# Vector space judgment

Define $(a_1,a_2)+(b_1,b_2)=(a_1+b_1,0)$ and $c(a_1,a_2)=(ca_1,0)$
With these operations, the following conditions
(1)There exists an element in $V$ denoted by $0$ such that $x+0=x$ for each $x$ in $V$
(2) For each element $x$ in $V$ there exists an element $y$ in $V$ such that $x+y=0$
fail.
Thus it is not a vector space.

I need more detailed explanation why these two rules do not hold.
Isn't it possible if I put $b_1=-a_1, b_2=-a_2$?

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Let the $0$ element be $(a_0,b_0)$

Then $(a_1,a_2)+(a_0,b_0)=(a_1,a_2)$ must hold for all $(a_1,a_2)\in V$

But $(a_1,a_2)+(a_0,b_0)=(a_1+a_0,0)$, We can choose $a_0=0$ to match first element of the tuple but second element of the tuple is $0$ regardless of what you choose $b_0$ and thus equality doesn't hold for all $(a_1,a_2)\in V$

Therefore, $\not\exists 0\in V$ such that $x+0=x$. Since there is no $0$, second condition doesn't mean anything

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Such a zero element doesn't exist, unless we are considering vectors only of the form $(a_1,0)$. To prove this, suppose $0=(z_1,z_2)$. Then $(a_1,a_2) + (z_1,z_2) = (a_1+z_1,0)$. But this must be equal to $(a_1,a_2)$. And this must be true for all $(a_1,a_2)$ (in particular, consider $(z_1,z_2) + (z_1,z_2)$). This gives $a_2=z_2=0$.

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