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How many Rational points in graph paper can a circle have? Conjecture then theorem
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Archimedes Plutonium
2017-06-06 10:29:08 UTC
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Alright the advancement on polygons makes my appetite more more. This time the circle.

Experiment:: get out a sheet of graph paper and compass. Place the center of a large circle on a intersection point-- a Rational number point and a second rational point as radius, draw the circle and as best you can eyeball and count how many Rational number point in your circle. I counted 12. Four from a square, eight from two rectangles.

Now i suspect the radius size has a role, dependent on the size of unit square, once past that determinant the Rational number intersects should be a constant number. I suspect the constant is 12.

So, we have a brand new conjecture to play with.

And the great importance of this would be to say that all circles are fully described by no more than 12 points in the plane and all circles have no more than 12 rational points and all other points of a circle are irrationals.

AP
Archimedes Plutonium
2017-06-06 11:54:41 UTC
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No i made a mistake in eyeball count. Apparenly all circles on a Grid have but 4 Rational points, if i bypass the center.

Now i wonder if the center is integral part of circle. The reason i ask is the interior is taken as not necessary. If the center is necessary then all circles are 5 Rational number points. And no dependence on size.

AP
bert
2017-06-06 12:03:50 UTC
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. . . all circles have no more than 12 rational points . . .
Not true. The circle with radius 65 has 20 rational points.
Relative to the origin (0, 0), they are (65, 0) and its three
rotations; (63, 16) and its seven rotations and reflections;
and (33, 56) and its seven rotations and reflections. Now,
really, I thought you knew enough about Pythagorean triangles
to deduce that when the radius is a product of N distinct
primes, all of which are of the form 4m+1, then the circle
has (4 + 2^(N+3)) rational points; that's 12 when N = 1,
20 when N = 2, 36 when N = 3, and so on.
--
b***@gmail.com
2017-06-06 12:18:48 UTC
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Also nomenclatur is wrong, I guess AP means integer point,
i.e. points from NxN or ZxZ, and not rational points, i.e.

points from QxQ, since there are infinitely many rational
points, for example on a unit circle, even densly...
Post by bert
. . . all circles have no more than 12 rational points . . .
Not true. The circle with radius 65 has 20 rational points.
Relative to the origin (0, 0), they are (65, 0) and its three
rotations; (63, 16) and its seven rotations and reflections;
and (33, 56) and its seven rotations and reflections. Now,
really, I thought you knew enough about Pythagorean triangles
to deduce that when the radius is a product of N distinct
primes, all of which are of the form 4m+1, then the circle
has (4 + 2^(N+3)) rational points; that's 12 when N = 1,
20 when N = 2, 36 when N = 3, and so on.
--
James Waldby
2017-06-06 21:51:53 UTC
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Post by bert
. . . all circles have no more than 12 rational points . . .
Not true. The circle with radius 65 has 20 rational points.
Relative to the origin (0, 0), they are (65, 0) and its three
rotations; (63, 16) and its seven rotations and reflections;
and (33, 56) and its seven rotations and reflections. Now,
really, I thought you knew enough about Pythagorean triangles
to deduce that when the radius is a product of N distinct
primes, all of which are of the form 4m+1, then the circle
has (4 + 2^(N+3)) rational points; that's 12 when N = 1,
20 when N = 2, 36 when N = 3, and so on.
There are infinitely many rational points on the circle with
radius 65, although there are only the 20 integer-lattice points
that you listed on that circle.

For example, (137/17, 4/17), (524/65, 7/65), (911/113, 8/113),
(2072/257, 1/257), etc. are members of different infinite families
of rational points on the circle with radius 65.
--
jiw
Archimedes Plutonium
2017-06-06 22:29:51 UTC
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Post by James Waldby
Post by bert
. . . all circles have no more than 12 rational points . . .
Not true. The circle with radius 65 has 20 rational points.
Relative to the origin (0, 0), they are (65, 0) and its three
rotations; (63, 16) and its seven rotations and reflections;
and (33, 56) and its seven rotations and reflections. Now,
really, I thought you knew enough about Pythagorean triangles
to deduce that when the radius is a product of N distinct
primes, all of which are of the form 4m+1, then the circle
has (4 + 2^(N+3)) rational points; that's 12 when N = 1,
20 when N = 2, 36 when N = 3, and so on.
There are infinitely many rational points on the circle with
radius 65, although there are only the 20 integer-lattice points
that you listed on that circle.
For example, (137/17, 4/17), (524/65, 7/65), (911/113, 8/113),
(2072/257, 1/257), etc. are members of different infinite families
of rational points on the circle with radius 65.
--
jiw
James could get away with that error of the past, because no-one ever looked at the situation of whether a Irrational number length exists in regular polygons.

Obvious that a 4-gon can and does exist in a Lattice of points, but does a 5-gon exist in that lattice of Rational points.


5


4 3



2 PP 1


Here I attempted to draw a 5-gon for I want to show the Polygon Perpendicular denoted by PP

We connect PP to 5 of the regular pentagon forming right triangle PP52

The PP in all cases of regular pentagon 10-gon, 20-gon are irrational number length.

That means regular polygons do not exist if they have an irrational number length-- exist only in imagination but never in Reality, because an irrational number length is two different numbers acting as one number. The sqrt2 in 1000 Grid is two different numbers 1.414 acting with 1.415 to produce exactly 2.000 in 1000 Grid. So a 5-gon or 20-gon have a point that is not Rational number point but composed of a two different numbers. In other words the tip of the vertex is missing. All regular polygons that are not 4, 8, 16, 32, etc are defective and missing parts or not having equal angles.

So, no, when Bert or James tell you that a Lattice of Rationals yields many circles with infinite Rationals contained in the circle, are preaching old decayed and error filled math.

James is an Old Fogey and let him walk into the western sunset,,, booed for his error filled math.

AP
Archimedes Plutonium
2017-06-06 22:45:54 UTC
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The Ancient Greeks were an extremely smart class of mathematicians, compared to the modern mathematicians of the past 100 years look like a class of dunces in comparison.

One of the Ancient Greek inventions was the Rule that you can only use Straightedge and Compass. Well, that is old now, but it sure solved many a mystery of math.

The Rule today, though, which would be absolutely foreign to James Walby and his like minded modern day failures of math, the new rule, the Rule of 2017 going forward is.

The Graph paper or the Coordinate System of Math can only use a Lattice Point Graph Paper, where each point is a Rational Number point of some specific Grid.

So like the Ancient Greeks insisting on only compass and straightedge. Today, in modern math, we insist on only a Lattice Graph Paper for doing Calculus, doing geometry, you name it.

And James Walby is out of touch with reality of math.

AP
Archimedes Plutonium
2017-06-06 22:06:45 UTC
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Post by bert
. . . all circles have no more than 12 rational points . . .
Not true. The circle with radius 65 has 20 rational points.
Relative to the origin (0, 0), they are (65, 0) and its three
rotations; (63, 16) and its seven rotations and reflections;
and (33, 56) and its seven rotations and reflections. Now,
really, I thought you knew enough about Pythagorean triangles
to deduce that when the radius is a product of N distinct
primes, all of which are of the form 4m+1, then the circle
has (4 + 2^(N+3)) rational points; that's 12 when N = 1,
20 when N = 2, 36 when N = 3, and so on.
--
Hi, Bert, I appreciate your answer, but I think this is much tougher than you think.

Admittedly, both of us have gone through life having been brainwashed about rationals, irrationals, geometry. Brainwashed like that of having to deal with metric, only we grew up on inches, feet, mile.

Admittedly I started off with -- let me explore this -- not with here is a well formed conjecture and so, what I find in a first few posts will not be what I find out later on.

Now Wally in another post mentions the word "lattice" and I need that concept. A concept of finite Rational points of a Grid of Lattice Rationals.

So, I have a Graph paper and focused on the intersection points, and we call each of those points a Rational Number point. The holes in between are irrationals and cannot use them

Now a graph paper is like this

____________
____________
____________

with lines going like this |||||| also

So where the lines
_____
_____
_____ and the lines ||||||||

intersect are points and those points are the only Rationals that exist

Do they make graph paper that has no lines but only points like this

...........
...........
...........

Those points are intersection points of lines that go ___ and lines that go |

So I want graph paper that is only Lattice points.

Now, let us say we drop down to the 10^604 Grid where the spacing between the Rationals is 1^10^-604. We cannot draw anymore points of Rationals in between two points because we are at infinity.

So, Bert, being here at infinity, and not able to ever draw more points of Rationals, we ask the question. Given a Rational number point for center and a radius that is rational number and then draw that circle, how many Rational points will that circle ensnare, not counting the center.

Now Bert thinks it varies according to radius. Bert thinks that a circle can achieve 20 Rational Points.

Now I look up the figure of a 20-gon and ask if there is a irrational number involved. In fact, I know that the 20-gon is the same answer as 10-gon and then 5-gon. I apply the POLYGON PERPENDICULAR method. Where I do a perpendicular from one side to the "opposite side where I have to include at minimum two vertices forming a right triangle. Is the perpendicular a rational number or irrational? In the case of 5-gon, 10-gon, 20-gon the perpendicular is irrational. That tells me automatically that Bert could not possibly have a circle with 20 Rational Lattice points.

Now, I can achieve a circle with 4, 8, 16, 32, 64, etc etc Rational number points, by going to the 10^603 Grid where I call a unit Square having ten by ten smaller squares and what I do then is form a octagon from out of a Square of the 10^603 Grid. Now, to get a circle with 16 Rational points, I have to go to the 10^602 Grid in order to reduce a square in 10^602 to achieve a 16-gon.

Bert, this is tough, this is not easy, for we have to shuck all the brainwash we were taught and try to think straight. It is difficult in shucking miles when we are to use kilometers.

AP
b***@gmail.com
2017-06-06 23:24:03 UTC
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Thats not rational points on a grid, thats only integer points.

A rational point has the form (x,y) where x and y are rational numbers.

A rational number has the form p/q where p and q are integer numbers.
Post by Archimedes Plutonium
Now, I can achieve a circle with 4, 8, 16, 32, 64, etc etc Rational number points, by going to the 10^603 Grid where I call a unit Square having ten by ten smaller squares and what I do then is form a octagon from out of a Square of the 10^603 Grid. Now, to get a circle with 16 Rational points, I have to go to the 10^602 Grid in order to reduce a square in 10^602 to achieve a 16-gon.
b***@gmail.com
2017-06-06 23:25:55 UTC
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https://en.wikipedia.org/wiki/Integer

https://en.wikipedia.org/wiki/Rational_number
Post by b***@gmail.com
Thats not rational points on a grid, thats only integer points.
A rational point has the form (x,y) where x and y are rational numbers.
A rational number has the form p/q where p and q are integer numbers.
Post by Archimedes Plutonium
Now, I can achieve a circle with 4, 8, 16, 32, 64, etc etc Rational number points, by going to the 10^603 Grid where I call a unit Square having ten by ten smaller squares and what I do then is form a octagon from out of a Square of the 10^603 Grid. Now, to get a circle with 16 Rational points, I have to go to the 10^602 Grid in order to reduce a square in 10^602 to achieve a 16-gon.
b***@gmail.com
2017-06-07 00:09:26 UTC
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Denoting these numbers by sets, we usually use:

For the integers: Z, from German "Zahlen", more precisely
they are called "Ganzzahlen" in German
https://de.wiktionary.org/wiki/Ganzzahl

For the rationals: Q, from Italian "Quoziente", like in
La somma, la differenza, il prodotto e il quoziente,
letter used Giuseppe Peano (ref ??)
Post by b***@gmail.com
https://en.wikipedia.org/wiki/Integer
https://en.wikipedia.org/wiki/Rational_number
Post by b***@gmail.com
Thats not rational points on a grid, thats only integer points.
A rational point has the form (x,y) where x and y are rational numbers.
A rational number has the form p/q where p and q are integer numbers.
Post by Archimedes Plutonium
Now, I can achieve a circle with 4, 8, 16, 32, 64, etc etc Rational number points, by going to the 10^603 Grid where I call a unit Square having ten by ten smaller squares and what I do then is form a octagon from out of a Square of the 10^603 Grid. Now, to get a circle with 16 Rational points, I have to go to the 10^602 Grid in order to reduce a square in 10^602 to achieve a 16-gon.
konyberg
2017-06-07 05:13:53 UTC
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Post by Archimedes Plutonium
Post by bert
. . . all circles have no more than 12 rational points . . .
Not true. The circle with radius 65 has 20 rational points.
Relative to the origin (0, 0), they are (65, 0) and its three
rotations; (63, 16) and its seven rotations and reflections;
and (33, 56) and its seven rotations and reflections. Now,
really, I thought you knew enough about Pythagorean triangles
to deduce that when the radius is a product of N distinct
primes, all of which are of the form 4m+1, then the circle
has (4 + 2^(N+3)) rational points; that's 12 when N = 1,
20 when N = 2, 36 when N = 3, and so on.
--
Hi, Bert, I appreciate your answer, but I think this is much tougher than you think.
Admittedly, both of us have gone through life having been brainwashed about rationals, irrationals, geometry. Brainwashed like that of having to deal with metric, only we grew up on inches, feet, mile.
Admittedly I started off with -- let me explore this -- not with here is a well formed conjecture and so, what I find in a first few posts will not be what I find out later on.
Now Wally in another post mentions the word "lattice" and I need that concept. A concept of finite Rational points of a Grid of Lattice Rationals.
So, I have a Graph paper and focused on the intersection points, and we call each of those points a Rational Number point. The holes in between are irrationals and cannot use them
Now a graph paper is like this
____________
____________
____________
with lines going like this |||||| also
So where the lines
_____
_____
_____ and the lines ||||||||
intersect are points and those points are the only Rationals that exist
Do they make graph paper that has no lines but only points like this
...........
...........
...........
Those points are intersection points of lines that go ___ and lines that go |
So I want graph paper that is only Lattice points.
Now, let us say we drop down to the 10^604 Grid where the spacing between the Rationals is 1^10^-604. We cannot draw anymore points of Rationals in between two points because we are at infinity.
So, Bert, being here at infinity, and not able to ever draw more points of Rationals, we ask the question. Given a Rational number point for center and a radius that is rational number and then draw that circle, how many Rational points will that circle ensnare, not counting the center.
Now Bert thinks it varies according to radius. Bert thinks that a circle can achieve 20 Rational Points.
Now I look up the figure of a 20-gon and ask if there is a irrational number involved. In fact, I know that the 20-gon is the same answer as 10-gon and then 5-gon. I apply the POLYGON PERPENDICULAR method. Where I do a perpendicular from one side to the "opposite side where I have to include at minimum two vertices forming a right triangle. Is the perpendicular a rational number or irrational? In the case of 5-gon, 10-gon, 20-gon the perpendicular is irrational. That tells me automatically that Bert could not possibly have a circle with 20 Rational Lattice points.
Now, I can achieve a circle with 4, 8, 16, 32, 64, etc etc Rational number points, by going to the 10^603 Grid where I call a unit Square having ten by ten smaller squares and what I do then is form a octagon from out of a Square of the 10^603 Grid. Now, to get a circle with 16 Rational points, I have to go to the 10^602 Grid in order to reduce a square in 10^602 to achieve a 16-gon.
Bert, this is tough, this is not easy, for we have to shuck all the brainwash we were taught and try to think straight. It is difficult in shucking miles when we are to use kilometers.
AP
Google Geoboard.

KON
Markus Klyver
2017-06-15 01:06:09 UTC
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The cardinality of the rational unit circle is aleph-null.

Wally W.
2017-06-06 12:15:48 UTC
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Post by Archimedes Plutonium
Alright the advancement on polygons makes my appetite more more. This time the circle.
Experiment:: get out a sheet of graph paper and compass. Place the center of a large circle on a intersection point-- a Rational number point and a second rational point as radius, draw the circle and as best you can eyeball and count how many Rational number point in your circle. I counted 12. Four from a square, eight from two rectangles.
Now i suspect the radius size has a role, dependent on the size of unit square, once past that determinant the Rational number intersects should be a constant number. I suspect the constant is 12.
So, we have a brand new conjecture to play with.
And the great importance of this would be to say that all circles are fully described by no more than 12 points in the plane and all circles have no more than 12 rational points and all other points of a circle are irrationals.
AP
See discussions about lattice points here:

Pi hiding in prime regularities
3Blue1Brown
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