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...

  • 1
    $\begingroup$ When they say "for any $(x_1,x_2)\in \Bbb R^2$, it really means for any. So yes, whenever $y_1$ and $y_2$ are real numbers, $(y_1,y_2)$ is in $\Bbb R^2$, and thus $a(y_1,y_2)=(ay_1,ay_2)$ for any $a\in\Bbb R$. $\endgroup$
    – pancini
    Apr 25, 2020 at 19:57
  • $\begingroup$ @ElliotG oh, so the first axiom holds then... $\endgroup$
    – user707245
    Apr 25, 2020 at 20:11
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    $\begingroup$ Try just checking if $(x_1,x_2)+(y_1,y_2)=(y_1,y_2)+(x_1,x_2)$. $\endgroup$
    – pancini
    Apr 25, 2020 at 20:17
  • $\begingroup$ ye, you are right. It isn't and my O.P is wrong! $\endgroup$
    – user707245
    Apr 25, 2020 at 20:30
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    $\begingroup$ You wrote $(y_1,y_2)+(x_1,x_2)=(y_1+x_1,y_2+x_2)$, which is not correct. You cannot use two definitions of $+$ at the same time here. $\endgroup$
    – pancini
    Apr 25, 2020 at 20:31

1 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)


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

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.


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