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 andprofessionalcriticism.

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.