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12.4: Axiom I

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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.

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 $$[ST]$$.

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.