Is this a vector space and is this the correct way to approach the problem?

Question

See if, based on the operations presented, it is a vector space in the real numbers.

For any $$(x_1, x_2), \space (y_1, y_2) \in \Bbb R^{2}; \space a \in \Bbb R$$ The operations presented are: $$(x_1, x_2) + (y_1, y_2) = (x_1, x_2-y_2)$$ and $$a(x_1, x_2) = (ax_1, ax_2)$$

My first question is if, based on them saying $$a(x_1, x_2) = (ax_1, ax_2)$$ does it mean that $$a(y_1, y_2) = (ay_1, ay_2)?$$ and would this apply if the multiplication was defined in any other way, like: $$a(x_1, x_2) = (0, ax_1)$$ then we could assume that $$a(y_1, y_2) = (0, ay_1)?$$

I tried to verify the first axiom based on their definition and my thought was as follows: $$(x_1, x_2) + (y_1, y_2) = (x_1, x_2-y_2) $$ $$(y_1, y_2) + (x_1, x_2) = (y_1+x_1, y_2+x_2)$$ This means that it doesn't fit the criteria to be a vector space... Is this the correct way to prove it? Also, I'm not sure if the order of the elements in the sums is relevant or not... I'm assuming it is based on their definition, but since we are dealing with elements in the real numbers I could also assume they aren't. In that case we could get something like: $$(y_1, y_2) + (x_1, x_2) = (y_1+x_1, y_2+x_2) = (x_1+y_1, x_2+y_2) = (x_1, x_2-y_2)$$ In this case, we have commutation and the first axiom holds...

Answer 1

It is given that $(x_1, x_2)+(y_1, y_2) = (x_1, x_2-y_2)$ for any $(x_1, x_2), (y_1, y_2) \in \mathbb{R}^2$ -- (equivalently, for any $x_1,x_2,y_1,y_2 \in \mathbb{R}$).

Let $x_1=1, x_2=2, y_1=3, y_2=4$. Clearly, $(x_1, x_2), (y_1, y_2) \in \mathbb{R}^2$ as $x_1, x_2, y_1, y_2 \in \mathbb{R}$.

Then we have

$(x_1, x_2)+(y_1, y_2) = (x_1, x_2-y_2)=(1, 2-4)=(1, -2)$.

So $(x_1, x_2)+(y_1, y_2)=(1, -2)$. (i)

Conversely,

$(y_1, y_2)+(x_1, x_2)=(y_1, y_2-x_2)=(3, 4-2)=(3, 2)$.

So $(y_1, y_2)+(x_1, x_2)=(3, 2)$. (ii)

From (i) and (ii), $(x_1, x_2)+(y_1, y_2) \neq (y_1, y_2)+(x_1, x_2)$. We have identified vectors in the space which do not satisfy the first vector space axiom. It follows that this is not a vector space.