Let $ u$ and $ v$ be vectors in $ \mathbb R^n$ . The exercise is to prove that if $ ||u+tv|| \ge ||u||$ for all real $ t$ , then $ u\cdot v=0$ ($ u$ and $ v$ are perpendicular).
I tried writing $ v$ as $ (n+xu)$ , where $ u\cdot n=0$ , and then try to prove that $ x$ must be zero, but was unable to develop this solution.
I have been stuck with this problem for a while now. I have a proof that letting $ U$ be an ultrafilter, exactly one of $ x,x^*$ belongs to $ U$ for all $ x$ in $ B$ , I did this by showing that both belong to $ U$ implies that $ U=B$ which cannot be the case, and if neither long to $ U$ then a contradiction can be derived by de-Morgan’s laws. However, I’m stuck with the reverse implication, I need to prove that:
“If $ U$ is a subset of a Boolean algebra $ B$ such that $ \forall x \in B$ , exactly one of $ x \in U$ and $ x^* \in U$ is true.
Is this even the case? Any help would be greatly appreciated, thanks in advance.