Betweenness Propositions

    A purpose of the Hilbert Betweenness Axioms (Hilbert's Order Axioms) is to give meaning to the undefined term between; seeing as between is undefined, we have only the axioms to guarantee that the points in this geometry behave in a way that is consistent with our interpretation of "betweenness". This topic points out that there are exactly two sides to a line in the Euclidean plane (Half-Planes Proposition), and that lines are not circular (Betweenness Property), and how to decompose a line into its parts (Linear Decomposition and Line Separation Propositions). The interior of an angle and the Crossbar Proposition are also detailed. Some of these results were taken for granted by Euclid.

Definition (Side Of A Line) Let be any line and let and be any points that do not lie on If or if segment contains no points lying on we say and are on the same side of   , whereas if and segment does intersect we say that and are on opposite sides of A side of a line is the set of all points that are on the same side of as some particular point not lying on and is denoted by

    The Linearity Axiom makes clear that betweenness does imply collinearity and between and is the same as  between and The Extension Axiom ensures that there are points between two distinct points, that a line does not end at and and that a ray is larger than a segment. The Order Axiom ensures that a line is linear and not circular. The Seperation Axiom is not a valid axiom in three-dimensional space and thus indirectly guarantees two-dimensional geometry. Further, if and are on the same side of and if and are not on the same side of , then by the Separation Axiom and law of the excluded middle, and are not on the same side of

Proposition (Half-Planes) Every line bounds exactly two half-planes and these half-planes have no point in common.

    Proof. It needs to be shown that there are at least two sides to a line, that every point in the plane not on is on one of the sides, and that no point is on both sides.
    By the Point Not On Line Proposition, there is a point not lying on a line By the Points On  Line Axiom, there is a point lying on By the Extension Axiom there is a point such that Then the Side Of A Line Definition, and are on opposite sides of so has at least two sides.
    To prove that every point in the plane is on one of the sides, let be a point not lying on By the Law Of The Excluded Middle and the Separation Axiom, if and are not on the same side of then and are on the same side of So the set of points not on is the union of the side of and the side of
    By the Separation Axiom, if were on both sides, then and would be on the same side, which contradicts that and are on opposite sides. Therefore, there is no point in common.

Proposition (Three-Point Property) Given If is a point not on the line determined by and then
    (i) and are on the same side of and
    (ii) and are on opposite sides of
    
        Proof. (i): Assume that and are on opposite sides of Then by the Side Of A Line Definition, follows because intersects in a unique point which must be by the Point Uniqueness Proposition. However, contradicts the hypothesis   by the Order Axiom.  Thus, and are on the same side of
    (ii): The points and are on opposites sides of because  by the Linearity Axiom and are distinct and the segment meets line at a point, namely because intersecting lines and have a unique point in common, by the Point Uniqueness Proposition.
    

Proposition (Betweenness Property) Given points and if either
    (i) and or
    (ii) and
then and are distinct and collinear. Further,
    (iii) if and then
    (iv) if and then
    (v) if and then and
    (vi) if and then
    
    Proof.  (i): By the Linearity Axiom, and are distinct and and are distinct. It remains to show that and are distinct. Suppose that then is which by the Order Axiom, contradicts Therefore,   and are distinct. By the Linearity Axiom, and are collinear, say line and and are collinear, say line By the Line Uniqueness Axiom, there is a unique line through and , and thus so that and are collinear.
    (ii): By the Linearity Axiom, and are distinct and and are distinct. It remains to show that and are distinct. Suppose that then is which by the Order Axiom, contradicts Therefore,   and are distinct. By the Linearity Axiom, and are collinear, say line and and are collinear, say line By the Line Uniqueness Axiom, there is a unique line through and , and thus so that and are collinear.
    (iii): By part (i), and are distinct. Let be a point not on line determined by and By the Three Point Proposition, points and are on the same side of and and are on the opposite sides of Thus, by the Separation Axiom, and are on opposite sides of By the Side Of A Line Definition to it follows
    (iv): By part (i), and are distinct. By the Three Point Proposition, points and are on the opposite sides of and and are on the same side of Thus, by the Separation Axiom, and are on opposite sides of By the Side Of A Line Definition to it follows
    (v): By part (i), and are distinct. By the Three Point Proposition, points and are on the opposite sides of By part (iii) and the Three Point Proposition, and are on the same side of Thus, by the Separation Axiom, and are on opposite sides of By the Side Of A Line Definition to it follows
    (vi): By part (i), and are distinct. By the Three Point Proposition, points and are on the same side of and and are on the opposite sides of Thus, by the Separation Axiom, and are on opposite sides of By the Side Of A Line Definition to it follows

Proposition (Linear Decomposition) For any two points and ,
    (i)  
    (ii)   and
    (iii) if is a line containing and not then every point of the ray (except ) is on the same side of as
Further, given
    (iv) and and
    (v) and
    
        Proof.  (i): By the Ray Definition and Segment Definition, and also, by the Linearity Axiom, By definition of intersection, it follows To prove let be any point in , it will be shown that belongs to If or then is an endpoint of and so then belongs to otherwise, by the Ray Definition and the Linearity Axiom, and are collinear. By the Order Axiom either, or Since belongs to , neither nor can hold. Therefore, must hold which implies that belongs to
    (ii): Let be any point in the set of points on the line If or then If and then by the Order Axiom and the Ray Definition, either or Thus, by definition of union, Conversely, by the Ray Definition, and therefore equality holds.
    (iii): Let be any point on that is on the opposite side of as and thus However, by the Ray Definition, or thus by the Order Axiom, cannot be on the opposite side of as
    (iv):  First, it is shown that If then because and the Betweenness Property implies, Thus, Similiarily, and thus by the definition of union Conversely, if then by the Three Point Property, points and are on opposite sides of Suppose Then by the Three Point Property, and are on the same side of By the Seperation Axiom, and are on opposite sides of so by the Side Of A Line Definition, By definition of union,   To prove that note that by the Segment Definition and definition of intersection, Assume for a contradiction that and By the Segment Definition and definition of intersection, and By the Betweenness Property, and yields The Order Axiom yields a contradiction because and and thus no such exists; whence
    (v):  By the Betweenness Property, with and it follows implies By definition of subset, Suppose By the Order Axiom either or The Separation Axiom and the Three Point Proposition imply that and cannot both hold. Therefore,  if then either or By the Ray Definition, To show conversely, let Assume for a contradiction that Then and by the Betweenness Property, imply which contradicts by the Order Axiom. Therefore, To show note that by the Ray Definition and definition of intersection Let and By the Ray Definition and definition of intersection, and By the Betweenness Property, and yields The Order Axiom yields a contradiction because and and thus no such exists; whence   
    
    

Proposition (Line Seperation Property) If and is the line through and then for every point lying on lies either on ray or on the opposite ray

    Proof.  Suppose that does not lie on By the Order Axiom and the Ray Definition, if does not lie on then If then lies on so assume then by the Order Axiom exactly one of the relations or holds. Assume for a contradiction that holds, then by the Order Axiom again, exactly one of the relations or holds. If then combining this with gives contradicting the hypothesis of . If then combining this with gives contradicting If then combining this with and the Order Axiom gives contradicting the hypothesis of Therefore, or which means that lies on the opposite ray     

Proposition (The Pasch Theorem)  If and are distinct noncollinear points and is any line intersecting in a point between and then also intersects  either or If does not lie on then does not intersect both and

    Proof.  Either lies on or it does not. Assume does not lie on , because otherwise the proposition holds by the Line Segment Definition. By the Side Of A Line Definition, points and lie on opposite sides of , because and do not lie on and the segment does intersect By the Half-Planes Proposition either is on the same side of as or on the same side of as If is on the same side of as then is on the opposite side of from which means that intersects and does not intersect similiarily if is on the same side of as then intersects and does not intersect by the Seperation Axiom.

Definition (Interior Of An Angle) Given an angle define a point to be in the interior of  the angle if is on the same side of as and if is also on the same side of as

Proposition (Interior Angle Property) Given an angle
    (i) If the point is lying on line then is in the interior of if and only if
Further, if is in the interior of then
    (ii) so is every other point on ray except
    (iii) no point on the opposite ray to is in the interior of and
    (iv) if then is in the interior of
    
        Proof. (i): Suppose is lying on and is in the interior of Then, by the Order Axiom either or By the Three Point Property, implies and are on opposite sides of contrary to the hypothesis that is interior to By the Three Point Property, implies and are on opposite sides of contrary to the hypothesis that is interior to Therefore, by the Order Axiom. Conversely, suppose Then by the Three Point Property and are on the same side of and and are on the same side of By the Interior Of An Angle Definition, is in the interior of
    (ii): By the Ray Definition and Interior Of An Angle Definition, is not in the interior of the angle By the Linear Decomposition Proposition, every point on the ray is on the same side of as and every point of the ray is on the same side of as By the Interior Of An Angle Definition, every point not on is in the interior of angle
    (iii): Suppose is a point on the opposite ray to Then because otherwise, by the Ray Definition, would contradict the Order Axiom. Assume for a contradiction that is in the interior of Then, and are on the same side of and are on the same side of and and are on the same side of However, by the Three Point Property, and are on opposite sides. Therefore, by the Law Of The Excluded Middle, is not in the interior of that is, no point on the opposite ray to is in the interior of
    (iv): Since is in the interior of is on the same side of as By assumption and are on the same side of   By the Three Point Property, and are on opposite sides of . Then, by the Separation Axiom, and are on opposite sides of . Therefore, by the Linear Decomposition Property, and do not meet. By part (ii), and do not meet where is any point on the opposite ray of which exists by the Extension Axiom. By the Linear Separation Property, and it follows that does not meet and so by the Interior Of An Angle Definition, is in the interior of

    
    

Definition (Ray Between) Two Rays Ray is between rays and if and are not opposite rays and is interior to

Proposition (Crossbar Theorem) If is between and then intersects segment

    Proof. Assume for a contradiction that and do not intersect. By the Extesnison Axiom there exists a point such that By the Line Separation Property, and is the only point in common to and By the Interior Angle Property every point on except and is in the interior of and also no point on is in the interior of So, by the Law Of The Excluded Middle, does not meet Since does not meet and does not meet Therefore, and are on the same side of
    However, the following argument shows that and must be on opposite sides of By the Extension Axiom there exists a point such that and by the Interior Angle Property, is in the interior of