Ingeometry,Playfair's axiom is anaxiom that can be used instead of the fifth postulate ofEuclid (theparallel postulate):
In aplane, given a line and a point not on it, at most one lineparallel to the given line can be drawn through the point.[1]
It is equivalent to Euclid's parallel postulate in the context ofEuclidean geometry[2] and was named after the ScottishmathematicianJohn Playfair. The "at most" clause is all that is needed since it can be proved from the first four axioms that at least one parallel line exists given a lineL and a pointP not onL, as follows:
The statement is often written with the phrase, "there is one and only one parallel". InEuclid's Elements, two lines are said to be parallel if they never meet and other characterizations of parallel lines are not used.[3][4]
This axiom is used not only in Euclidean geometry but also in the broader study ofaffine geometry where the concept of parallelism is central. In the affine geometry setting, the stronger form of Playfair's axiom (where "at most one" is replaced by "one and only one") is needed since the axioms ofneutral geometry are not present to provide a proof of existence. Playfair's version of the axiom has become so popular that it is often referred to asEuclid's parallel axiom,[5] even though it was not Euclid's version of the axiom.
Proclus (410–485 A.D.) clearly makes the statement in his commentary on Euclid I.31 (Book I, Proposition 31).[6]
In 1785William Ludlam expressed the parallel axiom as follows:[7]
This brief expression of Euclidean parallelism was adopted by Playfair in his textbookElements of Geometry (1795) that was republished often. He wrote[8]
Playfair acknowledged Ludlam and others for simplifying the Euclidean assertion. In later developments the point of intersection of the two lines came first, and the denial of two parallels became expressed as a unique parallel through the given point.[9]
In 1883Arthur Cayley was president of theBritish Association and expressed this opinion in his address to the Association:[10]
WhenDavid Hilbert wrote his book,Foundations of Geometry (1899),[11] providing a new set of axioms for Euclidean geometry, he used Playfair's form of the axiom instead of the original Euclidean version for discussing parallel lines.[12]
Euclid's parallel postulate states:
If aline segment intersects two straightlines forming two interior angles on the same side that sum to less than tworight angles, then the two lines, if extended indefinitely, meet on that side on which the angles sum to less than two right angles.[13]
The complexity of this statement when compared to Playfair's formulation is certainly a leading contribution to the popularity of quoting Playfair's axiom in discussions of the parallel postulate.
Within the context ofabsolute geometry the two statements are equivalent, meaning that each can be proved by assuming the other in the presence of the remaining axioms of the geometry. This is not to say that the statements arelogically equivalent (i.e., one can be proved from the other using only formal manipulations of logic), since, for example, when interpreted in thespherical model ofelliptical geometry one statement is true and the other isn't.[14] Logically equivalent statements have the same truth value in all models in which they have interpretations.
The proofs below assume that all the axioms of absolute (neutral) geometry are valid.
The easiest way to show this is using the Euclidean theorem (equivalent to the fifth postulate) that states that the angles of a triangle sum to two right angles. Given a line and a pointP not on that line, construct a line,t, perpendicular to the given one through the pointP, and then a perpendicular to this perpendicular at the pointP. This line is parallel because it cannot meet and form a triangle, which is stated in Book 1 Proposition 27 inEuclid's Elements.[15] Now it can be seen that no other parallels exist. Ifn was a second line throughP, thenn makes an acute angle witht (since it is not the perpendicular) and the hypothesis of the fifth postulate holds, and so,n meets.[16]
The fifth postulate can "apparently" be proven in this manner. If the lines in Euclidean construction meet on the side where interior angles sum to greater than two right angles, we would have a triangle whose two angles sum to an angle greater than two right angles, which is contrary to Proposition 1.17. Therefore, they must meet on the side where interior angles sum to less than two right angles.
However, this assumes that the lines will meet. What we have proven is that "If lines meet, then they must meet on the side where interior angles sum to less than two right angles." Playfair's axiom guarantees that the lines must meet by assuming beforehand that there be only one parallel line from a given point (namely, the perpendicular of a perpendicular). All other lines through P, are not parallel, and hence, must meet.
The classical equivalence between Playfair's axiom and Euclid's fifth postulate collapses in the absence of triangle congruence.[17] This is shown by constructing a geometry that redefines angles in a way that respects Hilbert's axioms of incidence, order, and congruence, except for the Side-Angle-Side (SAS) congruence. This geometry models the classical Playfair's axiom but not Euclid's fifth postulate.
Proposition 30 of Euclid reads, "Two lines, each parallel to a third line, are parallel to each other." It was noted[18] byAugustus De Morgan that this proposition islogically equivalent to Playfair’s axiom. This notice was recounted[19] byT. L. Heath in 1908. De Morgan’s argument runs as follows:LetX be the set of pairs of distinct lines which meet andY the set of distinct pairs of lines each of which is parallel to a single common line. Ifz represents a pair of distinct lines, then the statement,
is Playfair's axiom (in De Morgan's terms, NoX isY) and its logically equivalentcontrapositive,
is Euclid I.30, the transitivity of parallelism (NoY isX).
More recently the implication has been phrased differently in terms of thebinary relation expressed byparallel lines: Inaffine geometry the relation is taken to be anequivalence relation, which means that a line is considered to beparallel to itself. Andy Liu[20] wrote, "LetP be a point not on line 2. Suppose both line 1 and line 3 pass throughP and are parallel to line 2. Bytransitivity, they are parallel to each other, and hence cannot have exactlyP in common. It follows that they are the same line, which is Playfair's axiom."