We need still to check that the two planes α and β do not have any point outside the line a in common. In other words, we get the inclusion a⊆α∩β 46 Reasoning about the same points A and B and line a, axiom (I.6) yields that every point of line a lies in the plane β, too. By axiom (I.6), every point of line a lies in the plane α. By axioms (I.1) and (I.2), there exists a unique line a through points A and B. By axiom (I.7), the two planes have a further point B = A in common. Assume that the two planes have point of intersection A in common. Hence two different lines intersect at most in one point. Hence l = m, contradicting the assumption that l and m are two different lines. By axiom (I.2), there exists at most one line through these two points. Assume towards a contradiction they intersect at two different points A = B. Use definite names for the objects, explain the steps, and state which axioms are involved. Convince yourself of part (a), and give a short explanation. Two different lines have at most one point in common.If two lines have two or more points in common, they are equal.Any two different lines which are not parallel, have a unique point of intersection.Give at least two further useful formulations of Proposition 2.1 part (a). (c) A plane and a line not lying in this plane have either no or exactly one point in common. (b) Any two different planes have either no point in common, or they have one line and no further points in common. (a) Any two different lines have either no or exactly one point in common.
Proposition 2.1 (Hilbert’s Proposition 1). Among the consequences of the axioms of incidence, Hilbert spells out two Propositions.
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