This proof will be far more detailed than what you’ll see in later sections. (We usually do not bother to take note of the axioms, one by one.) We will also try to explain our reasoning as we go, to help you get used to the sort of careful reasoning involved in a proof. Lines that should actually be included in the proof will be set aside in block quotes.
First, we are proving a “for all” (universally quantified) statement. This means we should be careful not to assume anything about the vector we choose, so that our argument can apply to any vector we want:
Let \(\vv\) be any vector in \(V\text{.}\)
Next, you might want to remind yourself of the goal: we want to show that \((-1)\vv=-\vv\text{.}\) You can state that this is what you want to show, but it’s not absolutely necessary to do this. A common trick that shows up in a lot of mathematical proofs is a simple bit of arithmetic: \(a=b\) is the same thing as \(a-b=0\text{!}\) So really, what we want to show is that \((-1)\vv+\vv=\zer\text{.}\)
Now, remember that \(1\vv=\vv\text{,}\) and \(0\vv=\zer\text{,}\) so what we want to show is equivalent to \((-1)\vv+1\vv=0\vv\text{.}\) Remove the \(\vv\text{,}\) and we’re left with \(-1+1=0\text{,}\) and that we definitely know is true! (Of course, we can’t just “remove \(\vv\)”, but we can use the distributive property!)
This is basically the proof, but we need to state all the axioms we use, and best practice in logical arguments is that we should begin with our assumptions, and statements we agree are true, and proceed from those to the desired conclusion.
Since we know that \(-1+1=0\text{,}\) it follows that
\begin{equation*}
(-1+1)\vv=0\vv\text{.}
\end{equation*}
On the left hand side, we can use the distributive property S3 to get
\((-1+1)\vv=(-1)\vv+1\vv\text{.}\) On the right hand side, we can use
Theorem 1.2.1 to get
\(0\vv=\zer\text{.}\) Therefore,
\begin{equation*}
(-1)\vv+1\vv=\zer\text{.}
\end{equation*}
By axiom S5, \(1\vv=\vv\text{,}\) so we have \((-1)\vv+\vv=\zer\text{.}\)
OK, that’s more or less where we said we wanted to get to, and then we can just move \(\vv\) to the other side as \(-\vv\text{,}\) and we’re done. But we want to be careful to state all axioms! The rest of the proof involves carefully stepping through this process.
Another way to proceed, which shortcuts this whole process, is to use
Part d of
Exercise 1.2.2: since the additive inverse of
\(\vv\) is the unique vector
\(-\vv\) such that
\(-\vv+\vv=\zer\text{,}\) and
\((-1)\vv+\vv=\zer\text{,}\) it must be the case that
\((-1)\vv=-\vv\text{.}\)
This approach is completely valid, and you are free to use it, but we will take the long route to demonstrate further use of the axioms.
Since \((-1)\vv+\vv=\zer\text{,}\) we can add \(-\vv\) to both sides of the equation, giving
\begin{equation*}
((-1)\vv+\vv)+(-\vv)=\zer+(-\vv)\text{.}
\end{equation*}
By the associative property (axiom A3), \(((-1)\vv+\vv)+(-\vv)=(-1)\vv+(\vv+(-\vv))\text{,}\) and by the identity axiom A5, \(\zer+(-\vv)=-\vv\text{.}\) This gives us
\begin{equation*}
(-1)\vv+(\vv+(-\vv))=-\vv\text{.}
\end{equation*}
By axiom A5, \(\vv+(-\vv)=\zer\text{,}\) so \((-1)\vv+\zer=-\vv\text{.}\) Finally, we use axiom A4 one last time, and we have our result: \((-1)\vv=-\vv\text{.}\)