Let $V$ be a vector space and for $u,v\in{}V$ let $\langle{}u,v\rangle$ define an inner product on $V$. If {$u,v$} is an orthonormal set in $V$, then $\|u-v\|=\sqrt{2}$
Somehow I'm getting $\sqrt{-2}$ and not $\sqrt{2}$...
Here's what I have:
$$ \begin{align} \|u-v\| &= \sqrt{(u-v)(u-v)}\\ & = \sqrt{\|u\| - 2u\cdot{}v + \|v\|} \end{align}$$ Since {$u,v$} is an orthonormal set, $\|u\|$ and $\|v\|$ are zero. $$ \begin{align} &= \sqrt{-2u\cdot{}v}\end{align}$$ Since {$u,v$} is an orthonormal set, $u\cdot{}v = 1$ $$\begin{align} &= \sqrt{-2} \end{align}$$
Where have I gone wrong?
You've not applied the definition of orthonormal correctly: You cannot conclude that $\|u\| = 0$, but rather that $\|u\| = 1$. The only term that is zero is the product $u \cdot v$.
Hence, you get $\sqrt{1 - 2 \cdot 0 + 1}$.
Another red flag that you should look for: Any time you conclude a vector has norm zero, you can conclude that the vector is zero.