Chapter 4 - Vector Spaces - 4.2 Definition of Vector Spaces - True-False Review - Page 262: f

True

Let $v$ be a vector in vector space $V(F)$ Suppose that $u\:,\:w$ are additive inverse of $v$ and $u\neq w$ Because $u$ is additive inverse of $v$ By definition $u+v=0\;\;\;$_____$(1)$ Here $0$ is additive identity of $V(F)$ Because $w$ is additive inverse of $v$ By definition $w+v=0\;\;\;$_____$(2)$ Here $0$ is additive identity of $V(F)$ By (1) and (2) $u+v=w+v$ By right cancellation law $u=w$ Which is contradiction to the fact that $u\neq w$ $\Rightarrow $ Our assumption was wrong $\Rightarrow $ Additive inverse of $v$ is unique Because $v$ was arbitrary Hence additive inverse of vectors in vector space is unique .