[b]Metric Axiom 1: If A and B are points, then d(A[/b],B[b]) is greater than or equal to zero and d(A[/b],B[b])=0 [/b][i]iff [/i][b]A=B. [/b][br][i]Proof: [/i]The distance formula in the Poincare disk model is [math]d\left(A,B\right)=\left|ln\left(\frac{AM\cdot BN}{AN\cdot BM}\right)\right|[/math]By nature of the absolute value signs, d(A,B) will always be nonnegative. [br]To show that d(A,B)=0 [i]iff[/i] A=B, first assume that d(A,B)=0. We know that only ln(1)=0. Thus, [br][math]\left(\frac{AM\cdot BN}{AN\cdot BM}\right)=1[/math]. This can only occur when A=B. Similarly, if we assume that A=B, we can show that [br][math]d\left(A,B\right)=\left|ln\left(\frac{AM\cdot BN}{AN\cdot BM}\right)\right|=\left|ln\left(\frac{AM\cdot AN}{AN\cdot AM}\right)\right|=\left|ln\left(1\right)\right|=0[/math]. Thus, the second qualification of Metric Axiom 1 is also satisfied. So, Metric Axiom 1 holds for the Poincare distance. [br][br][b]Metric Axiom 2: If A and B are points, then d(A[/b],B[b])=d(B[/b],A[b]). [/b][br][i]Proof:[/i] We know that [math]d\left(A,B\right)=\left|ln\left(\frac{AM\cdot BN}{AN\cdot BM}\right)\right|[/math]. Consider, [br][math]d\left(B,A\right)=\left|ln\left(\frac{BM\cdot AN}{BN\cdot AM}\right)\right|=\left|-ln\left(\frac{AM\cdot BN}{AN\cdot BM}\right)\right|=\left|ln\left(\frac{AM\cdot BN}{AN\cdot BM}\right)\right|=d\left(A,B\right)[/math]. Thus, Metric Axiom 2 holds for the Poincare distance.