I need a place for ideas to be reviewed in some way and heard this may be a good place to start.
I have always been interested in math and science, and would like to share some ideas I have about transcendental numbers. Pi is what got me interested, until I found out that almost all numbers share its transcendental properties, and that it is hard to prove that a certain number is transcendental, specifically the sums, products and powers of transcendental numbers such as Pi and e. It is a work in progress, and I would like to get some input on if I am stating everything correctly, and what I can do to iron out any errors.
I appreciate any comments and professional criticism.
Transcendental Slope
Theorem
The slope of a straight line may be transcendental.
Proof
The linear slope form of any number x may be produced by:
m=(x/1)
m=x
If x is transcendental, then the slope of a line m is transcendental.
Example
Pi is proven to be transcendental by the Lindemann-Weierstrass Theorem
m=(Pi/1)
m=Pi
Slope m is transcendental.
Points on a linear line with Transcendental Slope
For points lying on a linear line that has transcendental slope:
1. No more than one algebraic point exists on a linear line with transcendental slope.
- An algebraic point can be chosen as the origin of a line with transcendental slope.
(An algebraic point is a point in which both its x and y coordinates are algebraic numbers.)- Two algebraic points on a line will determine an algebraic slope.
(If the algebraic point is unknown, a formula is needed to determine if any algebraic point lies on a line with transcendental slope m, and what its coordinates may be.)
2. For all points not algebraic, at least one of its coordinates must be transcendental.
- Given points P1 and P2 where P1x ne P2x or P1y ne P2y, If P1 is algebraic, the x and/or y coordinate of P2 is transcendental.
- Two algebraic points on a line will determine an algebraic slope, a contradiction to transcendental slope.
3. For all points not algebraic, if one coordinate is algebraic, the other must be transcendental.
- Given points P1 and P2 where P1x ne P2x or P1y ne P2y, If P1 is algebraic, the x and/or y coordinate of P2 is algebraic, the other must be transcendental.
- Two algebraic points on a line will determine an algebraic slope, a contradiction to transcendental slope.
Pi+e or -(e/Pi) is Transcendental
Proof
Using the linear equation of a straight line where x and y are coordinates of a point, m is the slope of the line, and b is the y-intercept:
y=mx+b (Slope Intercept Form)
Both Pi and e are proven to be transcendental by the Lindemann-Weierstrass Theorem.
If m=Pi, x=1, and b=e we can solve for y:
y=(Pi/1)*1+e
y=(Pi+e)
The point x=1, y=(Pi + e) lies on a line with the transcendental slope of Pi.
From "Points of a Transcendental Slope":
3. For all points not algebraic, if one coordinate is algebraic, the other must be transcendental.
We imply the coordinate y=(Pi + e) is transcendental because coordinate x=1 is algebraic.
Under Investigation
From Points of a Transcendental Slope:
1. No more than one algebraic point exists on a line with Transcendental Slope.
To imply "y=(Pi + e) is transcendental because x=1 is algebraic" one of the following conditions must be satisfied:
- There is no algebraic point that lies on this line.
- A point on this line other than x=1, y = (Pi + e) is algebraic.
Further Applications
Using slope intercept form, if m = Pi, y = 0, and b=e we can solve for x:
0= (Pi/1)*x+e
x= -(e/Pi)
And thus:
The point x= -(e/Pi), y=0 also lies on a line with the transcendental slope of Pi.
Of the two points on the line with transcendental slope, coordinates x=1 and y=0 are algebraic.
From Points of a Transcendental Slope:
1. No more than one algebraic point exists on a line with Transcendental Slope.<br>
3. For all points not algebraic, if one coordinate is algebraic, the other must be transcendental.
One or both of the following coordinates:
x= -(e/Pi) or y= Pi+e
is transcendental.


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