[bonanova]

So I met this girl named Lisa, and we've gone out a couple times.

But, being a true puzzle geek, only a couple of times: basically I still have no social life.

Which probably explains why I spent last night doodling on a sheet of paper.

I started off with zeroes, since those poor fellows sometimes do need a little affirmative action.

And I crammed as many zeroes as I could onto my paper.

But, even for a puzzle geek, that soon became boring.

So, I glued two of them together and made an eight!

I thought briefly, but ultimately rejected the notion, of writing up a patent disclosure.

And I began to imagine cramming my paper full of figure eights.

Then I wondered: can I fit as many eights on the paper as I can zeroes?

Now I ask you the same question.

Assume your pencil is infinitely sharp: that the lines which draw the figures are mathematical lines - they have zero width.

Edit: The figures are disjoint: the lines may not intersect nor touch.

Edited September 20, 2008 by bonanova
Specifying the disjointness requirement.

[Guest]

Hey, Bona, been a while hasn't it?

Still dominating the "new puzzles" sections, I see.

Well, time to get used to my cluelessness again.

well. If your lines have zero width, then really, couldn't you make an infinite number of zeros on the paper? And with figure eights, they'd take up twice as much space, and if the lines had any real width, there would be half as many figure eights.

Well, 0*2 is still zero. And 2/0.. well.. Only Chuck Norris can divide by zero.

I just realised how little sense I'm making right now. But, the way I saw this question, it's kind of like asking, "what is infinity divided by two?" and the answer is... INFINITY!

I hope I didn't look at the problem wrong.

I think I did.. But whatever.

[Guest]

Hey, Bona, been a while hasn't it?

Still dominating the "new puzzles" sections, I see.

Well, time to get used to my cluelessness again.

well. If your lines have zero width, then really, couldn't you make an infinite number of zeros on the paper? And with figure eights, they'd take up twice as much space, and if the lines had any real width, there would be half as many figure eights.

Well, 0*2 is still zero. And 2/0.. well.. Only Chuck Norris can divide by zero.

I just realised how little sense I'm making right now. But, the way I saw this question, it's kind of like asking, "what is infinity divided by two?" and the answer is... INFINITY!

I hope I didn't look at the problem wrong.

I think I did.. But whatever.

If you can't use a zero for more than one eight, then

"8" = "0" /2

[bonanova]

Hey Ploper - good to see you again.

taliesin,

Yes, I meant disjoint figure eights: their lines may not intersect nor touch.

[andromeda]

Hm...

It depends on how you stack them, and this is my idea. Just looking with my naked eye it seems like I could stack more 8 than 0, but you probably had something else in mind

[Guest]

I guess the salient question is: do all the eights and zeroes have to be the same size?

You can fit an infinite number of each on the same paper, so the answer is, yes, you can fit the same number.

EDIT: typo

Edited September 20, 2008 by Chuck Rampart

[Guest]

...is that of a fractal. You can continuously break it down into smaller components for infinity. In these "0's" you can fit as many disjointed "8's" as you so desire. After you do that, though, you can fit as many "0's" in the holes of the "8's" as you want. Oh, but wait, there is more space in the "0's" for more "8's!" Ok, my head is spinning now.

[unreality]

When I read the puzzle, my first impression was a grid of 0, connected at one of four corners perpindicular to the sides of the paper - like a grid of zeroes. If you had a 10x10 grid, you have 100 zeroes. But then for each row you can make 9 figure-eights, 9x10 = 90 in the rows, and 90 in the columns, so you can make 180 eights but only 100 zeroes.

The formula for a*b dimensions is:

(ab) = number of zeroes

(a)(b-1) + (b)(a-1) = (ab - a + ab - b) = (2ab - a - b) = figure eights

But then largeneal pointed out that it's basically an infinitely regressing fractal

[bonanova]

The question really has to do with filling the entire plane with zeroes or eights.

I made it more visual, with a sheet of paper using and infinitely "sharp" pencil point,

allowing the dimensions to be arbitrarily small, as largeneal points out.

It doesn't matter whether you think of the paper or plane, so feel free to couch

your explanation for either case.

Is it possible that the counts for zeroes and eights are not the same?

[Guest]

At first I had my "8"s be the same size as my "0"s. But then I wanted to make the zeroes smaller, and made them half as big. And then I shrunk my eights to half of their original size. But then I saw that my eights were really just two half-sized zeroes (essentially, I worked backwards from the prompt). So however small my eights were, I could fit two zeroes in the same space as one eight. And since there were two zeroes for every eight, my count for zeroes was always twice as big as my count for eights. If it were possible to reach infinity, my zeroes would get there twice as fast as my eights as I made each digit infinitely small. Although infinity and infinity*2 may technically be the same thing, I think I've got more zeroes.

It reminds me of taking the limit of something like (x^2 - 1)/(x+5) as x approaches infinity. Both the top and bottom are infinite when x=infinity, but the top gets there faster, so the limit is infinity (or doesn't exist), instead of 1 (anything over itself = 1 = infinity/infinity). Then again, that's a higher power of x on the top, while the zero/eight discussion is just a higher coefficient.

[Guest]

At first I had my "8"s be the same size as my "0"s. But then I wanted to make the zeroes smaller, and made them half as big. And then I shrunk my eights to half of their original size. But then I saw that my eights were really just two half-sized zeroes (essentially, I worked backwards from the prompt). So however small my eights were, I could fit two zeroes in the same space as one eight. And since there were two zeroes for every eight, my count for zeroes was always twice as big as my count for eights. If it were possible to reach infinity, my zeroes would get there twice as fast as my eights as I made each digit infinitely small. Although infinity and infinity*2 may technically be the same thing, I think I've got more zeroes.

It reminds me of taking the limit of something like (x^2 - 1)/(x+5) as x approaches infinity. Both the top and bottom are infinite when x=infinity, but the top gets there faster, so the limit is infinity (or doesn't exist), instead of 1 (anything over itself = 1 = infinity/infinity). Then again, that's a higher power of x on the top, while the zero/eight discussion is just a higher coefficient.

I thought there was supposed to be a gap between the zeros. So an eight is not exactly two half sized zeros. it is two half sized zeros without the gap between them. Will that make a difference?

[Yoruichi-san]

Umm...I guess I'm kind of confused, as the act of drawing 0's and 8's seems to depend on what sizes you decide to make the 0's and 8's...but mathematically, I was thinking...

Assuming each 0 is a circle, and all circles are the same size. The optimal packing for circles (way of getting the most possible circles per unit area) is hexagonal packing, i.e. each circle is bordered by 6 other circles, which gives an efficiency of pi/sqrt(12) = 90.69%, independent of the size of the circle.

It seems to me, from the way the OP was written, the the 8's are just two circles attached, so it would seem like there would be half as many of them as circles...

But since the OP stated they can't touch, then you would have to separate the circles by a small increment from the 6 other circles around it, but you only have to separate each circle of the 8 from 5 other circles...I'm not sure exactly how this effects it though...gotta think some more...

[andromeda]

This not an attempt to untangle the mystery, but just a silly remark

8 is nothing but a 0 with a belt!

[bonanova]

Consider that the plane can be partitioned into unit squares.

We could place a 0 in each square or we could place an 8 in each square.

The squares have a 1-1 correspondence with the natural numbers.

So, as Chuck Rampart points out, at least a countably infinite number of 0's

and a countably infinite number of 8's can be placed in the plane.

But is that it? Or could there be more?

The title alludes to knowledge of how many 0's can be placed in the plane.

Let's see what that is.

Consider circles centered at the origin with radius r.

If r is restricted to rational values [p/q where p and q are integers],

this becomes another way to place a countably infinite number of 0's in the plane.

But let r take on real number values, and we have uncountably infinite 0's.

So now the question is whether you can do that with 8's.

Can an uncountably infinite number of 8's be placed in the plane?

[Guest]

Consider that the plane can be partitioned into unit squares.

We could place a 0 in each square or we could place an 8 in each square.

The squares have a 1-1 correspondence with the natural numbers.

So, as Chuck Rampart points out, at least a countably infinite number of 0's

and a countably infinite number of 8's can be placed in the plane.

But is that it? Or could there be more?

The title alludes to knowledge of how many 0's can be placed in the plane.

Let's see what that is.

Consider circles centered at the origin with radius r.

If r is restricted to rational values [p/q where p and q are integers],

this becomes another way to place a countably infinite number of 0's in the plane.

But let r take on real number values, and we have uncountably infinite 0's.

So now the question is whether you can do that with 8's.

Can an uncountably infinite number of 8's be placed in the plane?

Are you suggesting something like

[bonanova]

Are you suggesting something like

Sure.

how densely can 8's be nested?

1. Uncountably infinitely dense -> cardinality of the reals - like concentric circles?
   
2. Countably infinitely dense -> cardinality of the rationals?
   
3. Finitely dense -> only a finite number?
   

[bonanova]

Not much action on this so let's close it out.

Countably infinite instances of 0 or 8 can be placed by putting one of either in the unit squares on the plane.

Uncountably infinite 0's [with the cardinality of the real numbers] can be placed on the plane: center them [say] at the origin, with real radii r.

However, uncountably infinite 8's cannot be placed on the plane [without lines crossing.]

So you can place more 0's than 8's in the plane.

Proof:

Place an 8 anywhere.

Inside each loop of the 8 you can find a point having rational x and y coordinates.

This is because the rationals are dense in the plane.

No other 8 can enclose both of these points [without intersecting the first 8 somewhere].

Therefore each pair of points on the plane with rational coordinates corresponds to no more than one 8.

Therefore the 8's that can be placed on the plane have the cardinality of the rational numbers, i.e. are countably infinite.

Cardinality is a way of talking about how many members a set has.

Greater cardinality = more members.

For finite sets, cardinality is intuitive: you just count to a finite number and stop.

For infinite sets, you might be able to count all the members, but not be able to stop at any finite number.

The natural numbers 1, 2, 3, ... are clearly countable.

Or, you might not be able to count them at all. The real numbers are uncountable.

So two infinite sets may not have the same cardinality, as in this case.

You can read more about cardinality here.