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?


1 Answer 1


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.

  • 1
    $\begingroup$ Of course! Dividing by zero, the classic mistake! I didn't think to look for problems in $R$. Thanks a bunch. $\endgroup$
    – notadoctor
    Mar 3, 2022 at 19:35

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