Show that $\mathbb R^2$ is not a vector space under these operations The operations are

*

*$(a,b) + (c,d) = (ac, bd)$

*$k(a,b) = (ka, kb)$, where $k$ belongs to $\mathbb R$.

Now I know that it's not a vector space because it's not closed under vector addition; and that to prove this I should make a numerical example that defies the rule.
Now I assumed random values of
$$u = (4,5)$$
$$v = (6,8) $$
but I'm not sure how to continue from there? I wanted to sum them to prove they won't yield the same result but of course they would..I'm kind of confused on how to formulate my counterexample.
 A: In fact isn´t a vector space, and the contradiction is with the sum  and the product defined.
It´s easy check that the problem is with the scalar product or with the sum , but for be illustrative I´m gonna check $1$ by $1$ property of vector space.Let us check:
Abelian group
Let $x=(x_1,x_2),y=(y_1,y_2),z=(z_1,z_2)\in V$ and $k\in \mathbb{R}$ Notice that:
Closure
$$x+y=(x_1y_1,x_2y_2)\in V$$
Conmutative
$$x+y=(x_1y_1,x_2y_2)=(y_1x_1,y_2x_2)=y+x$$
Neutral element
There are $e=(1,1)\in V$ such that for any $x\in V$ is true
$$x+e=(x_1\cdot 1, x_2 \cdot 1)=x$$
Inverse element
Given $x\in V$ such that $x\neq 0$ there exists $x^{-1}=(\frac{1}{x_1},\frac{1}{x_2})\in V$ such that
$$x+x^{-1}=(x_1\frac{1} \cdot {x_1},x_2 \cdot \frac{1}{x_2})=(1,1)=e$$
Hence $(V,+)$ is not a abelian group.
Otherwise $\left(V\setminus (0,0)+\right)$ is abelian group
From this part you can also argue that $V$ is not  a vector space since it isn´t a abelian group and as Bungo notice use that for $(0,0)$ we can´t find a inverse with the sum.
Now let us check that for $k\in \mathbb{R}$ $V$ is a semigroup.
Closure
$$kx=(kx_1,kx_2)\in V$$
Distributive
Here there are other new problem wich is because for one hand
$$k(x+y)=k(x_1y_1,x_2y_2)=(kx_1y_1,kx_2y_2)$$ but for the other hand
$$kx+ky=(kx_1,kx_2)+(ky_1,ky_2)=(k^2x_1y_1,k^2x_2y_2)$$
and hance
$(V,+,\cdot)$ is not a Vector space over $\mathbb{R}$
For the numerical contour example as above we show is enough  take a $k$ such that $k^2\neq k$ for it considere
$$x=(1,2),y=(1,1),k=2$$
And occur that
$$k(x+y)=k(1,2)=(2,4)$$ but $$kx+ky=(2,4)+(2,2)=(4,8)$$
which are different and then $V$ is not  a vector space.
I hope that it help you, and thanks to Bungo for fix my post.
