Question

[2 marks] Let V be the set of all ordered pairs of real numbers (u1, uv) with uj > 0. Consider the following addition and scalar multiplication operations on u = (u1, u2) and v = (v1, v2): u + v = (421, uz + v2), ku = (kuq, kuz) If the set V with the above operations satisfies Axiom 5 of a vector space (i.e., the existence of a negative element), what would be the negative of the vector u = (6,3)? Enter the components of the negative of u (in order) into the answer box below, separated with a comma. e.g., '-25,-28' (without the quotes).

Answer 1

If  , Then

u = (41, 42), v = (v1, 02)

U101 u+v=( 2,42 + 02)

So if $\mathbf{v}$ has to be the negative of $\mathbf{u}$ , then
u + v= e
where
e= (e1, C2)
is identity (zero).

2) -,u2 + ) = a +n=n

So and

Hei -= ulei = 4

0 = ta = in=ta + in
. So
e= (4,0)

Hence

,

U11 u+v=( U2+U2) = (4,0)

= 4 了 4
  
0 = a + in

,

16 > U1= -,V2 = -2

So negative of
u = (41, U2)
is
(in-:-
. Applying this to
u = (6,3)
, we get its negative to be

$(\frac{16}{6},-3)=(\frac{8}{3},-3)$ Answer.