# Given the operation in RR^2: (x_1,y_1)+(x_2,y_2)=(x_1x_2,y_1y_2) I would like to find whether this is a vector space in RR. Looking at the Additive Zero Axiom, we get: (x_1,y_1)+0=(x_1(0),y_1(0))=0 To satisfy the Additive Zero Axiom, (x_1,y_1)+0=(x_1,y_1) must be true. For this to be true, 0 would have to be (1,1) Is this possible, or would we be able to say this is not a vector space?

Given the operation in ${\mathbb{R}}^{2}$:
$\left({x}_{1},{y}_{1}\right)+\left({x}_{2},{y}_{2}\right)=\left({x}_{1}{x}_{2},{y}_{1}{y}_{2}\right)$
I would like to find whether this is a vector space in $\mathbb{R}$. Looking at the Additive Zero Axiom, we get:
$\left({x}_{1},{y}_{1}\right)+\mathbf{0}=\left({x}_{1}\left(0\right),{y}_{1}\left(0\right)\right)=\mathbf{0}$
To satisfy the Additive Zero Axiom, $\left({x}_{1},{y}_{1}\right)+\mathbf{0}=\left({x}_{1},{y}_{1}\right)$ must be true. For this to be true, 0 would have to be (1,1)
Is this possible, or would we be able to say this is not a vector space?
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Toby Barron
Is not a vector space, for instance let us try find the zero element in ${\mathbb{R}}^{2}$ with the given operation.
Let $P=\left(x,y\right)\in {\mathbb{R}}^{2}$
$\left(x,y\right)+\left({e}_{1},{e}_{2}\right)=\left(x,y\right)$ implies $\left(x{e}_{1},y{e}_{2}\right)=\left(x,y\right)$ and then ${e}_{1}=1$ and ${e}_{2}=1$ it is the zero element must be (1,1).
But in this case (0,0) isn´t invertible since (0,0)+(a,b)=(1,1) implies (0,0)=(1,1) which is a contradiction.
###### Did you like this example?
priscillianaw1
The additive identity is indeed $\left(1,1\right)$
Let's check for inverse of $\left(0,0\right)$
For any $x,y\in \mathbb{R}$
$\left(0,0\right)+\left(x,y\right)=\left(0,0\right)\ne \left(1,1\right).$
Hence it can't be a vector space.