Deeper than primes

[zooterkin]

There is no exact location on a line where there is no point, and ≠ is such non-exact location on a line, which exists between any given arbitrary pair of distinct exact locations (arbitrary pair of distinct points).

Back to the gibberish, I see.

 

[doronshadmi]

Back to the gibberish, I see.

Never gives up of ignorance, I see.

 

[doronshadmi]

That's not true. The function f(x) = (x² - 4)/(x - 2) is not defined for x = 2, so you can't see the point drawn at that exact location.

[qimg]http://img138.imageshack.us/img138/5445/missingpoint1.png[/qimg]

The OM seems to be a very different computational method. Some drawings would be helpful . . .

Epix, you can use any function that you wish.

It does not change the fact that there is ≠, which exists between any given arbitrary pair of distinct exact locations (arbitrary pair of distinct points) that are the strict results of the given function.

≠ is definitely free of any strict results, which gives it the ability to be used both as differentiator AND integrator of the concept of Pair.

Without, for example, 1(0(x)≠0) there is no pair.

 

[epix]

Epix, you can use any function that you wish.

It does not change the fact that there is ≠, which exists between any given arbitrary pair of distinct exact locations (arbitrary pair of distinct points) that are the strict results of the given function.

Who says that it isn't so? Any x_i that is a part of R is distinct, such as x_i-1 < x_i < x_i+1. That's why you can see a line. You claim that point [x = 2, y = 4] doesn't have an exact location. Did you hack the Hubble telescope and see something not round "at the end" of 2.000... and 4.000...?

Eventually, there has to be point p > 2 that you express as p = 2.000...1. That means the "strict" number 2 is not immortal. Or is it?

You are applying your everyday experience into an abstract world that is not physical, and that creates a conflict. But that's normal:

a⁴ = a * a * a * a

a³ = a * a * a

a² = a * a

a¹ = a

a⁰ = ?

a⁰ seems to be the same case as a/0.

12/4 = 12 - 3 - 3 - 3

12/3 = 12 - 4 - 4

12/2 = 12 - 6

12/1 = 12

12/0 = ?

Both analogies are not identical, as far as the result is concerned, even though your intuition tells you that they are.

 

[doronshadmi]

You claim that point [x = 2, y = 4] doesn't have an exact location.

Wrong, 1(0(2)≠0(4)) is 1() (notated by ≠) between strict numbers.

Eventually, there has to be point p > 2 that you express as p = 2.000...1. That means the "strict" number 2 is not immortal. Or is it?

2.000...1() is a non strict number (where the concept of Number is a measurement tool, whether the measured is a strict or non-strict level of existence or a given strict or non-strict location), for example:

0(2) is a valid expression, 0(2.000...1) is an invalid expression, 2.000...1(0(2)) is a valid expression.

Again, the general form is X(x), where X is strict or non-strict level of existence, and x is strict or non-strict location, such that X=0 has only strict locations (also X=0 does not have sub-levels of existence), and X>0 has (strict OR non-strict locations) OR (strict OR non-strict sub-levels of existence).

You are applying your everyday experience into an abstract world that is not physical, and that creates a conflict. But that's normal:

a⁴ = a * a * a * a

a³ = a * a * a

a² = a * a

a¹ = a

a⁰ = ?

a⁰ seems to be the same case as a/0.

12/4 = 12 - 3 - 3 - 3

12/3 = 12 - 4 - 4

12/2 = 12 - 6

12/1 = 12

12/0 = ?

Both analogies are not identical, as far as the result is concerned, even though your intuition tells you that they are.

No, 0 < a⁰ = 1 < a/0 = ∞ < ∞

 

[epix]

Epix, you can use any function that you wish.

It does not change the fact that there is ≠, which exists between any given arbitrary pair of distinct exact locations (arbitrary pair of distinct points) that are the strict results of the given function.

Arbitrary points are not "the strict results of the given function," as they cannot be given their definition. The result of a given function is not an arbitrary point -- that point is defined by the function. An arbitrary point is the independent variable, and the value of the dependent variable... well, depends on the way the independent variable is treated by the function f(x). In other words, f(x) = y, where x is the independent, or arbitrary, variable and y is the dependent variable. So, for x = 2, the function

f(x) = (x² - 4)/(x - 2) = 0/0

and the corresponding dependent variable y doesn't exist on the y-axis, coz the function doesn't define it (division by zero). That means there can't be no point drawn on the line that visualizes the f(x) = y relationship. Remember that integer 2 and real number 2 are not the same: in the latter case, 2 is a short for 2.000.... Unlike in a = 2.333... where a is approaching limit 7/3, the integers of real numbers are not approaching any limit and are therefore exact numbers.

2.33333 - 2.333 > 0

but

2.00000 - 2.000 = 0

That's why there is one and only one point missing in the straight line that represents the function f(x) above and that's why algebra uses fractional representation of non-integer numbers, which is the limit.

Since OM doesn't recognize the function as an instrument that graphically displays values of the modified independent variable, there is no way to prove that assertions made by OM are true or false.

 

[The Man]

Yes, any given exact location along 1() is 0().

So then there are no locations on your line that are not or can not be covered by points as you assert “any given exact location” is a point.

Still your 0()-only reasoning can comprehend http://www.internationalskeptics.com/forums/showpost.php?p=6472238&postcount=12096.

Again, stop simply trying to posit aspects of your own failed reasoning onto others.

You are a total 0()-loss and there is no use to continue the dialog with you on this interesting subject.

You can stop any time. Or can you?

There is no exact location on a line where there is no point, and ≠ is such non-exact location on a line, which exists between any given arbitrary pair of distinct exact locations (arbitrary pair of distinct points).

“≠” still isn’t a location Doron and you simply inserting “exact” and “non-exact into your dichotomist terminology doesn’t change that.

Please identify any location “between any given arbitrary pair of distinct exact locations (arbitrary pair of distinct points)” that is not or can not be covered by points.

 

[doronshadmi]

So then there are no locations on your line that are not or can not be covered by points as you assert “any given exact location” is a point.

Again,

Traditional Math can't comprehend that ≠ is the non-local property of 1() w.r.t 0(), such that ≠ is the existence of 1() between any given arbitrary pair of distinct 0() ( as expressed by 1(0(x)≠0) ), or simply the fact that 1() is at AND beyond 0() ( as expressed by 1(0()) ).

≠ is definitely free of any 0(), which gives it the ability to be used both as differentiator AND integrator of the concept of 1(0(x)≠0) Pair.

Moreover, Traditional Math's claim that 1() is completely covered by 0() is equivalent to the claim that (for example) variable x ( where x is any arbitrary distinct 0() of [0,1] ) is both ≤ 1 OR both ≥ 0, which is definitely a contradiction.

The Man, your 0()-only reasoning is too weak in order to deal with 1(0()) or 1(0(x)≠0).

You can write as many replies as you wish, which does not change the fact that you are closed under 0()-only reasoning, which is weaker than 1(0()) reasoning.

 

[zooterkin]

Again,

Traditional Math can't comprehend that ≠ is the non-local property of 1() w.r.t 0(), such that ≠ is the existence of 1() between any given arbitrary pair of distinct 0(), which expressed as 1(0(x)≠0).

≠ is definitely free of any 0(), which gives it the ability to be used both as differentiator AND integrator of the concept of 1(0(x)≠0) Pair.

Moreover, Traditional Math's claim that 1() is completely covered by 0() is equivalent to the claim that (for example) variable x ( where x is any arbitrary distinct 0() of [0,1] ) is both ≤ 1 OR both ≥ 0, which is definitely a contradiction.

So, you still can't identify a location on a line where there is no point? Got it. Do you think you could stop claiming that you can?

 

[doronshadmi]

Arbitrary points are not "the strict results of the given function," as they cannot be given their definition. The result of a given function is not an arbitrary point -- that point is defined by the function. An arbitrary point is the independent variable, and the value of the dependent variable... well, depends on the way the independent variable is treated by the function f(x). In other words, f(x) = y, where x is the independent, or arbitrary, variable and y is the dependent variable. So, for x = 2, the function

f(x) = (x² - 4)/(x - 2) = 0/0

and the corresponding dependent variable y doesn't exist on the y-axis, coz the function doesn't define it (division by zero). That means there can't be no point drawn on the line that visualizes the f(x) = y relationship. Remember that integer 2 and real number 2 are not the same: in the latter case, 2 is a short for 2.000.... Unlike in a = 2.333... where a is approaching limit 7/3, the integers of real numbers are not approaching any limit and are therefore exact numbers.

2.33333 - 2.333 > 0

but

2.00000 - 2.000 = 0

That's why there is one and only one point missing in the straight line that represents the function f(x) above and that's why algebra uses fractional representation of non-integer numbers, which is the limit.

Since OM doesn't recognize the function as an instrument that graphically displays values of the modified independent variable, there is no way to prove that assertions made by OM are true or false.

You are simply still missing the fact that f(x) is equivalent to 1(0()), such that f() is equivalent to 1() and x (if it is strict) is equivalent to 0().

Remember that integer 2 and real number 2 are not the same:

Wrong, 2 or 2.000... is exactly the same 0().

You are still missing 2.000...1(0(2)), such that 2.000...1() - 0(2) = 0.000...1(), which can be expressed as 2.000...1(0.000...1())

 

[doronshadmi]

So, you still can't identify a location on a line where there is no point? Got it. Do you think you could stop claiming that you can?

So, you can't comprehend 1(0()) or 1(0(x)≠0)! Got it.

Do you think you could stop claiming that (for example) variable x ( where x is any arbitrary distinct 0() of [0,1] ) is both ≤ 1 OR both ≥ 0, which is definitely a contradiction?

 

[epix]

You are simply still missing the fact that f(x) is equivalent to 1(0()), such that f() is equivalent to 1() and x (if it is strict) is equivalent to 0().

This is the reason why OM cannot display its own version of a f(x) graph, coz the definitions, based on the OM terminology, are the insurmountable obstacle. The traditional math understands that f() doesn't make sense without the indication which independent variables are to be modified by the function, and so f() never appears in the text, except in the case of non-mathematical applications where f() may substitute the traditional f... as a brief comment that summarizes certain unexpected outcomes.

In OM, f() = 1() means that a function without independent variables equals a one-dimensional space confused. You decided to notate a one-dimensional space as 1(), coz you didn't worry about a possible conflict with f(x), coz you never use it. Following the meaning of 1(), it's inevitable that f() implies an f-dimensional space and not a function where the independent variable(s) are not present.

I think you need to come up with correcting syntax ideas and edit all the posts that you have written.

 

[epix]

Wrong, 2 or 2.000... is exactly the same 0().

I didn't refer to the value of 2 and 2.000... Just read again.

 

[epix]

Moreover, Traditional Math's claim that 1() is completely covered by 0() is equivalent to the claim that (for example) variable x ( where x is any arbitrary distinct 0() of [0,1] ) is both ≤ 1 OR both ≥ 0, which is definitely a contradiction.

Your conclusion is contradictory by itself and the reason is that you can't still grasp the meaning and the usage of intervals despite all the explaining links provided.

Suppose that an insurance analyst wants to know the probability that a person dies due to a specific illness between the age of 65 and 75 including, where the age of the person upon expiring is defined as X. He collects sample data, does some number crunching and finds out that the probability P is

P(65 ≤ X ≤ 75) = .68 [68%]

That person who gets unlucky doesn't die twice at the age of 65 and 75; it dies only once in the age that is within the range indicated. That person dies only once, coz he or she lives only once. Just don't get confused by some high-powered theoretical claims based on the latest research in genetics done by Prof. Sinatra.

 

[doronshadmi]

In OM, f() = 1() means that a function without independent variables equals a one-dimensional space confused.

In OM X() is a level of strict of non-strict existence, which exists independently of any sub-levels. X can be number, function, dimensional space, or any measurable strict or non-strict level of existence.

I think you need to come up with correcting syntax ideas and edit all the posts that you have written

You still do not get the generalization of X(), such that a given X's existence is independent of any other given X.

 

[doronshadmi]

Your conclusion is contradictory by itself and the reason is that you can't still grasp the meaning and the usage of intervals despite all the explaining links provided.

epix, you can't grasp that strict 0(x) value can't be = AND ≠ to strict 0 value.

In other words, no given level of existence is completely covered by any collection of previous levels of existence, whether the previous levels are non-strict ( like 0.999...[base 10]() ) or strict ( like 0(x) ).

You are still missing, for example: 1(0.999...[base 10]()), 1(0(x)≠0), 0.999...[base 10](0(x)), etc ...

 

[epix]

In OM X() is a level of strict of non-strict existence, which exists independently of any sub-levels. X can be number, function, dimensional space, or any measurable strict or non-strict level of existence.

Can you provide an example of some function that you notate "f()" -- function with a certain level of strictness, as it exists non-strictly?

Where does the level of strictness appear in the term X()? Is it inside the parenthesis, or is it outside? Does f(2) mean that the function f() has the second level of strictness? You never mentioned that there are certain levels of strictness.

Just provide an example of two functions of the second and the third level of strict existence, so the difference could be seen. Follow the example of the traditional function: For x = 5

f(x) = 2x + 3 = 2*5 + 3 = 13

 

[epix]

epix, you can't grasp that strict 0(x) value can't be = AND ≠ to strict 0 value.

No one has ever argued anything like that. You are still confused by the meaning of the term explained here:
http://www.internationalskeptics.com/forums/showpost.php?p=6485624&postcount=12134

Once again . . .
Let x be a positive integer. Then

65 ≤ x ≤ 75 means that x can take on values from 65 to 75.

65 < x < 75 means that x can take on values from 66 to 74.

65 < x ≤ 75 means that x can take on values from 66 to 75.

65 ≤ x < 75 means that x can take on values from 65 to 74.

 

[The Man]

Again,

Traditional Math can't comprehend that ≠ is the non-local property of 1() w.r.t 0(), such that ≠ is the existence of 1() between any given arbitrary pair of distinct 0() ( as expressed by 1(0(x)≠0) ), or simply the fact that 1() is at AND beyond 0() ( as expressed by 1(0()) ).

Doron if you are trying to claim that two points define a line segment, well, that is exactly what “Traditional Math” and the geometry claims.

Once again your assertion that “1() is at AND beyond 0()” simply shows that you can’t even agree with yourself.

≠ is definitely free of any 0(), which gives it the ability to be used both as differentiator AND integrator of the concept of 1(0(x)≠0) Pair.

Once again Doron, “≠“ is simply an assertion of inequality. However, in a continuous space there is always at least another point between any two points.

Again please identify any location(s) on a line segment that is not and can not be covered by points.

Moreover, Traditional Math's claim that 1() is completely covered by 0() is equivalent to the claim that (for example) variable x ( where x is any arbitrary distinct 0() of [0,1] ) is both ≤ 1 OR both ≥ 0, which is definitely a contradiction.

No Doron your continued misrepresentation of the symbols “≤ “ “≥” only makes your claim “equivalent” to your own misrepresentation.

To try to clarify it again for you, any point included in the interval [0,1] is > 0 AND < 1, OR = 1, OR = 0. Your “both ≤ 1 OR both ≥ 0” claim is just nonsense and only shows that you do not understand the “≤” and “≥” symbols. “Both” < OR = (“both ≤ “) as well as “both” > OR = (“both ≥”) is a contradiction and it remains simply yours.

The Man, your 0()-only reasoning is too weak in order to deal with 1(0()) or 1(0(x)≠0).

You can write as many replies as you wish, which does not change the fact that you are closed under 0()-only reasoning, which is weaker than 1(0()) reasoning.

Again, stop simply trying to posit aspects of your own failed reasoning onto others.

 

[The Man]

Do you think you could stop claiming that (for example) variable x ( where x is any arbitrary distinct 0() of [0,1] ) is both ≤ 1 OR both ≥ 0, which is definitely a contradiction?

Do you think you can actually learn what the symbols “≤” and “≥” represent, evidently you simply do not want to.

As you're the only one making that ridiculous and nonsensical claim Doron, only you can stop making it and stop trying to attribute it to others. Or can you?