Aren’t different size infinities called alephs or something?
Also doesn’t this puzzle make a jump in considering the universe to be continuous up until time zero when the assumption becomes that its discrete?
The reason the OP couldn’t answer his own question, was that he missed the class lab sessions where the infinity experiment is being done.
Never fear, even though he missed all the labs so far, he hasn’t missed even .00000001% of them.
Nope, I think the day before was the Xeno’s Paradox lecture and his professor told him to try the old “keep coming half way to class” trick. I believe he’s still realllllly cloooose to the classroom.
I say that there are zero chips in the dish, because, given any particular chip, that chip was removed at some time before the 2 hour mark (to use your coordinates). Hence, no chip can still be in the bowl by the time 2 hours rolls around. Therefore, at precisely 2 hours, the bowl is completely empty. In particular, it never has infinitely many chips in it.
A fun discussion, IMO:
http://www.friesian.com/calculus.htm
I got as far as that. When I stopped grimacing, I closed the window.
You should give that page another chance, ultrafilter. That paragraph is just expressing the apparent difficulty with calculus that mathematicians struggled with before the limit definitions. It then goes on to discuss how, in the 20th century, infinitesimals were given a respectible footing, and outlines a particular technique for doing this. The so-called non-standard analysis that studies these objects does actually have applications in other fields.
Ultrafilter quoted from Friesian.com:
I don’t buy that.
12/3 = 4 can be understood to mean that 3 can be subtracted from 12 4 times before exhausting the value of 12. But how many times for 12/0 can 0 be subtracted from 12 before exhausting the value of 12? Not even an “infinite number” (if there were such a thing) of times will do.
Nice puzzle, by the way, Owlofcreamcheese. Is it original?
I actually did go back and read the whole thing. My first impression was that the author was advocating the position she described, but apparently I was wrong.
Lib: That’s the reason I was grimacing. “undefined” doesn’t mean “infinite”, it means “undefined”.
Even assuming an infinite number of objects - chips- are being placed in a bowl…
If each one is placed in half the time of the one before, and the first one is in an hour, the eleventh will be given 3.5 seconds, the fourteenth gets .44 seconds, etc., until as the 2 hour mark nears, billions will be placed every second, but because they are infinite, there will never be an end and 2 hours will never be reached. It will be approached and to human perception will be reached, but…not quite, and never, and the chips will never all be in the bowl, because there will always be more. That’s the thing about infinity. There’s always more.
The way I was taught about division by zero was by dividing 1 by progressively smaller fractions. 1/.1 = 10, 1/.01 = 100, 1/.001 = 1000, and so on. The closer the denominator gets to zero, the greater the product, so that it approaches infinity, but never reaches it. So, any number divided by zero is effectively = infinity.
What about the sequence 1 / -0.1 = -10, 1 / -0.01 = -100, 1 / -0.001 = -1000, …? Doesn’t this imply 1 / 0 is negative infinity?
In order that we agree that 1/0 is infinity in the limit, it would need to be the case that f(x)/g(x) goes to infinity whenever there is a c such that, as x goes to c, f(x) goes to 1 and g(x) goes to zero.
But that isn’t the case. Consider f(x) = cos(x) and g(x) = x. For x > 0, f(x)/g(x) > 0, but for x < 0, cos(x)/x < 0. So if this function has a limit as x -> 0, it must be zero. But that’s not the case–in fact, |cos(x)/x| goes to infinity as x -> 0. So cos(x)/x goes to -infinity as we approach from the left, and infinity as we approach from the right. No limit.
btw, my reasoning is that if, as x -> c, f(x) -> a and g(x) -> b (b != 0), then f(x)/g(x) -> a/b as x -> c.
See, this is why I should never get involved in math questions.
It is possible, in certain circumstances, to evaluate limits of the form f(x)/g(x) where f and g go to zero at the limit in question (or infinity over zero, or zero over infinity, or blah blah blah). L’Hopital’s rule, right? Right. So. Are there any functions for which, no matter how many times one applies the rule (since it allows for recursion), one always comes up with 0/0 (or whatever)? I understand it can fail, but the only times I’ve seen the rule unable to be applied was when a derivative no longer yielded a continuous function.
Total hijack there, sorry, but that rule fascinates me.
Ah there is so much confusion over infinity. First place it is not a number, so talking about 1/0 = infinity is nonesense. That said, it it is certainly possible that some limits exist, such as f(x)/g(x) as x approaches some place that both f(x) and g(x) are 0 (or even if they both approach–not reach–infinity). Second place, there is no problem in positing and working with infinite and infinitesimal numbers. And 1/infinitesimal will be an infinite number and all the rules of arithmetic will be satisfied. But you still cannot divide by 0. All of which has nothing to do with the OP, which is one of the apparent paradoxes of the infinitel
If you have two exponential functions, L’Hopital’s rule won’t help. But there are other means for that case.
To return to the o.p.
The “god” that you speak of cannot be the Judeo-Christian God, for He is infallible, while this god erroneously says “well I bet 1 billion is in there.” So he must be a pagan god. Nothing that your jesus says is erroneous, so it seems that he might be the true Jesus Christ. However, he is speaking with this pagan god, which is clearly forbidden by the commandments (the Bible is pretty clear that there is only one true god). So he must be a pagan jesus.
To return to the original question,
You agree that there are infinitely many chips in the dish in the first example, right? All the odd numbers = infinitely many chips. Imagine that I was watching him put chips in the bowl, and you were watching him take them out. I would note “he put two in…two more in…two in…2…2…2…” and pretty soon I’d be talking kinda fast. You would note “he took one out…one more out…one out…1…1…1…” and pretty soon you’d be talking fast. We would be giving all of the information relevant to the number of chips currently in the dish, since that number is completely determined by 1) the # of chips in the dish to begin with (zero), 2) the number of chips that go in the dish (and maybe when they go in), and 3) the number of chips that come out (and maybe when they come out). But we’d be saying the exact same thing in the first case, when all the odd numbered chips stayed in the dish, as in the second case, when I-don’t-know-what chips end up in the dish. Since all relevant information is identical, the # of chips in the dish must be identical, i.e., infinite.
Here’s a related problem. Say you have a sequence of functions. The first function f(1) is equal to one on the interval from 1 to 2, including 2 but not 1. That is, it equals 1 for x such that 1 < x (< or =) 2. It is 0 everywhere else. I’ll write this as f(1) = 1 on (1,2] and 0 elsewhere. The second function is f(2) = 1 on (2,4] (and equals 0 elsewhere), the third f(3) = 1 on (3,6] (and is 0 elsewhere), the 647th f(647) = 1 on (647,1294] (and 0 elsewhere), and the nth is f(n) = 1 on (n,2n] (and f(n) = 0 elsewhere).
Note that the area under the nth function is equal to the number of chips in the dish after God’s nth “move” of putting chips in & taking them out, and that the numbers on the chips in the dish after the nth move are the integers in the interval where f(n) equals 1. As n approaches infinity (increases without bound), the area under the function (equivalent to the number of chips in the dish) approaches infinity (since it always equals n). Now, you can’t name any number where the function is nonzero (maybe you could say that it’s nonzero on the interval from infinity to twofinity, but that would be nonsense), but the area does not suddenly become 0 (i.e. no chips in the dish).
We’re looking at this problem in two different ways - I (like god) am looking at the number of chips in the dish (or the area under the function) for finite n, and taking the limit of that as the number of moves approaches infinity. You (like jesus) are looking at the particular chips in the dish (or the nth function) and taking a limit of that as n goes to infinity, and then looking at the area under that. In that case, the function approaches the function f that is equal to 0 everywhere, so the area under it is equal to zero. But when you switch between these perspectives, you’re interchanging limits, and you aren’t always allowed to do that without changing the answer. This is one of the cases where you can’t do it.
So, in a way, we’re both right. But, though it may be ok to just say that the two different limits have different values when you’re looking at functions, it’s harder to say that in the case of a concrete situation like this. You have to wonder, in this case, what interpretation fits the problem posed (i.e. which limit to take first). Since we’re only looking at the number of chips in the dish, I see no reason to go through the intermediary of the numbers on the chips - just count. And when you count, God has infinitely many chips in the dish after 2 hours.
Name one.
Cabbage, you are simply the best!