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Show that the set V of positive real numbers with operations Addition: $x + y = xy$ Scalar Multiplication: $kx = x^k$ is a vector space where x, y are any two real numbers and k is any scalar.
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Axioms of vector spaces:

  • Closure Axioms (C)
  • Addition Axioms (A)
  • Scalar Multiplication Axioms (S)

1. Closure (C)

$C_{1}$: Since x + y = xy and x, y are real numbers

$\therefore$ V is closed under addition.

$C_{2}$: Since $kx = x^k$, where k is given as any scalar.

Hence $x^k$ is a real number.

$\therefore$ V is closed under scalar multiplication.

2. Addition (A)

$A_{1}$: Commutativity

$$x + y = xy = yx = y + x$$

$A_{2}$: Associativity

$$(x + y) + z = xy + z = x + yz = x + (y + z)$$

$A_{3}$: Existence of additive identity

$$x + 1 = x \\ 1 = x$$

for all x.

Here, 1 is zero for this operation of addition

$A_{4}$: Existence of additive inverse

$$x + \frac{1}{x} = x * \frac{1}{x} = 1$$

3. Scalar Multiplication

$M_{1}$: Distributivity of scalar multiplication

$$k(x + y) = kxy = (xy)^k \\ = x^k y^k = x^k + y^k \\ = kx + ky$$

$M_{2}$: Distributivity of scalars

$$(k + l)x = x^{k + l} = x^k * x^l \\ = x^k + x^l = kx + kl$$

$M_{3}$: Associative law of scalars

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

$M_{4}$: Existence of multiplicative identity

$$1 * x = x^1 = x \\ \therefore 1 *x = x$$

for every x

Hence V is a vector space

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Huihuihui mazaakk kar rahi thii , upper wale aadmi ne ghalat kara hai .. ye vector space form nahi kar rha :)

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