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Mathematics LibreTexts

12.4: Axiom I

  • Page ID
    23657
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    Evidently, the h-plane contains at least two points. Therefore, to show that Axiom I holds in the h-plane, we need to show that the h-distance defined in 12.1 is a metric on h-plane; that is, the conditions (a) - (d) in Definition 1.3.1 hold for h-distance.

    The following claim says that the h-distance meets the conditions (a) and (b)

    Claim \(\PageIndex{1}\)

    Given the h-points \(P\) and \(Q\), we have \(PQ_h \ge 0\) and \(PQ_h=0\) if and only if \(P=Q\).

    Proof

    According to Lemma 12.3.1 and the main observation (Theorem 12.3.1), we may assume that \(Q\) is the center of the absolute. In this case

    \(\delta(Q,P)=\dfrac{1+QP}{1-QP}\ge 1\) 

    and therefore

    \(QP_h=\ln[\delta(Q,P)] \ge 0.\) 

    Moreover, the equalities holds if and only if \(P=Q\).

    The following claim says that the h-distance meets the condition

    Claim \(\PageIndex{2}\)

    For any h-points \(P\) and \(Q\), we have \(PQ_h=QP_h\).

    Proof

    Let \(A\) and \(B\) be ideal points of \((PQ)_h\) and \(A,P,Q,B\) appear on the circline containing \((PQ)_h\) in the same order.

    截屏2021-02-24 上午9.53.00.png

    Then

    \(\begin{array} {rcl} {PQ_h} & = & {\ln \dfrac{AQ \cdot BP}{QB \cdot PA} =} \\ {} & = & {=\ln \dfrac{BP \cdot AQ}{PA \cdot QB}=} \\ {} & = & {QP_h} \end{array}\)

    The following claim shows, in particular, that the triangle inequality (which is Definition 1.3.1d) holds for \(h\)-distance.

    Claim \(\PageIndex{3}\)

    Given a triple of h-points \(P\), \(Q\), and \(R\), we have

    \(PQ_h+QR_h \ge PR_h.\) 

    Moreover, the equality holds if and only if \(P\), \(Q\), and \(R\) lie on one h-line in the same order.

    Proof

    Without loss of generality, we may assume that \(P\) is the center of the absolute and \(PQ_h \ge QR_h >0\).

    Suppose that \(\Delta\) denotes the h-circle with the center \(Q\) and h-radius \(\rho=QR_h\). Let \(S\) and \(T\) be the points of intersection of \((PQ)\) and \(\Delta\).

    By Lemma 12.3.3, \(PQ_h\z\ge QR_h\). Therefore, we can assume that the points \(P\), \(S\), \(Q\), and \(T\) appear on the h-line in the same order.

    According to Lemma Lemma 12.3.4, \(\Delta\) is a Euclidean circle; suppose that \(\hat Q\) denotes its Euclidean center. Note that \(\hat Q\) is the Euclidean midpoint of \([ST]\).

    截屏2021-02-24 上午9.59.23.png

    By the Euclidean triangle inequality

    \[PT = P\hat{Q}+\hat{Q} R \ge PR\]

    and the equality holds if and only if \(T=R\).

    By Lemma  Lemma 12.3.2,

    \(\begin{array} {l} {PT_h = \ln \dfrac{1 + PT}{1 - PT},} \\ {PR_h = \ln \dfrac{1 + PR}{1 - PR}.} \end{array}\)

    Since the function \(f(x)=\ln\frac{1+x}{1-x}\) is increasing for \(x\in[0,1)\), inequality 12.4.1 implies

    \(PT_h\ge PR_h\)

    and the equality holds if and only if \(T=R\).

    Finally, applying Lemma 12.3.3 again, we get that

    \(PT_h=PQ_h+QR_h.\)

    Hence the claim follows.