Let V denote the set of ordered pairs of real numbers. For $(a_1, a_2), (b_1,b_2) \in V, (a_1, a_2)+(b_1,b_2)=(a_1+b_1, a_2b_2)$.

I am trying to prove that a set V is (or is not) a vector space and I am stuck at proving the additive identity works and at showing the additive inverse doesn't. I have found the zero element to be (0,1) but I'm not sure how to work that into a proof. I am also not sure how to logically prove that if $a=(a_1,0)$, there is no inverse.

  • $\begingroup$ Update: check edit $\endgroup$ – p3ngu1n Jan 14 '16 at 0:41
  • $\begingroup$ To perhaps clarify Jonas's question, I'll add the alternative, or is $V = \mathbb{R}^2 \setminus \{(a, 0) \mid a \in \mathbb{R}\}$? $\endgroup$ – Brian Tung Jan 14 '16 at 0:42

$0$ element just means an element $x$ such that $\forall y \in V$ $x+y = y$.

So take $(a_1,a_2)$. Then

$$(0,1) + (a_1,a_2) = (0 + a_1,1 \cdot a_2) = (a_1,a_2)$$

Meanwhile, additive inverse for some $x$ means an element $y$ such that $x + y = (0,1)$.

Let $x = (a_1,0)$ and suppose $y = (b_1,b_2)$ existed. Then,

$$(0,1) = (a_1,0) + (b_1,b_2) = (a_1 + b_1,0)$$

So assuming that $V = R \times R$ is a Cartesian product, then this would imply in particular that $0 = 1$.


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