EventHorizon

Inspired by I thought I would give my own. 2,4,3,1,6,5,7 3,7,5,6,1,8,2,4 2,7,6,3,9,4,8,5,1 1) What would it be for 10 numbers? What method am I using? Here's another one (but it is (hopefully) harder, so I'll give all of them up to 9): 1 1,2 1,2,3 4,1,3,2 3,1,4,5,2 6,3,2,5,4,1 1,4,3,2,6,5,7 5,7,8,6,4,1,3,2 4,5,8,6,2,7,3,1,9 2) What is the order for 10 numbers? What is the method?

And another excerpt from the page for "0.999..." So since decimal places are the digits after the decimal point, it seems there is one pair of equal decimal expansion representations for each positive integer and each negative integer. And this is exactly the set of all the numbers that fit the OP. Any other terminating decimal expansion will have the pair of representations equal at one or more decimal places. It is interesting that 0 fails in this regard. The method of decimal representation we use has no way of approaching 0 with repeating 9's due to the fact that adding value in a decimal place always increases the distance from 0. If you change the OP to require a different digit in each place (not just decimal place, and can't simply not have a value explicitely put in a place (ie, both 0's in hundreds place of 34 and 67)), you find that at least one number needs to be infinite. You obviously cannot have an infinite number equal to a finite one. But what about two infinite numbers? Sure, if you subtract the decimal representations of two infinite numbers created by an infinite decimal expansion, you could end up with a finite decimal expansion. But it seems it isn't possible in the case where the digits are different at every position. If you try the "trailing 9's" approach you end up with ....99999.9999... = ...00000.0000..., but the number on the right is 0, while the other is infinite. But are all infinite numbers created by this decimal expansion method equal anyway? For instance, is ...22222.2222... = ...11111.1111... ? You can keep subtracting 1 from both and never reach a negative number (or even a finite number). I'm thinking every infinite number created using this method of decimal expansion is equal (and equal to aleph-naught), and so the subtraction of the decimal expansions doesn't really apply since all are the same number/concept... and the difference is simply undefined. Sure, there are higher degrees of infinity, but they are essentially 2^(one of these infinite number) or more in "value." Hyperreals sound interesting. I can't, at the moment, understand how there can be a non-zero infinitesimal. I'll quickly look that up. It seems that they handle it essentially like the different degrees of infinity, and just call all the infinitesimals nonzero. It seems to me like they are kept as two separate concepts, and when you order values you order them first by its "standard part" and then by the infinitesimals included. Anyone have thoughts on any of this or disagree vehemently on a point?

I found a mistake in the derivation, so my equation is wrong. I accidently omitted something which made it much easier to solve, but incorrect.

(Hint: don't look for the curve, just examine the distance between them and how it changes with time) I'll try (again) to find the equation for the path in a bit. Edit: I thought I'd include the last equation I got, but kept failing to verify. So I don't know it if is correct.

So now we just need to determine if it matters if the frictionless tunnel goes through the center of the earth or not.

This means that you can completely ignore the parts of the earth further than you from the center of the earth, so effectively you feel the gravity of standing on a smaller earth. The effective mass of this smaller earth would be d*(4/3)pi*r2^3, where d is the density of the earth (assumed in the OP as homogeneous) and r2 is your distance from the center of the earth. Notice that this effective mass is equal to the mass of the earth multiplied by r2^3/r^3. Plugging the new mass and radius into the equation to calculate gravitation gives f2 = GmE(r2^3/r^3)/r2^2 = (r2/r)*GmE/r^2 = (r2/r)*9.8m/s^2 So now we've got the gravity we can plug into some calculus. Here's the setup for a tunnel through the center of the earth. d2x/dt2= -9.8*x/r dx/dt (0) = 0 x(0) = r where r = (40*10^6)/2pi. (time to pull out my old diff eq textbook... since I remember essentially nothing from that class.) x''+(9.8/r)x=0 roots of the characteristic equation are +/- sqrt(9.8/r) * i. x = c1* cos(sqrt(9.8/r)*t) + c2* sin(sqrt(9.8/r)*t) The equation goes through half an oscillation (the answer we're looking for) in pi/(sqrt(9.8/r)) =pi*sqrt®/sqrt(9.8). =pi*sqrt(40*10^6/(2pi)))/sqrt(9.8) =2532.0769 seconds. =42.2 minutes I think I'll wait a bit to look at a tunnel not going through the center of the earth.... because I'm not sure I have much confidence in what I just did. But if it is right, falling through a vacuum to the center of the earth would take 20 minutes. That's kinda interesting. Does that sound plausible to anyone?

So it looks like I had paradoxes on the mind when drafting that post. What I meant (in each of the three mentions of the word paradox) was contradiction. I hate when I read through one of my posts and find misused words. Too bad I'm no longer a VIP or I'd be able to fix them. Edit: Hey, I'm a VIP again, thanks rookie1ja.

Didn't notice psykomakia's post. That's what i get for making dinner while puzzling.

Edit: Fixed terminology.

Third, fold again using a fan-like method as shown in the diagram below, alternating forwards and backwards folds. Now all the perforation and only the perforation is along that diagonal... so letta rip.

If you cannot assume...