Vector Augend plus Addend equals Augend implies Addend is Zero

From ProofWiki
Jump to: navigation, search

Theorem

Let $V$ be a vector space over a field $F$.

Let $\mathbf a, \mathbf b \in V$.

Let $\mathbf a + \mathbf b = \mathbf a$.


Then:

$\mathbf b = \bszero$

where $\bszero$ is the zero vector of $V$.


Proof

\(\displaystyle \mathbf a + \mathbf b\) \(=\) \(\displaystyle \mathbf a\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {-\mathbf a} + \paren {\mathbf a + \mathbf b}\) \(=\) \(\displaystyle \paren {-\mathbf a} + \mathbf a\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {\paren {-\mathbf a} + \mathbf a} + \mathbf b\) \(=\) \(\displaystyle \paren {-\mathbf a} + \mathbf a\) $\quad$ Vector Space Axiom $V \, 2$: Associativity $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \bszero + \mathbf b\) \(=\) \(\displaystyle \bszero\) $\quad$ Vector Space Axiom $V \, 4$: Inverses $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \mathbf b\) \(=\) \(\displaystyle \bszero\) $\quad$ Vector Space Axiom $V \, 3$: Identity $\quad$

$\blacksquare$


Sources