Question: Check whether the set of all pairs of real numbers of the form (1, x) with operations:
0

(1, y) + (1, x) = (1, y+x) and

k(1, y) = (1, ky) is a vector space.

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0

Solution:

(a) Closure:

$C_{1}:$ If u=(1, x), v=(1, y) are two element then,

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

Therefore, v is closed under addition.

$C_{2}:$ If u=(1, x) then,

ku = k(1, x) = (1, kx)

Therefore, v is closed under multiplication.

$A_{1}:$ u+v = (1, x)+(1, y) = (1, x+y) = (1, y+x) = (1, y)+(1, x) = v+u

$A_{2}:$ (u+v)+w = [(1, x)+(1, y)]+(1, z)=(1, x+y)+(1, z)

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

$A_{3}:$ u+0 = (1, x)+(1, 0) = (1, x+0) = (1, x) = u

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

$A_{4}:$ u+(-u) = (1, x)+(1, -x) = (1, x-x) = (1, 0)

$\therefore$ (1, -x) = -u is additive inverse of (1, x) = u

(c) Scalar Multiplication:

$M_{1}:$ k(u+v) = k[(1, x)+(1, y)] = k(1, x+y) = (1, k(x+y)) = (1, kx+ky) = (1, kx)+(1, ky) = k(1, x)+k(1, y) = ku+kv

$M_{2}:$ (k+l)u = (k+l)(1, x) = (1, (k+l)x) = (1, kx+kl) = (1, kx)+(1, lx) = k(1, x)+l(1, x) = ku+lu

$M_{3}:$ (k.l)u = (k.l) (1, x) = (1, klx) = k(1, lx) = k[l(1, x)] = k(lu)

$M_{4}:$ 1.u = (1, 1.x) = (1, x) = u

Hence, 1 is the multiplicative identity for this space.

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