1=0.999... infinities and box of chocolates..Phliosophy of Math...

[hansda]

The above post assumes there can be a "smallest positive number" and that this allows one to have consecutive points in geometry. However consecutive points are not a feature of either Euclidean geometry or set theoretical constructions equivalent to Euclidean geometry. Thus both the novel geometry and novel definition of number would have to be developed before a case for the proposition being sound could be developed.

What i meant to say is that distance between two consecutive points in geometry has to be the smallest, so that no third point can be placed in between these two consecutive points.

If 1/2 is a number and ϵ is the "smallest positive number" described by hansda above, then is 0 ≤ (1/2) � ϵ ≤ ϵ < 1/2 < 1 true? Then is 0 < (1/2) � ϵ < ϵ false so that ϵ really is the smallest positive number? In that case, either 0 = (1/2) � ϵ or ϵ = (1/2) � ϵ so that 2 � ( (1/2) � ϵ ) = (1/2) � ϵ + (1/2) � ϵ â‰* ϵ. Does this not seem like a very strange property for a number to have?
Also ϵ − (1/2) � ϵ â‰* (1/2) � ϵ follows from the same assumption that ϵ is the smallest positive number.

As there can not be any third point, between the two consecutive points, the distance between these two consecutive points has to be non-zero, non-divisible and infinitesimal.

 

[someguy1]

Like I said, twice as many!

No, exactly as many. If x is an infinite cardinal there is an obvious bijection between x and 2x. Just think about how you would correspond the natural numbers and the integers.

0, 1, -1, 2, -2, 3, -3, etc.

You can biject an uncountable set x with 2x also although you'd need a different proof.

 

[Motor Daddy]

No, exactly as many. If x is an infinite cardinal there is an obvious bijection between x and 2x. Just think about how you would correspond the natural numbers and the integers.

0, 1, -1, 2, -2, 3, -3, etc.

You can biject an uncountable set x with 2x also although you'd need a different proof.

No, twice as many!

If a length of 1 unit has a quantity of points, then 2 units has twice as many points!

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

In mathematics, a bijection (or bijective function or one-to-one correspondence) is a function between the elements of two sets. Every element of one set is paired with exactly one element of the other set, and every element of the other set is paired with exactly one element of the first set. There are no unpaired elements. In formal mathematical terms, a bijective function f: X → Y is a one to one and onto mapping of a set X to a set Y.

See, the difference is that the length of the radius is not mapped equally to the length of the diameter. It just happens to be a fact of nature that a diameter is twice as long as the radius, or a radius is 1 part of 2, 0.5, 50%, half as many.

If you have an option to pick a point on a radius does that mean you have access to every point on the diameter?

 

[someguy1]

No, twice as many!

If a length of 1 unit has a quantity of points, then 2 units has twice as many points!

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



See, the difference is that the length of the radius is not mapped equally to the length of the diameter. It just happens to be a fact of nature that a diameter is twice as long as the radius, or a radius is 1 part of 2, 0.5, 50%, half as many.

If you have an option to pick a point on a radius does that mean you have access to every point on the diameter?

The idea that you would try to disagree with my point by quoting the Wiki article on bijections is hilarious. If you'd take the trouble to read the article your misunderstandings would be corrected.

 

[Motor Daddy]

The idea that you would try to disagree with my point by quoting the Wiki article on bijections is hilarious.

What's even more hilarious is that you think every and any point on a diameter can be mapped to a point on the radius. Hint: There are points on the diameter that are not on the radius.

 

[someguy1]

What's even more hilarious is that you think every and any point on a diameter can be mapped to a point on the radius. Hint: There are points on the diameter that are not on the radius.

Well, let's examine this.

There are integers that are not natural numbers. For example, -3 is an integer that's not in the set of natural numbers.

But all the natural numbers are integers. The integers include all the natural numbers, plus all the negatives of all the natural numbers, plus zero.

Yet, we can biject the natural numbers to the integers, like this:

1 <-> 0

2 <-> 1

3 <-> -1

4 <-> 2

5 <-> -2

etc.

What do you think about that?

 

[Motor Daddy]

Well, let's examine this.

There are integers that are not natural numbers. For example, -3 is an integer that's not in the set of natural numbers.

But all the natural numbers are integers. The integers include all the natural numbers, plus all the negatives of all the natural numbers, plus zero.

Yet, we can biject the natural numbers to the integers, like this:

1 <-> 0

2 <-> 1

3 <-> -1

4 <-> 2

5 <-> -2

etc.

What do you think about that?

How about instead of coming up with more illusions and pulling more rabbits out of your hat, that instead you answer my question: If I ask you to pick a point on a radius that has an infinite quantity of points, are there any points along the diameter that you don't have access to?

 

[someguy1]

How about instead of coming up with more illusions and pulling more rabbits out of your hat, that instead you answer my question: If I ask you to pick a point on a radius that has an infinite quantity of points, are there any points along the diameter that you don't have access to?

I'm in complete agreement with you that the length of the diameter is twice the diameter of the radius. Two sets can have very different measure, yet the same cardinality.

The interval [0,2] has twice the length, or measure, as the interval [0,1]. But they have the same cardinality, as can be seen from the bijection x <-> 2x. They have the same number of points, yet different lengths.

I agree with you that this is counterintuitive. But it's been known for a long time.

It's even true that there are as many points in the unit interval [0,1] as there are in the unit square. You can go up in dimension and still have a bijection. Infinity's just funny like that.

 

[Motor Daddy]

I'm in complete agreement with you that the length of the diameter is twice the diameter of the radius. Two sets can have very different measure, yet the same cardinality.

The interval [0,2] has twice the length, or measure, as the interval [0,1]. But they have the same cardinality, as can be seen from the bijection x <-> 2x. They have the same number of points, yet different lengths.

I agree with you that this is counterintuitive. But it's been known for a long time.

It's even true that there are as many points in the unit interval [0,1] as there are in the unit square. You can go up in dimension and still have a bijection. Infinity's just funny like that.

You failed to answer the question. Why? Are you afraid of telling the truth? Afraid to admit that there are points on the diameter that are not on the radius?

 

[rpenner]

http://www.smbc-comics.com/?id=3161

 

[hansda]

Hansda, I feel you are drawing your points as mere points when the question is about the value of 1 and 0.999... and not just mere points.

A line consists of points. In a line, all the points correspond to real numbers. '1' and '0.999...' are also real numbers. So, these numbers will also correspond to some points in the line. I mean to say that, these two numbers and their corresponding points in the line will be two consecutive points; so that no third point can come in between these two points.

if you draw a point that indicates the sum value of something... say one point is made at 1/2 we are already implying a point at 1.


But we are not talking about merely two consecutive points. We are talking about a sum value of 1/2 of 1

The point at the 1/2 is the sum value of half of one.

When considering 0.999...= 1 the same issue presents. IMO

Which is why I drew the diagram below as posted earlier:

where by the sum total of 0.999... of 1 = 1

For 0.999... to equal 1 it can not follow 1 ... it it has to equal 1

What point you are trying to prove here?

 

[hansda]

Say for example you have a 3 dimensional volume that has an infinite number of points located with in it and the volume is filled with these points.

Qu: What is the distance between each point?

Is it zero or is it 1/infinity?

Distance between two points can not be zero. Distance between two points has to be non-zero. If the distance between two points is zero, then both these points are at the same location.

If it is zero then the volume of space can NOT be filled with points. [ no points can exist other than one zero point ]

As i explained above, this option is not true.

If it is 1/infinity (infinitesimal) the volume of space CAN be filled with zero points to make up the volume...

What do you mean by "zero points/point"? By definition all the points are dimension-less(ie radius zero).

Therefore my case is that a zero point must have the diameter of 1/infinity for it to exist in any form, as shown in this diagram:

If a point is having a non-zero diameter, it is no longer a point. It is either a circle or a sphere.

The problem [paradox] of zero dimensionality in a three dimensional space will not go away...

'Zero dimensionality' in a 3D space is a point(by definition). Where/What is the paradox/problem?

 

[Motor Daddy]

A line consists of points. In a line, all the points correspond to real numbers. '1' and '0.999...' are also real numbers. So, these numbers will also correspond to some points in the line. I mean to say that, these two numbers and their corresponding points in the line will be two consecutive points; so that no third point can come in between these two points.

What number corresponds to the point that precedes the point that corresponds with the number .999...?

 

[hansda]

What number corresponds to the point that precedes the point that corresponds with the number .999...?

It will be 0.999... - (1 - 0.999... ) .

 

[Motor Daddy]

It will be 0.999... - (1 - 0.999... ) .

What does (1-0.999...) equal?

 

[someguy1]

You failed to answer the question. Why? Are you afraid of telling the truth? Afraid to admit that there are points on the diameter that are not on the radius?

I agreed that there are points on the diameter not on the radius; but that nevertheless, the points on the radius and the points on the diameter are in 1-1 correspondence if you assume there are infinitely many points. I said this several times already.

 

[arfa brane]

A line consists of points. In a line, all the points correspond to real numbers. '1' and '0.999...' are also real numbers. So, these numbers will also correspond to some points in the line. I mean to say that, these two numbers and their corresponding points in the line will be two consecutive points; so that no third point can come in between these two points.

If no third point exists (can be found) between two points, then the two points cannot be consecutive, they must be the same point in that case.
You're confusing the integers with the reals: there is no integer between two consecutive integers and a finite number between any two distinct integers, the reals have an infinite number of points between any two points which are distinct.

What must hold, therefore, is that the integers are arbitrary 'distances' (or marks) on the real line; if there is no integer between any two of these then the "real" distance between them can be arbitrary, since none of these distances contains a number which is an integer.
Conventionally, unit distances are 'laid off' along a section of the real line, but this only corresponds to arranging the elements of a set (like, in a row) so they're easier to count.

 

[Motor Daddy]

I agreed that there are points on the diameter not on the radius; but that nevertheless, the points on the radius and the points on the diameter are in 1-1 correspondence if you assume there are infinitely many points. I said this several times already.

What you are saying contradicts itself. You agree there are points on the diameter that are not on the radius, and yet in the same breath say there is a 1-1 correspondence. There is not a 1-1 if there are points on the diameter that are not on the radius.

Imagine a blue line about 1 meter long. Now imagine a red line .5 meters long that's matched to half of the blue line. All the red points are matched to a blue point, but only half of the blues are matched to a red. If blue points were males and red points were females, all the females would have a partner but only half the males would have a partner.

WHY? Because there are TWICE AS MANY points on a diameter as there are on the radius!

 

[someguy1]

What you are saying contradicts itself. You agree there are points on the diameter that are not on the radius, and yet in the same breath say there is a 1-1 correspondence. There is not a 1-1 if there are points on the diameter that are not on the radius.

Imagine a blue line about 1 meter long. Now imagine a red line .5 meters long that's matched to half of the blue line. All the red points are matched to a blue point, but only half of the blues are matched to a red. If blue points were males and red points were females, all the females would have a partner but only half the males would have a partner.

WHY? Because there are TWICE AS MANY points on a diameter as there are on the radius!

Since what you say contradicts well-known and widely accepted modern mathematics, the burden's on you to refute or modify the current theories. I showed you the example with the natural numbers and the integers. I have already said the same things to you several times, so I guess I have to let this one go. You are entitled to keep repeating things that are at odds with well established results. I can't add anything to what I've said.

 

[Motor Daddy]

Since what you say contradicts well-known and widely accepted modern mathematics, the burden's on you to refute or modify the current theories. I showed you the example with the natural numbers and the integers. I have already said the same things to you several times, so I guess I have to let this one go. You are entitled to keep repeating things that are at odds with well established results. I can't add anything to what I've said.

Thanks for playing.