Let V 2 integral For u v element V and a element R define v

Let V = (2, integral). For u, v element V and a element R define vector addition by u v:= uv - 2(u + v) + 6 and scalar multiplication by a u:= (u - 2)^a + 2. It can be shown that (V, ) is a vector space over the scalar field R. Find the following: the sum: 5 5 = the scalar multiple: -3 5 = the additive inverse of 5: 5 = the zero vector: 0_v = the additive inverse of x:

Solution

5 + 5 = 5.5-2(5+5) +6 = 25-20+6 = 11

-3 . 5 = (5-2)^-3 +2 = 3^-3 + 2 = 1/27 + 2 = (54+1)/27 = 55/27

Now to find additive inverse of 5; we first try to find zero vector of V :

u + e = e => ue-2(u+e)+6 = e

=> ue -2u -2e +6 = e

=> ue -2u + 6 = 3e

=> (u-3)e = 2(u-3)

=> e=2

Hence zero of V is 2

Now we want : a + 5 = 2

=> a.5 - 2(a+5) + 6 = 2

=> 5a -2a-10+6 = 2

=> 3a = 6

=> a = 2

Hence - 5 = 2

Additive inverse of 5 is 2

0_V = 2

Additive inverse of x is :

a + x = 2

ax - 2(a+x) +6 = 2

=> ax - 2a -2x = -4

=> a(x-2) = -4+2x

=> a(x-2) = 2(x-2)

=> a=2

Hence additive inverse of x is 2