Toolboxnj Posted February 10, 2005 Report Share Posted February 10, 2005 Was in my Philosophy of Science class (good class on Logical Positism and Hume) and my prof had a good riddle to open class tonight. I'll try and post as accuratly as possible (didn't write it down). There's a dollar making machine that can make unlimited amount of dollars at an ever faster rate. At t=0, the man comes up to you and gives you two options: a/ I'll give you 10 dollar bills with serial numbers 1-9, but you must burn #1 b/ I'll give you only 1 dollar bill At t=1/2 he gives you the same option Again at 3/4, 7/8, 15/16, 31/32, 63/64.. and so on. The machine only works faster and faster to churn out the bills at a higher rate of speed. He didn't give an answer to the riddle, but gave a couple arguments. I was wondering how Objectivists would answer the question. I'll put my feet in and say that since there is no infinity, you'd choose A. Peikoff writes in OPAR that infinity means "larger than any specific quantity" and since A=A, everything is finite (page 31). Is it an unanswerable riddle to an Objectivist then? BTW, he has said that although he doesn't personally consider himself a logical positivist/empiricist he does lean that way (if that helps). It's not a quiz or anything, just something to chew over in our heads over the weekend. Quote Link to comment Share on other sites More sharing options...

Walker Posted February 10, 2005 Report Share Posted February 10, 2005 I don't really understand the riddle. From what you've written, I don't understand what role the money making machine plays. All I see is that a man offers to give you two different pay outs each time period. One payout equals $8 and one equals $1. Since it's written as a perpetuity (an annuity that gives pay outs forever), you'd take the present value of each perpetuity ( = PMT/(going interest rate). But, the one that pays $8 will obviously have a greater present value than the $1. But again, I think I'm misunderstanding the problem. What is the deal with 't' - what is the defined period? What is the deal with the serial numbers? Quote Link to comment Share on other sites More sharing options...

AndrewSternberg Posted February 10, 2005 Report Share Posted February 10, 2005 I would collect gold and then refuse to accept dollar bills in payment for any of my goods or services. Quote Link to comment Share on other sites More sharing options...

Toolboxnj Posted February 10, 2005 Author Report Share Posted February 10, 2005 All I see is that a man offers to give you two different pay outs each time period.Â One payout equals $8 and one equals $1.Â Since it's written as a perpetuity (an annuity that gives pay outs forever), you'd take the present value of each perpetuity ( = PMT/(going interest rate).Â But, the one that pays $8 will obviously have a greater present value than the $1. That's correct in some sense. Remember, you have to "burn" the lowest value dollar every time with option #1. But again, I think I'm misunderstanding the problem.Â What is the deal with 't' - what is the defined period?Â What is the deal with the serial numbers? "T' merely defines time as a variable. The serial numbers identify the bill so the lowest number can be burnt. Quote Link to comment Share on other sites More sharing options...

danielshrugged Posted February 10, 2005 Report Share Posted February 10, 2005 That's correct in some sense.Â Remember, you have to "burn" the lowest value dollar every time with option #1. "T' merely defines time as a variable. The serial numbers identify the bill so the lowest number can be burnt. I still don't understand the riddle. Perhaps you could give an example of the first few iterations? Quote Link to comment Share on other sites More sharing options...

Hal Posted February 10, 2005 Report Share Posted February 10, 2005 (edited) I think the idea is that if you pick option 1) then every dollar bill gets burnt? Imagine the first 10 bills were numbered 1 through 10, the next lot numbered 11 through 20 and so on. Youd initially have to burn the dollar bill marked 1, then after you got given the next lot you'd have to burn 2, then 3, and so on. If you imagined that the 'giving/burning' thing was carried out infinite times t, then every dollar bill would eventually be burned and you'd have nothing left (the nth bill would be burnt after n*10 iterations). I took a maths class once on analysis since it seemed interesting, and we done something quite similar to this - its basically a question about summing infinite series. In maths, an infinite series is defined to be a sum of infinitely many numbers, eg 1+1+1+1+1+..., or whatever. Even though you are adding infinitely many terms, some series eventually converge to a finite value (eg 0.9 + 0.09 + 0.009 + ... converges to 1, and 1 + 1/4 + 1/9 + 1/16 + 1/25 + 1/36 (ie 1/n_squared) ... converges to pi squared divided by 6). However a series such as 1+1+1+1+1... will never converge to a finite value since it keeps on getting bigger, nor will the series (1+1/2+1/3+1/4+...). Now, some series can have negative terms as well as positive terms, such as the series 1 + (-2) + 3 + (-4) + ... If you replace all the negative terms with positive terms (in this case it would be 1+2+3+4+...) and the series converges, then it is said to be absolutely convergent. The important part is that if a series is absolutely convergent, then it doesnt matter in what order you sum the terms - it will always converge to the same value. However if the series isnt absolutely convergent then you cant do this - you will get different answers depending upon how you carry out the infinite sums - in fact you can prove that you can make the series sum up to any value whatsoever. Therefore a non-absolutely convergent series with infinitely many negative terms has no limiting value - you can prove that you can get any value Your teacher is basically asking you to sum the series 10 - 1 + 10 - 1 + 10 - 1 + ... Since this series isnt absolutely convergent, this is impossible, and you can show that the sum can be equal to anything, depending on how you add the terms. Your teacher is basically arranging the terms in a way that the sum is equal to zero in the limit. It could just as well come out to be 27 or 2343, if they were arranged in a different way. For a clearer example of what I mean, consider doing the following sum: 1 - 1 + 1 - 1 + 1 - 1 + ... This series isnt absolutely convergent, so you can get different answers depending on how you do the sum. For instance: (1 - 1) + (1 - 1) + (1 - 1) + ... is going to be equal to zero, whereas 1 - (1 - 1) - (1 - 1) - (1 - 1) - ..., which is the exact same sum, is going to be equal to 1. This kind of thing caused difficulties for mathematicians before the theory of series was properly formalized, and led to people putting forward 'proofs' that 0 = 1 and so on. Edited February 10, 2005 by Hal Quote Link to comment Share on other sites More sharing options...

danielshrugged Posted February 10, 2005 Report Share Posted February 10, 2005 (edited) (1 - 1) + (1 - 1) + (1 - 1) + ...Â is going to be equal to zero, whereas 1 - (1 - 1) - (1 - 1) - (1 - 1) - ..., which is the exact same sum, is going to be equal to 1. But in the first case you have 6 numbers and in the second you have 7. You've stopped the two series at different times--that's why your claimed sum differs in each case. In other words, all you've said is that: 1 - 1 + 1 - 1 = 0 whereas 1 - 1 + 1 = 1 This isn't at all unusual. Obviously, if you keep going, you will have a sum of 0 if there are an even number of terms and a sum of 1 if there are an odd number, no matter how long a series one takes. ---- I still don't get the original riddle. It seems like in (A) one gets 9 additional dollars bills with each iteration, as opposed to 1 additional dollar bill with each iteration in (. Not a tough choice, if that's all that's going on. Edited: to change a smiley face into the letter B. Edited February 10, 2005 by danielshrugged Quote Link to comment Share on other sites More sharing options...

Aurelia Posted February 11, 2005 Report Share Posted February 11, 2005 But in the first case you have 6 numbers and in the second you have 7. You've stopped the two series at different times--that's why your claimed sum differs in each case. In other words, all you've said is that: 1 - 1 + 1 - 1 = 0 whereas 1 - 1 + 1 = 1 This isn't at all unusual. Obviously, if you keep going, you will have a sum of 0 if there are an even number of terms and a sum of 1 if there are an odd number, no matter how long a series one takes. What I think your excluding here is that the series is infinate. There is no such thing as odd/versus even amount of numbers in infinity. There is no stopping of the series. As far as the original riddle, I have no idea, I've never had a formal education in series and I haven't read about them in a while. Though, you may want to read up on Zeno's paradox to understand how a series works (and how they can be used fallaciously), as I recall, that helped me. ~Amanda Quote Link to comment Share on other sites More sharing options...

danielshrugged Posted February 11, 2005 Report Share Posted February 11, 2005 What I think your excluding here is that the series is infinate. There is no such thing as odd/versus even amount of numbers in infinity. There is no stopping of the series. ~AmandaÂ I didn't claim there was. I merely said that however long the string of numbers, you can determine the sum by whether there is an even or odd amount. As the amount of numbers approaches infinity, therefore, it approaches NEITHER 0 nor 1. Neither of those is the sum of the series at infinity. Quote Link to comment Share on other sites More sharing options...

Hal Posted February 11, 2005 Report Share Posted February 11, 2005 (edited) I didn't claim there was. I merely said that however long the string of numbers, you can determine the sum by whether there is an even or odd amount. As the amount of numbers approaches infinity, therefore, it approaches NEITHER 0 nor 1. Neither of those is the sum of the series at infinity. But it does approach either 0 or 1, depending upon how you are adding the terms. Let's say we take the terms in the above series and rearrange them to give: 1 + 1 + 1 + 1 + 1 + (1-1) + (1-1) + ... which appears to sum to 3 (the fact that some 1's are being 'borrowed' from the 'end' of the series doesnt matter - you have infinite 1's infinite -1's, so you arent going to run out of either "infinity minus infinity equals infinity" (scare quotes intended)). Since the series doesnt converge absolutely, the order in which you add the terms will determine the sum in the limit. More formally: Let S be the sum of the series (1 - 1) + (1 - 1) + ..., ie S = (1 - 1) + (1 - 1) + ... = 0 but then S = 1 - { (1 - 1) + (1 - 1) + (1 - 1) + ...} = 1 - S = 1 - 0 = 0 Hence 1 = 0. here is a slightly more indepth explanation. Edited February 11, 2005 by Hal Quote Link to comment Share on other sites More sharing options...

danielshrugged Posted February 11, 2005 Report Share Posted February 11, 2005 (edited) [url=http://courses.ncssm.edu/math/TCMConf/TCM2002/talks2002/Alt%20Series.pdf)</div><div class='quotemain'><!--QuoteEBegin-->hereis a slightly more indepth explanation. I'll say this much: whatever concept of limit being used here, it does not seem to be Newton's, and I find it hard to see how any of that math applies to the real world. That said, I have not considered whether there might be some mathematical use for this. Edited to fix quote. Edited February 11, 2005 by danielshrugged Quote Link to comment Share on other sites More sharing options...

Capitalism Forever Posted February 11, 2005 Report Share Posted February 11, 2005 If I understand it correctly, the riddle is asking the question, "What is more: 8 x Infinity or 1 x Infinity?" Quote Link to comment Share on other sites More sharing options...

Toolboxnj Posted February 13, 2005 Author Report Share Posted February 13, 2005 If I understand it correctly, the riddle is asking the question, "What is more: 8 x Infinity or 1 x Infinity?" Well, what about the 1 bill that you burn. Aren't you burning infinate bills, even if you burn one. So, wouldn't it be 8 infinity - 1 infinity? I'll know on Monday. Quote Link to comment Share on other sites More sharing options...

EC Posted February 13, 2005 Report Share Posted February 13, 2005 Theres no such thing as "8infinity" or "1infinity" and infinity - infinity is undefined. From the way you set the problem up it looked like T was approaching a limit of 1. From what I can tell or how you set the problem up it just looks like our real choice is with which choice is approaching infinity quicker. This would have to be choice b since it's growing 9 times quicker than choice a. So if the question is: which choice will get me rich quicker it would have to be choice b. If the question is as each choice approaches infinity which one is "bigger" (larger in number) it would always be the second choice because it grows quicker. But when your dealing with each choices infinities qua infinity that one set is not larger than the other. Infinity is Infinity. Maybe realizing each infinity is the same "size" is this assignments philosphical purpose? Quote Link to comment Share on other sites More sharing options...

coirecfox Posted February 13, 2005 Report Share Posted February 13, 2005 I just wanted to point out to a few people that infinity is not a number. It is a concept. It is used to represent the fact that the number series is non-terminating. You cannot count to infinity. If you say to yourself I am going to count to infinity, and then begin: 1,2,3,4,5,6,7--you have stopped at seven. In essence infinity has become seven. Not to say that infinity=seven, but that is where you stopped. "Infinity" is the potential number to which you could have gotten and "seven" is the actual. The concept 'infinity' bears no relation to things that exist, because everything that exists is finite. Quote Link to comment Share on other sites More sharing options...

softwareNerd Posted February 13, 2005 Report Share Posted February 13, 2005 Well, what about the 1 bill that you burn.Â Aren't you burning infinate bills, even if you burn one.Â So, wouldn't it be 8 infinity - 1 infinity? I'll know on Monday. I'm not the only one who has trouble understanding the problem. Well, actually, the answer is so obvious, that I assume I have misunderstood it. Working out the first few iterations from the two options: Option 1: Time 0: (10 - 1) = $9 Time 1/2: 9 + (10 - 1) = $18 Time 3/4: 18 + (10 - 1) = $27 Option 2: Time 0: 1 = $1 Time 1/2: 1 + 1 = $2 Time 3/4: 2 + 1 = $3 You see why I guess that I do not understand the problem? The way I've understood it, even the village idiot can tell which one is better. (Can't vouch for for college philosophy professors, though . ) Quote Link to comment Share on other sites More sharing options...

Capitalism Forever Posted February 14, 2005 Report Share Posted February 14, 2005 You see why I guess that I do not understand the problem? The way I've understood it, even the village idiot can tell which one is better. Try Time 1. Quote Link to comment Share on other sites More sharing options...

softwareNerd Posted February 14, 2005 Report Share Posted February 14, 2005 (edited) Try Time 1. How is that different? Either, Time 1 does not happen, in which case why worry about it? Or, more likely, I approach it and am still better off. (I say likely, because I do not know enough of the math to tell you what a mathematician would say. All I really care about is reality.) Edited February 14, 2005 by softwareNerd Quote Link to comment Share on other sites More sharing options...

demiac Posted March 29, 2005 Report Share Posted March 29, 2005 Either, Time 1 does not happen, in which case why worry about it? Time 1 happens when the machine runs for a second. At this point both options have given you an amount of money that are sums of diverging series: infinite. Infinity is not defined well enough, so it doesn't matter which do you choose, you get a undefined amount of money anyway. Quote Link to comment Share on other sites More sharing options...

TomL Posted March 29, 2005 Report Share Posted March 29, 2005 Time 1 happens when the machine runs for a second. And therein lies the error in the problem itself. Once the machine has run for one second, it cannot have made an infinite number of bills. Machines operating in this universe have a finite capacity for production, especially over a short time period. The whole problem is arbitrary and of no concern to anyone with a head for reality and for solving problems which actually exist in this universe. This one doesn't. Quote Link to comment Share on other sites More sharing options...

softwareNerd Posted March 29, 2005 Report Share Posted March 29, 2005 ... Infinity is not defined well enough, so it doesn't matter which do you choose, you get a undefined amount of money anyway.Since you seem to understand the question, let me ask this: at any particular point that is less than 1 second, which option is better? Also, at any point after one second, which option is better? Quote Link to comment Share on other sites More sharing options...

demiac Posted March 31, 2005 Report Share Posted March 31, 2005 And therein lies the error in the problem itself.Â Once the machine has run for one second, it cannot have made an infinite number of bills.Â Machines operating in this universe have a finite capacity for production, especially over a short time period. That is true. There are (at least) two impossible paths the events can go: we have infinite amount of money, or time stops at 1. The whole problem is arbitrary and of no concern to anyone with a head for reality and for solving problems which actually exist in this universe.Â This one doesn't. It has a concern to anyone with a head for reality who chooses so, and the solution these look for is that the problem is arbitrary. Since you seem to understand the question, let me ask this: at any particular point that is less than 1 second, which option is better? Also, at any point after one second, which option is better? I really don't know the answer to either. If a series of events has an impossible event (when t=1), do the other events really happen either? Quote Link to comment Share on other sites More sharing options...

Capitalism Forever Posted March 31, 2005 Report Share Posted March 31, 2005 Since you seem to understand the question, let me ask this: at any particular point that is less than 1 second, which option is better? At t=0, option a is better because it leaves you with $8, while option b would yield you $1. At t=1/2, option a would increase your earnings by $8, while option b by only $1; the same is true for any t=(n-1)/n. So, for n=1 (that is, t=0), option a is better; if option a is better for any n, it is better for n+1 as well; therefore, option a is better for any n >= 1. A nice, if pointless, example of induction. Also, at any point after one second, which option is better? That is a theology question. Quote Link to comment Share on other sites More sharing options...

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