V = {x,y $ \epsilon $ IR | x + y = xy and c.x = x$^c$}

(i) x + y = xy

x $ \epsilon $ IR and y $ \epsilon $ IR $ \implies $ xy $ \epsilon $ IR

Therefore, x + y $ \epsilon $ IR

Therefore, V is closed under addition

(ii) cx = x$^c$

c $ \epsilon $ IR, x $ \epsilon $ IR $ \implies $ x$^c$ $ \epsilon $ IR

Therefore, cx $ \epsilon $ V

Therefore, V is closed under scalar multiplication

(iii) (x+y) = xy = yx [Product of two real numbers is commutative]

Therefore, (x+y) = xy = yx = (y+x)

Therefore, '+' is commutative in V

(iv) (x+y) + z

= (xy) + z

= (xy)z

= x(yz) [Multiplication is associative in IR]

= x(y+z)

= x + (y+z)

Therefore, '+' is associative in V

(v) Consider 1 $ \epsilon $ IR

x + 1 = x.1 = x

Therefore, x + 1 = x and 1 $ \epsilon $ IR

'1' is additive identity in V

(vi) Consider $ \frac{1}{2} \epsilon $ IR [x $ \epsilon $ IR and x $ \neq $ 0 $ \implies \frac{1}{x} \epsilon $ IR]

x + $ \frac{1}{x} $ = x $ \frac{1}{x} $ = 1 [Additive identity]

Therefore, $ \frac{1}{x} $ is the additive inverse of 'x' in V

Therefore, additive inverse exists in V

(vii) Consider, k(x+y) = k(xy) = xy$^k$

= x$^k$.y$^k$

= x$^k$ + y$^k$

= kx + ky [x$^k$ = kx and y$^k$ = ky]

Therefore, Distributivity of scalar multiplication is satisfied.

(viii) (k+l)(x) = x$^{k+l}$ = x$^k$.x$^l$ = (kx).(lx) = kx + lx

Therefore, Distributivity of scalars is satisfied in V

(ix) (kl)(x) = x$^{kl}$ = (x$^l$)$^k$ = k(x$^l$) = k(lx)

Therefore, Associative law of Scalars is satisfied.

(x) Let, c = 1

1.x = x$^1$ = x

Therefore, '1' is multiplication identity in V

Therefore, V has multiplicative identity.

Therefore, all the ten axioms are satisfied.

Therefore, V is a vector space.