Proof
(related to Theorem: Existence and Uniqueness of a Straight Line Through Two Points)
- By hypothesis, $A, B$ are points of an $n$-dimensional number space $\mathbb R^n$ with $A\neq B.$
- "Existence":
- Let $O$ be the origin of $\mathbb R^n.$
- Then we can describe a straight line using the vectors as the set $$L:=\{P\in \mathbb R^n:\; \overrightarrow{OP}:=\overrightarrow{OA}+t\cdot \overrightarrow{AB}\}.$$
- Obviously, for $t=0,$ $A\in L,$ and for $t=1,$ $B\in L.$
- Thus, there is at least one straight line through $A$ and $B.$
- "Uniqueness":
- Let $C, B$ be two points of an $n$-dimensional number space $\mathbb R^n$ with $C\neq D$.
- Assume, $L':=\{Q\in \mathbb R^n:\; \overrightarrow{OQ}:=\overrightarrow{OC}+t\cdot \overrightarrow{CD}\}$ is a straight line described by these points containing also the points $A$ and $B.$
- Thus, since $A\in L',$ there exists a $t_A\in\mathbb R$ with $\overrightarrow{OA}=\overrightarrow{OC}+t_A\cdot\overrightarrow{CD}.$
- Similarly, since $B\in L',$ there exists a $t_B\in\mathbb R$ with $\overrightarrow{OB}=\overrightarrow{OC}+t_B\cdot\overrightarrow{CD}.$
- The difference of both last equations gives us $\overrightarrow{AB}=(t_B-t_A)\cdot\overrightarrow{CB}.$
- We have $(t_B-t_A)\neq 0,$ because $A\neq B$ by hypothesis.
- Altogether, we have found that $A\in L'$ and that $\overrightarrow{AB}=c\cdot\overrightarrow{CB}$ for some constant $0\neq c\in \mathbb R.$
- This means by the lemma about equivalence of different descriptions of a straight line using two vectors that $L'=L.$
- Thus, there is at most one straight line through $A$ and $B.$
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