PROOF 59
Prove.
Removing the parallel postulate
On Euclid's axiom system.
A way to read this proof
First separate what is assumed from what is to be proved. At each step, ask which definition or earlier result permits the move. If it is unclear, return to that premise rather than skipping to the conclusion.
Definition D1
Euclid builds plane geometry on five postulates.
Proof.
(1)Postulate 1. A segment can be drawn joining any two points.convention
(2)Postulate 2. A segment can be extended straight indefinitely.convention
(3)Postulate 3. Given a center and a radius, a circle can be drawn.convention
(4)Postulate 4. All right angles are equal to one another.convention
(5)Postulate 5 (original form). If two lines meet a line so that the two interior angles on the same side sum to less than two right angles, the two lines, extended on that side, meet.convention
(6)The fifth postulate is equivalent to the following: through a point off a line there is exactly one line not meeting it. This walk uses this form as A1.(5), A2
(7)The first four postulates are each one line long; only the fifth is long. This difference is the question of this walk.(1), (2), (3), (4), (5)
∎
Definition D2
A model of an axiom system is an assignment of concrete meanings to its undefined terms under which all the axioms become true.
Proof.
(1)The undefined terms of plane geometry are point and line. What is called a point and what a line is not fixed.convention, D1
(2)Two parallel lines means two lines that do not meet.convention
∎
Lemma L1
When two parallel lines are met by a line, the alternate angles are equal.
Proof.
(1)Suppose two parallel lines ℓ and m meet a line t at points Q, P respectively.assumption
(2)ℓ and m do not meet, so P is not on ℓ.(1), D2
(3)A line m′ can be drawn through P whose alternate angle with t equals ∠(t, ℓ).(1), (2), A4
(4)m′ and ℓ have equal alternate angles, so they are parallel.(3), A3
(5)There is only one line through P parallel to ℓ, so m′ = m(1), (2), (4), A1
(6)Therefore the alternate angle m makes with t equals ∠(t, ℓ).(3), (5)
(7)A3 is the converse direction and does not use A1. A1 is used at line (5) only.(4), (5), A3, A1
∎
Theorem T1
The three interior angles of a triangle sum to 180°.
Proof.
(1)In triangle ABC, take the line ℓ through A parallel to line BC. Such a line is unique.A1
(2)Take two points D, E on ℓ with A between D and E, with B and D on the same side of line AC, and C and E on the same side of line AB.(1), convention
(3)Line AB cuts the two parallel lines ℓ and BC, so ∠DAB = ∠ABC(1), (2), L1
(4)Line AC cuts the two parallel lines ℓ and BC, so ∠EAC = ∠ACB(1), (2), L1
(5)D, A, E are collinear, so ∠DAB + ∠BAC + ∠CAE = 180°(2), A5
(6)∠ABC + ∠BAC + ∠ACB = 180°(3), (4), (5), E
∎
Theorem T2
The fifth postulate and "every triangle has angle sum 180°" are equivalent.
Proof.
(1)With the fifth postulate, every triangle has angle sum 180°.T1, D1
(2)Conversely, if every triangle has angle sum 180°, the fifth postulate holds. This direction is given by A6.A6
(3)Each statement follows from the other, so they are equivalent.(1), (2), convention
(4)Hence the fifth postulate can be restated as a property of angles. Whichever is taken as the axiom, the same plane results.(3)
∎
Theorem T3
The fifth postulate cannot be proved from Postulates 1 through 4.
Proof.
(1)Read the interior of a circle as the plane, and the circular arcs meeting that circle at right angles, together with its diameters, as lines. Call this the hyperbolic model.D2
(2)Under this interpretation, Postulates 1 through 4 are all true.(1), D1, D2
(3)Under this interpretation, through a point off a line there are infinitely many lines not meeting it. Hence the fifth postulate is false.(1), D1, D2
(4)The axiom system of Postulates 1–4 plus the negation of the fifth has a model, so that system is consistent.(2), (3), A7
(5)If the fifth postulate were provable from Postulates 1–4, the system of (4) would prove both the fifth postulate and its negation, producing a contradiction.(4), convention
(6)Therefore the fifth postulate cannot be proved from Postulates 1–4.(4), (5)
(7)The fifth postulate is false here, so by the contrapositive of A6, not every triangle in this model has angle sum 180°.(3), A6, A8
(8)The spherical model exhibits another negation. Reading each pair of antipodal points on a sphere as one point and great circles as lines, any two lines always meet, so there are no parallel lines at all.D2
(9)In this model, however, Postulate 2 must be read as "a line has no end". Read as "a line has infinite length", Postulate 2 becomes false.(8), D1
(10)Independence is shown by the model of (1); the model of (8) shows that the negation is not of one kind.(6), (8), (9)
∎
Corollary C
Vertical angles are equal. And this proof does not use the fifth postulate.
Proof.
(1)Two lines meet at a point, forming four angles. Let ∠a, ∠b be two adjacent angles, and ∠c the angle adjacent to ∠b and opposite ∠a.assumption
(2)∠a and ∠b are two angles on one line, so ∠a + ∠b = 180°(1), A5
(3)∠b and ∠c are two angles on one line, so ∠b + ∠c = 180°(1), A5
(4)∠a + ∠b = ∠b + ∠c(2), (3), E
(5)∠a + ∠b = ∠c + ∠b(4), A10
(6)∠a = ∠c(5), A9
(7)Lines (2) through (6) never cite A1. Hence Proof 48 remains intact when the fifth postulate is removed.(2), (3), (4), (5), (6)
∎
Remark
Can A1 be removed from this axiom system?
This walk places the theorems that use A1 and those that do not side by side. Check which side survives the removal.