc.<br />Ms. Hentgeshttps://www.blogger.com/profile/00816175522118444479noreply@blogger.comtag:blogger.com,1999:blog-5212354357884270733.post-40091499352750134652017-06-04T00:22:49.134+01:002017-06-04T00:22:49.134+01:00I think I was persuaded by a variant of your secon...I think I was persuaded by a variant of your second argument when I was in school, some 40 years ago. However the argument was constructed using only rationals. First we learned that rationals give rise to recurring or terminating decimal expansions. Then we learned that the algorithm in the second algorithm would recover the original rational from any recurring or terminating decimal expansion (the 10 in the example should be 1 followed by one 0 for each digit in the recurring sequence). Finally we were introduced to reals by observing that there were digit sequences that neither recur nor terminate.Unknownhttps://www.blogger.com/profile/16254159107587416110noreply@blogger.comtag:blogger.com,1999:blog-5212354357884270733.post-54954151281773551552017-06-03T21:26:02.826+01:002017-06-03T21:26:02.826+01:00Which should get into an interesting discussion ab...Which should get into an interesting discussion about the structure of the reals...Rob Lowhttps://www.blogger.com/profile/01202748963636548483noreply@blogger.comtag:blogger.com,1999:blog-5212354357884270733.post-25770388950992742122017-06-03T21:23:17.978+01:002017-06-03T21:23:17.978+01:00For the skeptical students I often ask them "...For the skeptical students I often ask them "If zero point nine repeating is less than one, what number is between them?" Ms. Hentgeshttps://www.blogger.com/profile/00816175522118444479noreply@blogger.comtag:blogger.com,1999:blog-5212354357884270733.post-28190840377613506132017-06-03T18:44:00.636+01:002017-06-03T18:44:00.636+01:00Yes. I agree with a lot of this, and the observati...Yes. I agree with a lot of this, and the observation that it delves deeply into the definition of the reals is certainly part of my argument. The difficulty lies in the gap between the naive notion of the reals (which generally isn't precise enough to admit a precise argument) and the formal standard mathematics one (which is precise enough, but takes a lot of effort to come to terms with). <br /><br />There are certainly contexts in which non standard analysis seems closer to intuition than standard, but the relationship (transfer principle) there is also extremely subtle.<br /><br />Thanks for your comments!Rob Lowhttps://www.blogger.com/profile/01202748963636548483noreply@blogger.comtag:blogger.com,1999:blog-5212354357884270733.post-79953831135661087332017-06-03T18:26:07.382+01:002017-06-03T18:26:07.382+01:00I appreciate that this makes the important point t...I appreciate that this makes the important point that the proof that 0.9˙=1 is actually non-trivial, and that the commonly given arguments substitute mathematical trickery for addressing what's actually hard about the proof.<br /><br />But I think one should go further. The primary issue isn't so much the complexity of the proof as the fact that it digs deeper into the formal definition of the real numbers than many people are used to. It's very clear that most of the people who struggle with this are actually feeling their way towards an alternate axiomitization of the real numbers. With the benefit of a few centuries of work, we know to be skeptical of the naive theory of infinitesimals that people seem to find intuitive - there's no way to capture exactly the structure people seem to find intuitive.<br /><br />The accepted formal definition of the reals is a formalization imposed by mathematicians on an informal notion, and we should stop assuming that when non-experts talk about the reals, they automatically mean the same thing mathematicians do. In fact, they usually mean the naive notion, ill-defined as it is. (And some people do genuinely seem to mean something closer to what we call the hyperreals than to what we call the reals.)HPThttps://www.blogger.com/profile/12272184283940783175noreply@blogger.com