Question 1

1 Screenshot 2020-07-11 at 22.59.20

(a)

We have:

0_V = v - v = (v + z) - v

using commutativity:

(v + z) - v = (z + v) - v

and then using associativity we have:

(z + v) - v = z + (v - v) = z + 0_V = z

so z = 0_V as required.

(b)

Using the order from Definition 22.

Write v_1 \in \mathbb R \times \mathbb Z as v_1 = (x_1,k_1) and similarly v_2 \in \mathbb R \times \mathbb Z as v_2 = (x_2, k_2) and v_3 \in \mathbb R \times \mathbb Z as v_3 = (x_3, k_3).

[Axiom 1: Commutativity of Addition]

We have v_1 + v_2 = (x_1,k_1) + (x_2,k_2) = (x_1 + x_2, k_1 + k_2). Since real addition commutes we have x_1 + x_2 = x_2 + x_1 and k_1 + k_2 = k_2 + k_1 so (x_1 + x_2, k_1 + k_2) = (x_2 + x_1, k_2 + k_1) = (x_2,k_2) + (x_1, k_1) = v_2 + v_1 as required. So the commutativity axiom for addition is satisfied.

[Axiom 2: Associativity of Addition]

Using associativity of real addition:

\begin{align*}(v_1 + v_2) + v_3 & = ((x_1, k_1) + (x_2, k_2)) + (x_3,k_3) \\ & = (x_1 + x_2, k_1 + k_2) + (x_3, k_3) \\ & = (x_1 + x_2 + x_3, k_1 + k_2 + k_3) \\ & = (x_1 + (x_2 + x_3), k_1 + (k_2 + k_3)) \\ & = (x_1, k_1) + (x_2 + x_3, k_2 + k_3) \\ & = (x_1, k_1) + ((x_2, k_2) + (x_3, k_3)) \\ & = v_1 + (v_2 + v_3)\end{align*}

So the second axiom is satisfied.

[Axiom 3: Existence of Additive Identity]

Note that (0,0) \in \mathbb R \times \mathbb Z and we have:

\begin{align*}v_1 + (0,0) & = (x_1, k_1) + (0,0) \\ & = (x_1 + 0, k_1 + 0) \\ & = (x_1, k_0)\end{align*}

using the fact that 0 + x_1 = x_1 and 0 + k_1 = k_1. So (0,0) is an additive identity for V.

[Axiom 4: Existence of Additive Inverse]

We have that if x \in \mathbb R and k \in \mathbb Z, then -x \in \mathbb R and -k \in \mathbb Z. So if (x, k) \in \mathbb R \times \mathbb Z then (-x, -k) \in \mathbb R \times \mathbb Z, and so:

\begin{align*}v_1 + (-x_1, -k_1) & = (x_1, k_1) + (-x_1, -k_1) \\ & = (x_1 - x_1, k_1 - k_1) \\ & = (0,0)\end{align*}

this is the additive identity (-x_1, -k_1) is an additive inverse for v_1. (and since this was arbitrary, any element of V has an additive inverse)

[Axiom 5: Distributivity of Scalar Multiplication over Vector Addition]

Let \lambda \in \mathbb R. Then:

\begin{align*}\lambda (v_1 + v_2) & = \lambda (x_1 + x_2, k_1 + k_2) \\ & = (\lambda(x_1 + x_2), 0) \\ & = (\lambda x_1 + \lambda x_2, 0) \\ & = (\lambda x_1,0) + (\lambda x_2, 0) \\ & = \lambda(x_1, k_1) + \lambda (x_2, k_2) \\ & = \lambda v_1 + \lambda v_2\end{align*}

as required. So this axiom is also satisfied.

[Axiom 6: Distributivity of Scalar Multiplication over Field Addition]

Let \lambda, \mu \in \mathbb R. We have:

\begin{align*}(\lambda + \mu)v_1 & = (\lambda + \mu)(x_1, k_1) \\ & = ((\lambda + \mu)x_1, 0) \\ & = (\lambda x_1, 0) + (\mu x_1, 0) \\ & = \lambda(x_1, k_1) + \mu(x_1, k_1) \\ & = \lambda v_1 + \mu v_1\end{align*}

so this axiom is satisfied.

[Axiom 7]

Let \lambda, \mu \in \mathbb R. Then:

\begin{align*}(\lambda \mu) v_1 &= (\lambda \mu)(x_1, k_1) \\ & = ((\lambda \mu)x_1, 0) \\ & = (\lambda (\mu x_1), 0) \\ & = \lambda(\mu x_1, 0) \\ & = \lambda(\mu(x_1, k_1)) \\ & = \lambda(\mu v_1)\end{align*}

so this axiom is satisfied.

[Axiom 8: Identity for Scalar Multiplication]

The thing that stops V from being a vector space is the lack of an identity for scalar multiplication, because we have:

\begin{align*}1v_1 & = 1(x_1, k_1) \\ & = (1x_1, 0) \\ & = (x_1, 0) \\ & \ne v_1\end{align*}

if k \ne 0.

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