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Given the operation in $\mathbb{R}^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 $\mathbb{R}$. Looking at the Additive Zero Axiom, we get:

$$ (x_1,y_1) + \boldsymbol{0} = (x_1(0),y_1(0)) = \boldsymbol{0}$$

To satisfy the Additive Zero Axiom, $(x_1,y_1) + \boldsymbol{0} = (x_1,y_1)$ must be true. For this to be true, $\boldsymbol{0}$ would have to be $(1,1)$

Is this possible, or would we be able to say this is not a vector space?

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    $\begingroup$ By far nothing's gone wrong. Go ahead and check the other axioms $\endgroup$
    – Trebor
    Sep 16 at 3:13
  • $\begingroup$ Welcome to the site! If the issue has been resolved, accepting and/or upvoting answers is the best way to say "thanks!": it scores points, signals resolution, and prevents bumping and automatic deletion. $\endgroup$
    – Ryan G
    2 days ago
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Is not a vector space, for instance let us try find the zero element in $\mathbb{R}^2$ with the given operation.

Let $P=(x,y)\in \mathbb{R}^2$

$(x,y)+(e_1,e_2)=(x,y)$ implies $(xe_1,ye_2)=(x,y)$ 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.

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  • $\begingroup$ Thank you for your help! This now makes complete sense. $\endgroup$
    – Kevin G
    Sep 16 at 2:27
  • $\begingroup$ You´re welcome! $\endgroup$ Sep 16 at 2:28
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The additive identity is indeed $(1,1)$.

Let's check for inverse of $(0,0)$.

For any $x, y \in \mathbb{R}$,

$$(0, 0) + (x, y)= (0,0) \ne (1,1).$$ Hence it can't be a vector space.

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  • $\begingroup$ Thank you very much for your reply! This makes sense now, (0,0) does not have an additive inverse with (1,1) being the additive identity $\endgroup$
    – Kevin G
    Sep 16 at 2:26

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