90 Cartesian coordinates; applications Ch. 6Ifb>0, thenx 0 <x 0 +bandsoby6.1.1U 0 ≤mU 1. In this case we say that the
correspondence between≤land≤misdirect.
Ifb<0thenx 0 +b<x 0 and soU 1 ≤mU 0 .In this case we say that the correspon-
dence between≤land≤misindirect. In what follows we assume thatb>0sothat
the correspondence between≤land≤mis direct. The other case can be covered by
replacing≤mby≥min the following.U 2 U 0
Z 0
Z 2
U 3 ′ U 3 ′′
Z 3 ′
Z 3 ′′
U 0 U 2
Z 0 Z^2
U 3
Z 3
Suppose now that Z 2 ≤l Z 3 ;
we wish to show thatU 2 ≤m
U 3 whereU 2 =πm(Z 2 ),U 3 =
πm(Z 3 ). We subdivide into three
cases.
U 1 U 2
Z 1 Z^2
U 3
Z 3
Figure 6.5.
CASE 1. Suppose thatZ 2 ≤lZ 0 .ThenZ 2 ≤lZ 0 ≤lZ 1 so thatZ 0 ∈[Z 2 ,Z 1 ].Then
by 4.3.2,U 0 ∈[U 2 ,U 1 ].AsU 0 ≤mU 1 ,wethenhaveU 2 ≤mU 0. There are now two
possibilities, thatZ 3 ≤lZ 0 or thatZ 0 ≤lZ 3. In the first of these subcases,Z 3 ∈[Z 2 ,Z 0 ]
soU 3 ∈[U 2 ,U 0 ].AsU 2 ≤mU 0 we then haveU 2 ≤mU 3. In the second of these subcases
we haveZ 0 ∈[Z 2 ,Z 3 ]soU 0 ∈[U 2 ,U 3 ].AsU 2 ≤mU 0 we haveU 0 ≤mU 3 soU 2 ≤mU 3.
CASE 2. Suppose thatZ 0 ≤lZ 2 ≤lZ 1 .ThenZ 2 ∈[Z 0 ,Z 1 ]soU 2 ∈[U 0 ,U 1 ].As
U 0 ≤mU 1 thenU 0 ≤mU 2 ≤mU 1 .NowZ 2 ∈[Z 0 ,Z 3 ]soU 2 ∈[U 0 ,U 3 ].AsU 0 ≤mU 2 it
follows thatU 2 ≤mU 3.
CASE 3. Suppose thatZ 1 ≤lZ 2 .ThenZ 1 ∈[Z 0 ,Z 2 ]so thatU 1 ∈[U 0 ,U 2 ].AsU 0 ≤m
U 1 we then haveU 1 ≤mU 2 .ThenZ 2 ∈[Z 1 ,Z 3 ]soU 2 ∈[U 1 ,U 3 ].AsU 1 ≤mU 2 we have
U 2 ≤mU 3.
Now continuing with all three cases, we note thatU 2 ≡(x 0 +bt 2 , 0 ),U 3 ≡(x 0 +
bt 3 , 0 )and asU 2 ≤mU 3 by 6.1.1 we havex 0 +bt 2 ≤x 0 +bt 3 .Asb>0 this implies
thatt 2 ≤t 3.
We also have thatt 2 ≤t 3 impliesZ 2 ≤lZ 3. For otherwiseZ 3 ≤lZ 2 andsobythe
abovet 3 ≤t 2 , which gives a contradiction unlessZ 2 =Z 3.
Whenlis perpendicular toOIwe useπOJinstead ofπm. By a similar argument
we reach the same conclusion.
(iii) This follows directly from (ii) of the present theorem. It can also be proved
as follows. Note that in (6.4.2)t=0givesZ 0 andt=1givesZ 1. Then forZ≡(x,y)