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View Full Version : Awesome math thing I thought of

Yerushalmi
07-28-2013, 08:09 AM
Okay, so imagine you have a line segment:

--------------

Now divide it into n portions of equal size.

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|1|2|3|....|n|

Now take an identical line segment and divide it into n+1 portions of equal size (each of which is therefore slightly smaller than the first set of portions).

--------------
|1|2|....|n+1|
(ignore the fact that the n+1 portion is bigger because I had to fit more characters in it - in reality it's the same size as all the others)

Label each portion with its number, as I've already done above.

What percentage of the total length of the line segment has the SAME label under both numbering schemes?

(An alternative way to think about the problem: When you ask a computer's random number generator for a number between 1 and x, it doesn't actually pick a random number between 1 and x, instead it picks a random decimal between 0 and 1, multiplies by x, adds 1, and rounds down. So really, asking the computer for a random number is asking it to pick a random place on a line segment and tell you how far along the line segment it is. Imagine therefore you asked the computer to pick a random number between 1 and x, and then you suddenly realize that you made a mistake and you should've picked a random number between 1 and x+1. What are the chances that you would've gotten the same number?)

I thought of this question, and discovered the answer, two nights ago, but since it was Shabbat I couldn't write out the algebra and figure out why that's the answer. I only worked out the "why" this morning.

Anyway, the answer will be forthcoming after a little while for you folks to stew over it. If, of course, anybody here thinks math is awesome besides me :p

Elite Lord Sigma
07-28-2013, 09:09 AM
I believe the answer is 50%. For any one segment of the line with n segments, the portion that does not correspond to the same numbered segment in the line with n+1 segments is represented by the equation 1/n - 1/(n+1). For example, suppose n = 2. Segment 1 of each line would take up 1/2 and 1/3 of the length, respectively. The difference in length is 1/2 - 1/3 = 1/6.

However, as it's getting late, I need a bit of time to reflect on what the correct way to go about proving that my answer is correct for all values of n.

Yerushalmi
07-28-2013, 09:13 AM
For any one segment of the line with n segments, the portion that does not correspond to the same numbered segment in the line with n+1 segments is represented by the equation 1/n - 1/(n+1).
That's only true for the first line segment. For instance, the very last portion of the n line segment, which is of course labeled n, does not correspond to the n+1 line segment for almost the entirety of its length! In fact, there the portion that does correspond is represented by that same 1/n - 1/(n+1).

Raistlin
07-28-2013, 09:43 AM
Well for n=1, the answer is 50%. For n=2, the answer is... still 50%, if my math is correct (and it's 3 AM so it may well not be; I was using 1 = 1 ft). For n=3, it's... still 50%.

For n=2, if 1 foot (12 inches) is the unit of measurement, then line 2 (L2) would be divided into 3 segments of 8 inches each. The first 8 inches would match, the next 4 wouldn't, and then 4 would match, and then the last 8 wouldn't. For n=3, the matching-unmatching inches of each foot (one segment in L1) would be, as far as I can tell, 9-3, 6-6, 3-9.

So my rudimentary guess would have to be 50%, but I am way too exhausted to try to figure out exactly why.

Math is fun. We need more math threads.

Elite Lord Sigma
07-28-2013, 09:50 AM
For any one segment of the line with n segments, the portion that does not correspond to the same numbered segment in the line with n+1 segments is represented by the equation 1/n - 1/(n+1).
That's only true for the first line segment. For instance, the very last portion of the n line segment, which is of course labeled n, does not correspond to the n+1 line segment for almost the entirety of its length! In fact, there the portion that does correspond is represented by that same 1/n - 1/(n+1).

Poor phrasing/exhaustion on my part; you're right. Going back to what I said, 1/6 is the difference between segment 1 in the two lines. The difference between segment 2 is 1/6 + the previous segment, 2/6. 1/6 + 2/6 is 1/2. I'm sure there's a series of some sort that can account for any value of n, but it escapes me at the moment. I'll give it some more thought in the morning.

The Man
07-28-2013, 09:58 AM
111111111111111111111111111111111111111111111111
111111111111111111111111222222222222222222222222

111111111111111111111111222222222222222222222222
111111111111111122222222222222223333333333333333

111111111111111122222222222222223333333333333333
111111111111222222222222333333333333444444444444

1111111111222222222233333333334444444444
1111111122222222333333334444444455555555

111111222222333333444444555555
111112222233333444445555566666

111111122222223333333444444455555556666666
111111222222333333444444555555666666777777

Yeah I'm getting 50% as well, unless I'm counting incorrectly. Too lazy to divide it up further, and I definitely no longer have the maths knowledge to do the proof of why it would always be 50%, but I'm pretty sure that's right. Unless I'm just misreading the problem, which is entirely possible.

Yerushalmi
07-28-2013, 10:34 AM

Let's look entirely at the n+1 line segment.

The 1-label of the n+1 system is smaller than that of the n system, but it begins at the same place. Therefore the entire length of that segment matches. This length is 1/(n+1) of the entirety of the line.

Like every other label, the 2-label of the n+1 system takes up 1/(n+1) of the entirety of the line, but it starts while the n-system is still labeled 1. By how much? Well, the n-system's 1-label is 1/n long, and the n+1-system's 1-label is 1/(n+1) long, so the n-system's 1-label is (1/n-1/(n+1)) too long. Only then does the overlap begin, so the total overlap is:

1 - ( 1 - 1
n+1 n n+1)

The 3-label, again, starts too early in the n+1 segment, but this time it needs to cover the (1/n-1/(n+1)) overlap created by both the segments that came before it. Therefore its overlap is:

1 - 2( 1 - 1
n+1 n n+1)

And so forth.

Summing the series we get:

1 + 1 - ( 1 - 1 + 1 - 2( 1 - 1 + 1 - n( 1 - 1
n+1 n+1 n n+1) n+1 n n+1) ... n+1 n n+1)

Taking everything we get out of the parentheses, we get:

1 + 1 - 1 + 1 + 1 - 2 + 2 + 1 - n + n.
n+1 n+1 n n+1 n+1 n n+1 ... n+1 n n+1)

Now, if you notice, other than the first item (which really should have a 0/n and a 0/(n+1) next to it), each triplet has two fractions divided by (n+1) and one fraction divided by n. Group the (n+1)s within each triplet together and you'll notice that each triplet in succession adds 1/(n+1) to 2/(n+1) to 3/(n+1), etc. until you reach (n+1)/(n+1). The n's are easier: subtracting 1/n, then 2/n, then 3/n, etc. until you reach n/n.

So we now have:
(1+2+3+...+n+(n+1))/(n+1) - (1+2+3+...n)/n

Now, any (1+2+3...x)/x is equal to (x+1)/2 (http://www.wolframalpha.com/input/?i=%281%2B2%2B3%2B...%2Bx%29%2Fx). So in the first fraction, where x=n+1, you have (n+1+1)/2=(n+2)/2. In the second fraction, where x=n, you have (n+1)/2.

((n+2)-(n+1))/2 = 1/2.

So. Awesome.

Chris
07-28-2013, 02:03 PM
I feel terrible for not understanding a single thing written in this thread. Back to being the slutty waterboy with the nice rack. :(

Jinx
07-28-2013, 02:09 PM
-----------12-33-45-5------------5454--------

I can write lines and numbers too :D