Can someone help me know if the axioms of this set were done right?

Question

The set of all pairs of real numbers $(x,y)$ with the operations $(x,y) + (x',y') = (y + y' , x + x' )$ and $k(x,y) = (5kx, 5ky)$.

Determine if it is a vector space, and if not, list one axiom that fails.

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So what I did was:

${u = (x, y), v = (x', y'), w = (x", y") | u, v, w ∈ \mathbb{V}}$

1. If $u$ and $v ∈ \mathbb{V}$, then $u + v ∈ \mathbb{V}$

$$u + v =(x, y) + (x', y') = (x + x', y + y')$$

2. If $u$ and $v ∈ \mathbb{V}$, then $u + v = v + u$

$$u + v = (x + x', y + y')$$ $$v + u = (x' + x, y' + y)$$

3. If $u$, $v$ and $w ∈ \mathbb{V}$, then $u + (v + w) = (u + v) + w$

$$u + (v + w) = (x, y) + [(x', y') + (x", y")] = (x, y) + (x' + x", y' + y") = (x + x' + x", y + y' + y")$$

$(u + v) + w = (x + x', y + y') + (x", y") = (x + x' + x", y + y' + y")$

4. If $u ∈ \mathbb{V}$, then there exists a vector $-u ∈ \mathbb{V}$, such that $u + (-u) = 0$:

$$-u = (-1)(x, y) = (-5x, -5y)$$ $$u + (-u) = (x, y) + (-5x, -5y) = (-4x, -4y)$$

5. There exists a $0 ∈ \mathbb{V}$, such that $0 + u = u + 0 = u$:

$$0 = (0, 0)$$ $$0 + u = (0, 0) + (x, y) = (x, y)$$ $$u + 0 = (x, y) + (0, 0) = (x, y)$$ $$u = (x, y)$$

6. If $u ∈ \mathbb{V}$ and $k ∈ \mathbb{R}$, then $ku ∈ \mathbb{V}$

$$ku = k(x, y) = (5kx, 5ky)$$

7. If $u, v ∈ \mathbb{V}$, then $k(u + v) = ku + kv$

$$k(u + v) = k(x + x', y + y') = (5k(x + x'), 5k(y + y'))$$ $$ku = (5kx, 5ky)$$ $$kv = k(x', y') = (5kx', 5ky')$$ $$ku + kv = (5kx, 5ky) + (5kx', 5ky') = (5kx + 5kx', 5ky + 5ky')$$

8. If $u ∈ \mathbb{V}$ and $k, l ∈ \mathbb{R}$, then $(k + l)u = ku + lu$:

$$(k + l)u = (k + l)(x, y) = (5(k + l)x, 5(k + l)y)$$ $$ku = (5kx, 5ky). lu = l(x, y) = (5lx, 5ly)$$ $$ku + lu = (5kx, 5ky) + (5lx, 5ly) = (5kx + 5lx, 5ky + 5ly)$$

9. If $u ∈ \mathbb{V}$ and $k, l ∈ \mathbb{R}$, then $k(lu) = (kl)u$:

$$k(lu) = k[(5lx, 5ly)] = (25klx, 25kly)$$ $$(kl)u = (kl)(x, y) = (5klx, 5kly)$$

10. If $u ∈ \mathbb{V}$, then $1u = u$:

$$1u = 1(x, y) = (5x, 5y)$$ $$u = (x, y)$$

So if I did it right, axioms 4, 9 and 10 are not satisfied. And then this isn't a vector space.

Answer 1

You're basically correct.
Some comments.

• Axiom 4. does hold. The a priori notation $-u$, especially after the word 'exists', is misleading. Every vector has an additive inverse, that's true, but if other axioms fail, it might not be equal to $(-1)u$.
• Axioms 1-5. hold because it is the usual addition, which is very well known to have these properties.
• It is enough to point out one single axiom that fails (easiest now is Ax.10), then you can calmly say

  No, it's not a vector space.

Answer 2

Your analysis of $4$ is wrong.

It stipulates that for every $u$, there is a vector called $-u$, such that $u + (-u) = 0$. It does not say that $-u$ has to be $(-1)u$ (that is a consequence of the other axioms). For this case, $4$ is satisfied if you set $-u$ to be $(-x, -y)$ for $u = (x,y)$.

The other analyses are fine.