(a) Closure:

C1: If u=(1, y), v=(1, y') are two element then,

u+v = (1, y)+(1, y') = (1, y+y')

Therefore, v is closed under addition.

C2: If u=(1, y) then,

ku = k(1, y) = (1, ky)

Therefore, v is closed under multiplication.

(b) Addition:

A1: u+v = (1, y)+(1, y') = (1, y+y') = (1, y'+y) = (1, y')+(1, y) = v+u

A2: (u+v)+w = [(1, y)+(1, y')]+(1, z)=(1, y+y')+(1, z)

(u+v)+w = (1, (y+y')+z) = (1, y+(y'+z)) = (1, y)+[(1, y'+z)] = (1, y)+[(1, y')+(1, z)] = u+(v+z)

A3: u+0 = (1, y)+(1, 0) = (1, y+0) = (1, y) = u

0 = (1, 0) is zero for this space.

A4: u+(-u) = (1, y)+(1, -y) = (1, y-y) = (1, 0)

∴ (1, -y) = -u is additive inverse of (1, y) = u

(c) Scalar Multiplication:

M1: k(u+v) = k[(1, y)+(1, y')] = k(1, y+y') = (1, k(y+y')) = (1, ky+ky') = (1, ky)+(1, ky') = k(1, y)+k(1, y') = ku+kv

M2: (k+l)u = (k+l)(1, y) = (1, (k+l)y) = (1, ky+kl) = (1, ky)+(1, ly) = k(1, y)+l(1, y) = ku+lu

M3: (k.l)u = (k.l) (1, y) = (1, kly) = k(1, ly) = k[l(1, y)] = k(lu)

M4: 1.u = (1, 1.y) = (1, y) = u

Hence, 1 is the multiplicative identity for this space.

Since all the actions are satisfied so V is vector space.