Sec. 11.2 Sum of couples, multiplication of a couple by a scalar 189(v) For alla,b∈V,a+b=b+a.
(vi) Next,R×V→V is a function.(vii) For alla∈V and allt 1 ,t 2 ∈R,t 2 .(t 1 .a)=(t 2 t 1 ).a.
(viii) For alla,b∈V and allt∈R,t.(a+b)=t.a+t.b.(ix) For alla∈V and allt 1 ,,t 2 ∈R,(t 1 +t 2 ).a=t 1 .a+t 2 .a.
(x) For alla∈V, 1 .a=a.
We then have the following result.
(V(Π;O),+,.)is a vector space overR.
Proof.
(i) This has been covered already in 11.2.1.
(ii) Now(O,Z 1 )+(O,Z 2 )=(O,Z 4 )where(x 4 ,y 4 )=(x 1 +x 2 ,y 1 +y 2 ).Then
[(O,Z 1 )+(O,Z 2 )]+ (O,Z 3 )=(O,Z 4 )+(O,Z 3 )=(O,Z 5 ),
where
(x 5 ,y 5 )=(x 4 +x 3 ,y 4 +y 3 )=((x 1 +x 2 )+x 3 ,(y 1 +y 2 )+y 3 ).
Similarly(O,Z 2 )+(O,Z 3 )=(O,Z 6 )where(x 6 ,y 6 )=(x 2 +x 3 ,y 2 +y 3 ),andso
(O,Z 1 )+[(O,Z 2 )+(O,Z 3 )] = (O,Z 1 )+(O,Z 6 )=(O,Z 7 )
where(x 7 ,y 7 )=(x 1 +x 6 ,y 1 +y 6 )=(x 1 +(x 2 +x 3 ),y 1 +(y 2 +y 3 )). ClearlyZ 5 =Z 7.
(iii) For anyZ 1 ∈Π,(O,Z 1 )+(O,O)=(O,Z 2 )where(x 2 ,y 2 )=(x 1 + 0 ,y 1 + 0 )=
(x 1 ,y 1 ),sothatZ 2 =Z 1. Similarly(O,O)+(O,Z 1 )=(O,Z 3 )where(x 3 ,y 3 )=( 0 +
x 1 , 0 +y 1 )=(x 1 ,y 1 ),sothatZ 3 =Z 1.
(iv) Now(O,Z 1 )+(O,Z 2 )=(O,Z 3 ),(O,Z 2 )+(O,Z 1 )=(O,Z 4 )where(x 3 ,y 3 )=
(x 1 +x 2 ,y 1 +y 2 )and(x 4 ,y 4 )=(x 2 +x 1 ,y 2 +y 1 ). ClearlyZ 3 =Z 4.
(v) If(x 2 ,y 2 )=(−x 1 ,−y 1 ),then(O,Z 1 )+(O,Z 2 )=(O,Z 3 )where
(x 3 ,y 3 )=(x 1 −x 1 ,y 1 −y 1 )=( 0 , 0 ); henceZ 3 =O. Similarly(O,Z 2 )+(O,Z 1 )=
(O,Z 4 )where(x 4 ,y 4 )=(−x 1 +x 1 ,−y 1 +y 1 )=( 0 , 0 ); henceZ 4 =O.
(vi) This was covered in 11.2.1.
(vii) Fort 1 .(O,Z 1 )=(O,Z 2 )where(x 2 ,y 2 )=(t 1 x 1 ,t 1 y 1 ).Then
t 2 .(t 1 .(O,Z 1 )) =t 2 .(O,Z 2 )=(O,Z 3 )where(x 3 ,y 3 )=(t 2 (t 1 x 1 ),t 2 (t 1 y 1 )).Also
(t 2 t 1 ).(O,Z 1 )=(O,Z 4 )where(x 4 ,y 4 )=((t 2 t 1 )x 1 ,(t 2 t 1 )y 1 ). ThusZ 3 =Z 4.
(viii) For(O,Z 1 )+(O,Z 2 )=(O,Z 3 )andt.[(O,Z 1 )+(O,Z 2 )] =t.(O,Z 3 )
=(O,Z 4 )where(x 3 ,y 3 )=(x 1 +x 2 ,y 1 +y 2 ),(x 4 ,y 4 )=(t(x 1 +x 2 ),t(y 1 +y 2 )).Also
t.(O,Z 1 )=(O,Z 5 ),t.(O,Z 2 )=(O,Z 6 )where(x 5 ,y 5 )=(tx 1 ,ty 1 ),
(x 6 ,y 6 )=(tx 2 ,ty 2 ). Moreover(O,Z 5 )+(O,Z 6 )=(O,Z 7 )where(x 7 ,y 7 )=(x 5 +
x 6 ,y 5 +y 6 )=(tx 1 +tx 2 ,ty 1 +ty 2 ). HenceZ 4 =Z 7.
(ix) Fort 1 .(O,Z 1 )=(O,Z 2 ),t 2 .(O,Z 1 )=(O,Z 3 ),(t 1 +t 2 ).(O,Z 1 )=(O,Z 4 )and
(O,Z 2 )+(O,Z 3 )=(O,Z 5 )where(x 2 ,y 2 )=(t 1 x 1 ,t 1 y 1 ),(x 3 ,y 3 )=(t 2 x 1 ,t 2 y 1 )and
(x 4 ,y 4 )=((t 1 +t 2 )x 1 ,(t 1 +t 2 )y 1 ). Moreover(x 5 ,y 5 )=(x 2 +x 3 ,y 2 +y 3 )=(t 1 x 1 +
t 2 x 1 ,t 1 y 1 +t 2 y 1 ). ClearlyZ 4 =Z 5.
(x) For 1.(O,Z 1 )=(O,Z 2 )where(x 2 ,y 2 )=( 1 .x 1 , 1 .y 1 )=(x 1 ,y 1 ). ThusZ 2 =Z 1.