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.
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).
Never gives up of ignorance, I see.Back to the gibberish, I see.
Epix, you can use any function that you wish.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 . . .
Who says that it isn't so? Any xi that is a part of R is distinct, such as xi-1 < xi < xi+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...?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.
Wrong, 1(0(2)≠0(4)) is 1() (notated by ≠) between strict numbers.You claim that point [x = 2, y = 4] doesn't have an exact location.
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: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.
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 functionEpix, 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.
Yes, any given exact location along 1() is 0().
Still your 0()-only reasoning can comprehend http://www.internationalskeptics.com/forums/showpost.php?p=6472238&postcount=12096.
You are a total 0()-loss and there is no use to continue the dialog with you on this interesting subject.
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).
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(), 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.
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.
Wrong, 2 or 2.000... is exactly the same 0().epix said:Remember that integer 2 and real number 2 are not the same:
So, you can't comprehend 1(0()) or 1(0(x)≠0So, 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?
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.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().
I didn't refer to the value of 2 and 2.000... Just read again.Wrong, 2 or 2.000... is exactly the same 0().
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.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.
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.In OM, f() = 1() means that a function without independent variables equals a one-dimensional spaceconfused
.
You still do not get the generalization of X(), such that a given X's existence is independent of any other given X.I think you need to come up with correcting syntax ideas and edit all the posts that you have written
epix, you can't grasp that strict 0(x) value can't be = AND ≠ to strict 0Your 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.
Can you provide an example of some function that you notate "f()" -- function with a certain level of strictness, as it exists non-strictly?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.
epix, you can't grasp that strict 0(x) value can't be = AND ≠ to strict 0value.
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.
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?
Well, I really can't grasp it, but you should stick with the ethics and not mention certain taboos. There are social taboos, but there are also academic taboos. The traditional math is inconsistent and one of the cases that demonstrates it is the one that you keep complaining about and the essence of that case is that the traditional math asserts A = B and at the same time A ≠ B. More specifically, it is the case of identity and difference that you complain about and should not. Any ethical mathematician looks the other way, but you are an OM mathematician not fond of the traditional math, so that explains the breach of the taboo.epix, you can't grasp that strict 0(x) value can't be = AND ≠ to strict 0value.
In other words, no given level of existence is completely covered by any collection of previous levels 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
Does f(2) mean that the function f() has the second level of strictness?
No, it simply means that:Does f(2) mean that the function f() has the second level of strictness?
Your 0()-only reasoning simply can't get that the claim that "1-dim space is completely covered by 0-dim spaces" is equivalent to the claim that (for example) "variable x ( where x is any arbitrary distinct member of [0,1] ) is both ≤ 1 OR both ≥ 0".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?
However, in a continuous space there is always at least another point between any two points.
Wrong. 1-dim space exists independently of any sub-levels of existence along it, where 0-dim space is such sub-level of existence.epix said:Definition: A line is not continuous if at least one point on it is not defined by the function that creates such a line.
Whether it is N or R collection of x elements (where any arbitrary x is 0-dim space) , it does no change the fact that there is always ≠ between any given pair of 0-dim spaces along 1-dim space, which prevents form any amount of 0-dim spaces to completely cover 1-dim space.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.
It can't help you to avoid ≠ between some arbitrary distinct pair.
For example: the ...1 of 0.000...1[base 1] is equivalent to ≠ between 0.999...[base 10] and 1, if we avoid mixed bases.
It can't help you to avoid ≠ between some arbitrary distinct pair.
For example: the ...1 of 0.000...1[base 1] is equivalent to ≠ between 0.999...[base 10] and 1, if we avoid mixed bases.
Wrong. 1-dim space exists independently of any sub-levels of existence along it, where 0-dim space is such sub-level of existence.
OM expresses this notion as 1(0()).
This debate has so far gone on for 304 pages??
G.I.G.O.
For example: the ...1 of 0.000...1[base 1] is equivalent to ≠ between 0.999...[base 10] and 1, if we avoid mixed bases.
It can't help you to avoid ≠ between some arbitrary distinct pair.
For example: the ...1 of 0.000...1[base 1] is equivalent to ≠ between 0.999...[base 10] and 1, if we avoid mixed bases.
“Traditional Math”, which is the reasoning that you are using all along this thread, can't comprehend 1() that is 1-dimensional space with no sub-dimensional spaces along it.The Man said: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.
The Man said:Once again your assertion that “1() is at AND beyond 0()” simply shows that you can’t even agree with yourself.
See, after your couple of replies that again demonstrate your math illiteracy, I'm not sure if that "base 1" is a typo or not.
The Man, it is a typo, and your poor maneuvers around it simply demonstrate your no-motivation to understand OM.
Ah, yes, Traditional Math(s), which has given us things such as space travel, computers and many other things we see around us today.“Traditional Math”, which is the reasoning that you are using all along this thread, can't comprehend 1() that is 1-dimensional space with no sub-dimensional spaces along it.
Under OM, 1()epix is = AND 1()The Man, and so epix=The Man. Your 0()-only reasoning doesn't allow you to grasp a simple concept.I think you'll find you are replying to epix, not The Man.
OM application mainly concerns the time travel. Once are the mathematicians able to grasp that "1() that is 1-dimensional space with no sub-dimensional spaces along it," we are right back in the 15th century.Ah, yes, Traditional Math(s), which has given us things such as space travel, computers and many other things we see around us today.
Now, remind me what you can do with OM.
“Traditional Math”, which is the reasoning that you are using all along this thread, can't comprehend 1() that is 1-dimensional space with no sub-dimensional spaces along it.
1(0(x) ≠ 0) is a general expression of the concept of Segment, such that any given arbitrary segment 0(x) ≠ 0
is ( non-extendible to 1() ) AND ( irreducible to 0() ).
Once again your reasoning, where “1() is defined by 0()” simply shows that you can’t comprehend 1() as independent of any sub-spaces.
As e result you have no ability to comprehend 1(0(x)), where 1() is at AND beyond 0(x).
Also, your notation cannot prove that "0.999..." added to "0.000...1" equals 1, as f(x) + g(x) can, coz your notation lacks algebraic terms necessary for the proof.