Distance Function without Weighing Current Rules of Mathematics

[point]

straight line AB located at number line ( coordinate x ) , point A located on a number of number line, point B located on any number of number line , that this is a function of ?

 

[RUber]

Not sure I understand your question. A and B are points on the same number line? Where is x? What are you trying to find?

 

[point]

Not sure I understand your question. A and B are points on the same number line? Where is x? What are you trying to find?

yes - A and B are points on the same number line

 

[RUber]

What is your question about a function?
Normally, a function is an input/output relationship. You are describing two random points and a line connecting them.

 

[HallsofIvy]

If A is a fixed point on the number line and B a variable point, then, for example, the distance between them is a function of A, f(A). If both are variable points, then it is a function of both A and B, f(A, b). Other than that, I do not know what "this" refers to in your last sentence.

 

[point]

If A is a fixed point on the number line and B a variable point, then, for example, the distance between them is a function of A, f(A). .

or a little differently A=x , B=a , AB=y , y=|x-a|straight line AB located at number line ( coordinate x ) , point A located on any number of number line , point B located on any number of number line , that this is a function of ?

 

[point]

first solution
A=x1, B=x2 , AB=y ,y=|x _1-x_2|

second solution
A=x , B=y₁=f(x) , AB=y₂, y_2=|x-f(x)|

continuation - dynamic graphics, static graphics, partial graph y=|a-x| ?

 

[Mark44]

I suspect that you are not a native speaker of English. What you have written is very unclear, so it's hard to comprehend what you're trying to do.

first solution
A=x1, B=x2 , AB=y ,y=|x _1-x_2|

|x₁ - x₂| gives you the distance between two numbers on the x-axis.

Your notation is nonstandard. AB is normally used to indicate the line segment between points A and B. To represent the length of this line segment, you could write |AB| or length(AB).

second solution
A=x , B=y₁=f(x) , AB=y₂, y_2=|x-f(x)|

?
Second solution of what?
This is extremely unclear. You have way too many variables here: A, B, x, f(x), y₁, y₂

continuation - dynamic graphics, static graphics, partial graph y=|a-x| ?

Without more information, I don't think we'll be able to help you.

 

[point]

I suspect that you are not a native speaker of English. What you have written is very unclear, so it's hard to comprehend what you're trying to do.
.

native speaker of Serbian , that you read the google translator in English

Your notation is nonstandard. AB is normally used to indicate the line segment between points A and B. To represent the length of this line segment, you could write |AB| or length(AB).
.

AB=|A-B| or y=|x₁-x₂|, since the variables segment here and point the variable I used bookmarks (x ,y)

Second solution of what?
This is extremely unclear. You have way too many variables here: A, B, x, f(x), y₁, y₂

.

the first solution is provided with two variables (x_1 , x_2 )
second solution is given with variable ( x ) and function of variables ( f(x) = y_1 )
y_2=|x-f(x)| or y_2=|x-y_1| , y_1=f(x)
the current structure of the function is: y independent variable of n independent variables (x_n )
This structure functions (second solution) does not exist in the present mathematics
independent variable x
dependent variable y₁(dependent on x )
dependent variable y₂(dependent on x and y₁ )
There are functions that are specified via the geometric object (shown here is the simplest form, the segment (straight line), through its two points) whose structure is different from the current function, which you want to help introduce myself

 

[Mark44]

It would help us if you told us what the function is. From what you have written, it appears that you have an iterative function something like this:
$x_{n + 1} = f(x_n)$
and you're trying to see if $|x_n - f(x_n)| = |x_n - x_{n + 1}|$ is getting smaller.

 

[point]

y = a-x
The graph of the current solution:
x-coordinate represents all real numbers, when solved function we have two numbers (y, x) , introduces the new coordinates y perpendicular to the x-coordinate and cut the number 0 (plane), the number of y is transferred to the y-coordinate , line (which is parallel to the y-coordinate, and on it is a point that is the number x) is cut from the line (which is parallel to the x-coordinate and it is a point that is the number y) gets the point in the plane (x, y)
which means that the point (x, y) on the x-coordinate of the mapped into a point in the plane (x, y) points are merged to obtain a graph

y = | a-x |
Graph of my solution:
x-coordinate represents all real numbers, when solved function we have three numbers (a, y, x), introduces the new coordinates y perpendicular to the x-coordinate and cut the number 0 (plane), the number of y is transferred to the y-coordinates, lines (the first parallel to the y-coordinate, and on it is a point that is the number a , the second is parallel to the y-coordinate, and on it is a point that is the number x) is cut from the line (which is parallel to the x-coordinate and it is the point which is also the number y) gave the points in the plane (x, y) and (a, y) of the connecting point is obtained straight line
which means that the points (a, x, y) on the x-coordinates are mapped onto the straight line AB in the plane (A (x, y) B (s, y)), the straight line are merged to obtain the graph of

static graphics
Ap and Aq semi-line and surface between them
https://o9alca.bn1302.livefilestore...xFPijp-ESI0Mgb-3wertC-r-TqxXJRg/ss.png?psid=1

 

[Mark44]

y = a-x
The graph of the current solution:
x-coordinate represents all real numbers, when solved function we have two numbers (y, x) ,

Points are normally written in this order (x, y).

In any case, the graph of y = a - x is a straight line with slope -1, that crosses the y-axis at (0, a) and crosses the x-axis at (a, 0).

introduces the new coordinates y perpendicular to the x-coordinate and cut the number 0 (plane), the number of y is transferred to the y-coordinate , line (which is parallel to the y-coordinate, and on it is a point that is the number x) is cut from the line (which is parallel to the x-coordinate and it is a point that is the number y) gets the point in the plane (x, y)
which means that the point (x, y) on the x-coordinate of the mapped into a point in the plane (x, y) points are merged to obtain a graph

I'm having a hard time understanding what you're trying to say here. A coordinate is one of the numbers in an ordered pair. For example, the point (2, 1) has an x-coordinate (2) and a y-coordinate (1). It doesn't make any sense to talk about something being perpendicular to the x-coordinate. A line can be parallel to the y-axis or to the line x = 2.

y = | a-x |

The graph of this function has a V shape. Wherever a - x ≥ 0, the graph of y = |a - x| is identical to the graph of y = a - x. Where a - x < 0, the graph of y = |a - x| is the reflection across the x-axis of the graph of y = a - x.

Assuming for the moment that a is a positive number, the graph of y = |a - x| is a straight line that goes through (0, a) down to (a, 0) with a slope of -1. From then on, the line goes up with a slope of +1.

Graph of my solution:
x-coordinate represents all real numbers, when solved function we have three numbers (a, y, x)

The number a is considered to be a parameter. It might not be known, but it doesn't change, unlike x and y, which are variable.

, introduces the new coordinates y perpendicular to the x-coordinate and cut the number 0 (plane), the number of y is transferred to the y-coordinates, lines (the first parallel to the y-coordinate, and on it is a point that is the number a , the second is parallel to the y-coordinate, and on it is a point that is the number x) is cut from the line (which is parallel to the x-coordinate and it is the point which is also the number y) gave the points in the plane (x, y) and (a, y) of the connecting point is obtained straight line
which means that the points (a, x, y) on the x-coordinates are mapped onto the straight line AB in the plane (A (x, y) B (s, y)), the straight line are merged to obtain the graph of

static graphics
Ap and Aq semi-line and surface between them
https://o9alca.bn1302.livefilestore...xFPijp-ESI0Mgb-3wertC-r-TqxXJRg/ss.png?psid=1

In the vigilink graph (which takes a very long time to load), you are graphing y = |3 - x|. The V shape of this graph is what I was talking about above.

 

[point]

The graph of this function has a V shape. Wherever a - x ≥ 0, the graph of y = |a - x| is identical to the graph of y = a - x. Where a - x < 0, the graph of y = |a - x| is the reflection across the x-axis of the graph of y = a - x.

you're wrong x ≥ 0 ( should x ≥ a ) , x < 0 ( should x < a )

The number a is considered to be a parameter. It might not be known, but it doesn't change, unlike x and y, which are variable.

straight line from the x-coordinates of the projected area in the plane, and needs two points ( a ,x ) and the distance between them ( y )

 

[Mark44]

The graph of this function has a V shape. Wherever a - x ≥ 0, the graph of y = |a - x| is identical to the graph of y = a - x. Where a - x < 0, the graph of y = |a - x| is the reflection across the x-axis of the graph of y = a - x.

you're wrong x ≥ 0 ( should x ≥ a ) , x < 0 ( should x < a )

I didn't write x ≥ 0. Notice that I wrote a - x ≥ 0 and a - x < 0. The first inequality is equivalent to a ≥ x. The second inequality is equivalent to a < x.

The number a is considered to be a parameter. It might not be known, but it doesn't change, unlike x and y, which are variable.

straight line from the x-coordinates of the projected area in the plane, and needs two points ( a ,x ) and the distance between them ( y )

?
I don't know what you're saying.

 

[point]

I didn't write x ≥ 0. Notice that I wrote a - x ≥ 0 and a - x < 0. The first inequality is equivalent to a ≥ x. The second inequality is equivalent to a < x.

error in translation

I don't know what you're saying.

I'll explain using pictures
x=1 , red color straight , x- coordinates ( |3-1| , |3-x| ) is mapped to plane (˙y=|3-x| , |3-1|=2 )

x=4 , red color straight , x- coordinates ( |3-4| , |3-x| ) is mapped to plane (˙y=|3-x| , |3-4|=1 )

 

[Mark44]

I'll explain using pictures
x=1 , red color straight , x- coordinates ( |3-1| , |3-x| ) is mapped to plane (˙y=|3-x| , |3-1|=2 )

The pictures aren't that helpful, but what's worse is that the mathematics description you're writing doesn't make any sense. In the first drawing there are two red line segments. The upper line segment can be described as {(x, y) : 1 ≤ x ≤ 3, y = 2}. The lower segment can be described in a similar manner.

I have no idea what you mean by this phrase: "mapped to the plane (y = |3 - x|, |3 - 1| = 2)" How is what you have described here a plane?

View attachment 79736
x=4 , red color straight , x- coordinates ( |3-4| , |3-x| ) is mapped to plane (˙y=|3-x| , |3-4|=1 )
View attachment 79737

 

[point]

SERBIAN
Pošto sam otkrio nove mogućnosti , moguća rešenja
Na x-koordinati , postoji duž AB , tačka A je nepokretna na x-koordinati , tačka B se nalazi na bilo kojem mestu x-koordinate , opisati ovo funkcijom .
Rešenja : A=a , B=x , AB=y
a) y=|a-x|
b) y=-|a-x|
c) y=a-x
d) y=x-a

GOOGLE TRANSLATOR
Since I discovered new possibilities, possible solutions

On the x-coordinate, there is straight AB, point A is fixed on the x-coordinate, point B is located at any point x-coordinates, to describe this function.

Solutions : A=a , B=x , AB=y
a) y=|a-x|
b) y=-|a-x|
c) y=a-x
d) y=x-a

 

[Mark44]

On the x-coordinate, there is straight AB, point A is fixed on the x-coordinate, point B is located at any point x-coordinates, to describe this function.
Solutions : A=a , B=x , AB=y

Here's how I would write it:
On the x-coordinate axis, there is straight line segment AB, point A is fixed on the x-coordinate axis at (a, 0), point B is located at (x, 0) any point x-coordinates, to describe this function.
Solutions : A=a , B=x , AB=y

a) y=|a-x|
b) y=-|a-x|

Choice a gives the distance from A to B, and will be positive unless a = x. Choice b is the negative of the distance from A to B.

By definition of the absolute value,
|a - x| = a - x, if a >= x
|a - x| = -(a - x) = x - a, if a <= x

c) y=a-x
d) y=x-a

These two are the negatives of one another. If x > a, then a - x < 0, or equivalently, x - a > 0.
If x < a, then a - x > 0, or equivalently, x - a < 0.

 

[point]

SERBIAN
Preslikavanje funkcije iz x-koordinate u ravan ( dekartov koordinatni sistem )
y=x-a , x i a ostaju na x-koordinatu , y ide na y-koordinatu .
prati sliku
https://pkxnqg.bn1302.livefilestore...WSh5idkVdC-swrTkqYaXV8fmts9x7Ks/ii.png?psid=1
prave iz x i a paralelne sa y-koordinate
prava iz y paralelna sa x-koordinate
u preseku pravih nastaju tačke A i B
tačke A i B se spajaju i dobija se duž AB
dato je za x=4 , a=2 , y=2
ponovimo postupak za x=3.5 , a=2 , y=1.5 , prati sliku
u preseku pravih nastaju tačke C i D
tačke C i D se spajaju i dobija se duž CD

https://befwwg.bn1302.livefilestore...DKraCcJKIy-UHkR4VeCHL_PmPvJTSMeM/i.png?psid=1
spajaju se tačke AC ( BD ) duži AB i CD
tačke ABDC čine površinu za 4≥x≥3.5

GOOGLE TRANSLATOR
The mapping function from the x-coordinates of the plane (Cartesian coordinate system)
y = x-a, x and a remain on the x-coordinate, y goes to the y-coordinate.
view photo
https://pkxnqg.bn1302.livefilestore...WSh5idkVdC-swrTkqYaXV8fmts9x7Ks/ii.png?psid=1
the lines of x and a parallel to the y-coordinates
line of y parallel to the x-coordinate
formed at the intersection of real points A and B
points A and B are combined and gets straight line AB
is given by x = 4, a = 2, y = 2
Repeat for x = 3.5, a = 2, y = 1.5, view photo
formed at the intersection of real points C and D
points C and D are combined and received straight line CD

https://befwwg.bn1302.livefilestore...DKraCcJKIy-UHkR4VeCHL_PmPvJTSMeM/i.png?psid=1
connect the dots AC (BD) straight lines AB and CD
ABDC points form the surface of 4≥x≥3.5

 

[point]

How to look graphics functions

a) y=|a-x|
b) y=-|a-x|
c) y=a-x
d) y=x-a
e) y=\{|a-x|\} \cup \{-|a-x|\}

 

[point]

a) y=|2-x|
graph, the red surface
https://cfxpzq.bn1302.livefilestore...duf35TDz7kJFlvpinPGfiGmOhMAVbDw/01.png?psid=1

b) y=-|2-x|
graph, the red surface
https://nq6hfq.bn1302.livefilestore...xhUjRvcF6i6WXHS-qNoA1O50MwSx-Kw/02.png?psid=1c) y=2-x
graph, the red surface
https://0nivia.bn1302.livefilestore...qqqkUyEXwbMEkel843ZQ20Rq__x6V-A/03.png?psid=1

d) y=x-2
graph, the red surface
https://d6pekg.bn1302.livefilestore...0AJlHabqjfVBAgiGORjpymT7vzmMKCA/04.png?psid=1

e) y=\{|2-x|\}\cup\{-|2-x|\} or y=\{2-x\}\cup\{x-2\}
graph, the red surface
https://qhdsnq.bn1302.livefilestore...HeXxWALY-BwcLW4G5ObhnQSYfKmEaVQ/05.png?psid=1

which are geometric objects obtained for valuesx and y , shape a≥x≥b ( a≥y≥b ) ? , you have a graph

 

[Mark44]

a) y=|2-x|
graph, the red surface
https://cfxpzq.bn1302.livefilestore...duf35TDz7kJFlvpinPGfiGmOhMAVbDw/01.png?psid=1

No.
The graph of the equation y = |2 - x| is only the two lines that form the lower boundary. What you show in red would be the graph of the inequality $y \ge |2 - x|$.

b) y=-|2-x|
graph, the red surface
https://nq6hfq.bn1302.livefilestore...xhUjRvcF6i6WXHS-qNoA1O50MwSx-Kw/02.png?psid=1

No, again. The equation y = -|2 - x| is the two lines that make up the upper boundary. What you show in red is the graph of the inequality $y \le -|2 - x|$.

c) y=2-x
graph, the red surface

No again, and this graph is not a surface -- it's a straight line. This equation represents only a straight line, not a two-dimensional region. The line passes through (0, 2) on the y-axis and (2, 0) on the x-axis.

https://0nivia.bn1302.livefilestore...qqqkUyEXwbMEkel843ZQ20Rq__x6V-A/03.png?psid=1

d) y=x-2
graph, the red surface
https://d6pekg.bn1302.livefilestore...0AJlHabqjfVBAgiGORjpymT7vzmMKCA/04.png?psid=1

No again. This is only a line, not a two-dimensional region. It passes through (0, -2) and (2, 0).

e) y=\{|2-x|\}\cup\{-|2-x|\} or y=\{2-x\}\cup\{x-2\}
graph, the red surface
https://qhdsnq.bn1302.livefilestore...HeXxWALY-BwcLW4G5ObhnQSYfKmEaVQ/05.png?psid=1

which are geometric objects obtained for valuesx and y , shape a≥x≥b ( a≥y≥b ) ? , you have a graph

You seem to have a misunderstanding about what the graphs of equations look like. Any equation of the form Ax + By = C represents a straight line, not a two-dimensional region. A linear inequality, such as Ax + By ≥ C or Ax + By ≤ C does represent a two-dimensional region, a half plane.

Until you understand these basic concepts of analytic geometry, it will be hard to understand what you are trying to do.

 

[point]

You seem to have a misunderstanding about what the graphs of equations look like. Any equation of the form Ax + By = C represents a straight line, not a two-dimensional region. A linear inequality, such as Ax + By ≥ C or Ax + By ≤ C does represent a two-dimensional region, a half plane.

Until you understand these basic concepts of analytic geometry, it will be hard to understand what you are trying to do.

do not weigh the current rules of mathematics
--------------
I explained the rules
"
The mapping function from the x-coordinates of the plane (Cartesian coordinate system)
y = x-a, x and a remain on the x-coordinate, y goes to the y-coordinate.
view photo
https://pkxnqg.bn1302.livefilestore...WSh5idkVdC-swrTkqYaXV8fmts9x7Ks/ii.png?psid=1
the lines of x and a parallel to the y-coordinates
line of y parallel to the x-coordinate
formed at the intersection of real points A and B
points A and B are combined and gets straight line AB
is given by x = 4, a = 2, y = 2
Repeat for x = 3.5, a = 2, y = 1.5, view photo
formed at the intersection of real points C and D
points C and D are combined and received straight line CD

https://befwwg.bn1302.livefilestore...DKraCcJKIy-UHkR4VeCHL_PmPvJTSMeM/i.png?psid=1
connect the dots AC (BD) straight lines AB and CD
ABDC points form the surface of 4≥x≥3.5"
for this rule is the solution for graphics functions

 

[Mark44]

do not weigh the current rules of mathematics

The "current rules" of mathematics are what we have to work with. New rules would come under the heading of "personal theory," which is not allowed here.

I explained the rules

Your explanations are incomprehensible, at least by me and one other mentor.

Since we seem to be talking past each other to no avail, I am closing this thread.