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Statement

Let $V$ be any vector space and let $\mathbf{x}, \mathbf{y}, \mathbf{z}$ be in $V$. Prove that if $\mathbf{x} + \mathbf{z} = \mathbf{y} + \mathbf{z}$, then $\mathbf{x} = \mathbf{y}$.

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Proof

Suppose for full generality that $V$ is any vector space and $\mathbf{x}, \mathbf{y}, \mathbf{z}$ are any vectors in $V$. Suppose $\mathbf{x} + \mathbf{z} = \mathbf{y} + \mathbf{z}$. Show that $\mathbf{x} = \mathbf{y}$. Since $\mathbf{z}$ is in $V$, there exists a vector $-\mathbf{z}$ in $V$ such that

$\mathbf{z} + (-\mathbf{z}) = \mathbf{0}$	because .
Thus, $\mathbf{x} + \mathbf{z} = \mathbf{y} + \mathbf{z}$	implies
$(\mathbf{x} + \mathbf{z}) + (-\mathbf{z}) = (\mathbf{y} + \mathbf{z}) + (-\mathbf{z})$	and thus
$\mathbf{x} + (\mathbf{z} + (-\mathbf{z})) = \mathbf{y} + (\mathbf{z} + (-\mathbf{z}))$	because
$\mathbf{x} + \mathbf{0} = \mathbf{y} + \mathbf{0}$	because
and thus $\mathbf{x} = \mathbf{y}$	by properties of the zero vector.

Therefore, $\mathbf{x} + \mathbf{z} = \mathbf{y} + \mathbf{z}$ implies $\mathbf{x} = \mathbf{y}$, and $V$ has the cancellation property. $\Box$

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Allowed answers

V is a vector space, V is closed under addition, V is closed under scalar multiplication, addition is commutative in V, addition is associative in V, the zero vector in V is an additive identity, additive inverses exist in V, 1 is a multiplicative identity, scalar multiplication is associative, scalar multiplication distributes over vector addition, scalar multiplication distributes over scalar addition