# Why isn't the set V, as defined in the question body, a vector space?

Let $$V$$ denote the set of ordered pairs of real numbers and define our operations as follows:

For $$(a_1, a_2)$$ , $$(b_1, b_2)$$ $$\in$$ $$V$$ and $$c \in R$$,

$$(a_1, a_2) + (b_1, b_2) = (a_1 + b_1, a_2b_2)$$ and $$c(a_1, a_2) = (ca_1, a_2)$$

Now I've determined that this isn't a vector space, but I thought it was only because it fails one of the distributive rules, namely $$(a+b)x \neq ax + bx$$ for $$x \in V$$ and $$a,b \in R$$.

However, I'm told it also fails the additive inverse rule: $$\forall x \in V$$, $$\exists y \in V$$ such that $$x + y = 0$$.

The potential $$0$$ vector in this set would be $$(0, 1)$$, since $$\forall x \in V$$, $$x + (0, 1) = x$$. But now why can't we simply define an additive inverse as follows:

$$\forall x \in V$$, define $$y = (-x_1, \frac{1}{x_2})$$ $$\Rightarrow x + y = (x_1, x_2) + (-x_1, \frac{1}{x_2}) = (x_1 + (-x_1), x_2(\frac{1}{x_2})) = (0, 1) = 0$$

Are we not allowed to use division here, or something, since it's not defined as an operation in a simple vector space like this? But it is certainly defined on $$R$$, which is the field we are assuming for this potential vector space, right? What am I missing?

Your inverse is not defined for, say, the vector $$(1,0)$$. According to your formula, it should be $$(-1, \frac{1}{0})$$. But division by zero is, of course, undefined.
• Of course! Dividing by zero, the classic mistake! I didn't think to look for problems in $R$. Thanks a bunch. Mar 3, 2022 at 19:35