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Math Help - Transcendental Theory, Slope of a line, and how Pi+e and -(e/Pi) is transcendental

  1. #1
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    Transcendental Theory, Slope of a line, and how Pi+e or -(e/Pi) is transcendental

    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.
    Last edited by madgadjt; April 1st 2012 at 10:10 PM.
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  2. #2
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    Re: Transcendental Theory, Slope of a line, and how Pi+e and -(e/Pi) is transcendenta

    Hi,

    I don't know much about number theory, so my comment may not be "professional", but I think your third claim:

    3. For all points not algebraic, if one coordinate is algebraic, the other must be transcendental.

    may be correct for curves which have one algebraic point. But I believe that it is not known whether the curve y = pi*x + e has an algebraic point or not.

    jens
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  3. #3
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    Smile Re: Transcendental Theory, Slope of a line, and how Pi+e and -(e/Pi) is transcendenta

    Np, I meant professional as in don't flame me lol..

    All points (algebraic and transcendental) in a 2d plane have an infinite number of lines with transcendental slope that pass through it, and an algebraic point may lie on it. It is easy to find if you choose a point yourself such as x=2, y=1.
    The rules only apply to linear slopes (straight lines) as of this moment, and since the line equation y = pi*x + e is a linear (straight line), all points on a straight line with transcendental slope other than x=2 y=1 would then follow rule 3 in order to lie on the line otherwise its slope would be algebraic and contradict the transcendental nature of the line itself.

    --
    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.

    --

    *Because the set of algebraic numbers is countable, the set of lines with transcendental slope and one algebraic point is also countable. I am working on a way to decide if the line falls within this subset determined by one given point (could just be y intercept) and the slope. If it does, it implies there is an algebraic point somewhere on the line, and that it can be calculated. This calculation may be what determines if the line falls within this set.

    A line that falls within this set is such that m=(1-Pi), x=0, y=pi (essentially the y-intercept)
    The algebraic point that lies on this line is x=1,y=1

    --
    One or both of the following coordinates: x= -(e/Pi) or y= Pi+e is transcendental.
    --

    *Because both lie on the line, and only one could be algebraic.
    Last edited by madgadjt; April 2nd 2012 at 02:41 AM. Reason: updated description of algebraic point subset
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  4. #4
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    Re: Transcendental Theory, Slope of a line, and how Pi+e or -(e/Pi) is transcendental

    Quote Originally Posted by madgadjt View Post
    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.
    Kind of a trivial proof but, okay.

    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.[/quote]
    You proved before that a line with transcendental slope has no more than one algebraic point. How do you know (1, pi+ e) is not that one algebraic point?

    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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