Proof... = 4

[yhelothar]

 

[Leros]

I have no idea if this proof is right, but:

 

[daw123]

LOL, very good.

 

[eLiu]

I have no idea if this proof is right, but:

LOL that's fucking awesome.

That proof is correct. Some things he says in the explanation aren't quite true but they don't affect the argument. Like he's discussing countable vs uncountable infinity. The fact that dividing off midpoints means you never count 1/3 has nothing to do with irrational numbers; it has to do with the denseness of rational numbers in the real numbers (which is irrelevant here). The rational numbers (numbers that can be written as p/q, where p,q are integers) are countable. Which means that there is a counting scheme that will count 1/3 & all possible other fractions. But there is no way to count so that you hit say sqrt(2)/pi.

The "approximation" in OP's proof never becomes a circle.

 

[yhelothar]

I don't get the explanation... it's saying something like, in order for that proof to be true, there must be a number of corners greater than infinity, and that circle only have infinite number of corners, so thus it's not a circle.

How do you have a number greater than infinity?

 

[rocadelpunk]

I don't get the explanation... it's saying something like, in order for that proof to be true, there must be a number of corners greater than infinity, and that circle only have infinite number of corners, so thus it's not a circle.

How do you have a number greater than infinity?

I haven't read the thread, but there are different sizes of infinity.

Some infinitys are larger than others.

 

[Leros]

I haven't read the thread, but there are different sizes of infinity.

Some infinitys are larger than others.

Simple example:

The set of all integers is twice as large as the set off all even integers, although both are infinite.

Edit: According to Dr Pizza this is wrong.

 

[Leros]

I don't get the explanation... it's saying something like, in order for that proof to be true, there must be a number of corners greater than infinity, and that circle only have infinite number of corners, so thus it's not a circle.

How do you have a number greater than infinity?

I think that in the case of the proof: ∅ is an infinite uncountable set and ∞ is an infinite countable set.

In that sense ∅ > ∞ since ∞ could potentially be a subset of ∅ , although I'm not sure if that is even a valid thing to say...

 

[yhelothar]

I think that in the case of the proof: ∅ is an infinite uncountable set and ∞ is an infinite countable set.

In that sense &#8709; < &#8734;, although I'm not sure if that is a valid thing to say...

How can there ever be an infinite countable set? Infinity by definition means never ending, and thus means can never be counted to exhaustion.

 

[Leros]

How can there ever be an infinite countable set? Infinity by definition means never ending, and thus means can never be counted to exhaustion.

Countable just means you can have a method which will eventually count all the numbers at t = infinity.

I can count all the natural numbers by starting with i = 0, and then incrementing i by 1. With enough time, I will eventually reach any natural number you select. Therefore the natural numbers is a countable set. You cannot develop such a method to count all the irrational numbers. Therefore irrational numbers are uncountable.

Basically if you can map a set to the natural numbers, then that set is also countable.

The elements of a countable set can be counted one at a time &#8212; although the counting may never finish, every element of the set will eventually be associated with a natural number.

http://en.wikipedia.org/wiki/Countable_set

 

[timosyy]

Because it's math.

 

[datalink7]

.999999.... = 1

 

[IronWing]

Basically the original proof ends up with ever finer saw teeth but no circle.

 

[Leros]

Basically the original proof ends up with ever finer saw teeth but no circle.

Doesn't an integral result in the same thing? Infinitely fine saw teeth of rectangles?

 

[mjrpes3]

Doesn't an integral result in the same thing? Infinitely fine saw teeth of rectangles?

Integrals are the area underneath a series of lines created by a set of points. For one, the lines aren't just vertical or horizontal, but can be diagonal. Second, there would be no such thing as a vertical line in an integral, since that would be two points of the same domain (maybe in the higher maths do that sort of wizardry but I never got that far ).

EDIT: err maybe ironwing is right... been a while.

 

[IronWing]

Doesn't an integral result in the same thing? Infinitely fine saw teeth of rectangles?

Yep, and it equals 4. But it never changes the basic shape from a saw blade to a circle. The lengths of the sides of the teeth race to zero at precisely the rate the number of teeth races to infinity.

 

[yhelothar]

Yep, and it equals 4. But it never changes the basic shape from a saw blade to a circle. The lengths of the sides of the teeth race to zero at precisely the rate the number of teeth races to infinity.

Would that make pi = 0 as lim-> inf of number of teeth?
:awe:

 

[CZroe]

"Removing the corners" would actually mean having four new angled lines connecting the remaining pieces of the original four lines. The proof expects us to just ignore that as if it makes sense to keep adding as much as each step removes, which isn't possible if a line or a point occupies no area. If it were a physical wire, you're really just "scruntching it up" and it'll be ready to uncoil/spring back out to the full potential area of the square. Obviously, the area of the square is not equal to the circle.

 

[DrPizza]

Simple example:

The set of all integers is twice as large as the set off all even integers, although both are infinite.

That's not two different sizes of infinity. That's the same size... there aren't twice as many integers. For EVERY integer, there exists exactly 1 even number that corresponds to it. It's a perfect 1 to 1 mapping.

I love the OP and 1st response. Awesome! I think I'll start my calc & pre-calc classes with these proofs after Thanksgiving break.

 

[Leros]

That's not two different sizes of infinity. That's the same size... there aren't twice as many integers. For EVERY integer, there exists exactly 1 even number that corresponds to it. It's a perfect 1 to 1 mapping.

My bad. Its been a while since I've done any discrete math or set theory stuff.

 

[Howard]

My bad. Its been a while since I've done any discrete math or set theory stuff.

I said the same thing you did several years ago here, but nobody had the technical background to set me straight at the time.

 

[IronWing]

Would that make pi = 0 as lim-> inf of number of teeth?
:awe:

No, you never run out of pie.

So, is pi always equal to 3.14159....? Or does the value of pi change in different reference frames or under acceleration? Would the value be the same near a large star as it is out in the sticks? Would it be the same measured for a circle "parallel" to a massive star as it would be perpendicular to the star?

 

[Fenixgoon]

doesn't work, because you end up with an infinite number of corners on the reduced "circle"

the fundamental geometry is not the same - the reduced circle may *approximate* the real circle in shape, but you still have the corners, no matter how small a reduction you make. the appropriate method to drawing a circle (i suspect) is to make a regualr N-gon where N approaches infinity, and compute the perimeter from that (which should come out to be Pi).

included paint picture: http://pics.bbzzdd.com/users/fenixgoon/pi_proof.png
lets assume this is zoomed in 10000000000000000x, you still have a corner that fails to match the geometry of the circle due to the algorithm used, which yields the extra perimeter.

 

[Sumguy]

Cool.

At least now we know that the perimeter of a circle of d=1 made up of infinitely small rectangles is greater than that of a true circle by 4-pi.

Edit: Assuming that the rectangles lie on the outside of the curve, that is (forgetting the term for this). What if all the rectangles were on the inside of the circle, resulting in a somewhat smaller approximation?













Man I fucking hate proofs.

 

[Cheesetogo]

That's a pretty interesting concept. If that reasoning held true, we could also show that the sum of the two sides of a right triangle are exactly equal to the hypotenuse.