# Properties of Affine Spaces

## Theorem

Let $\EE$ be an affine space with difference space $V$.

Let $0$ denote the zero element of $V$.

Then the following hold for all $p,q,r \in \EE$ and all $u, v \in V$:

$(1): \quad p - p = 0$
$(2): \quad p + 0 = p$
$(3): \quad p + u = p + v \iff u = v$
$(4): \quad q - p = r - p \iff q = r$

## Proof

### $(1): \quad p - p = 0$

We have:

 $\ds \paren {p - p} + \paren {q - p}$ $=$ $\ds \paren {p + \paren {q - p} } - p$ $\ds$ $=$ $\ds q - p$
$p - p = 0$

$\Box$

### $(2): \quad p + 0 = p$

Using $(1)$ we see that:

 $\ds p + 0$ $=$ $\ds p + \paren {p - p}$ $\ds$ $=$ $\ds p$

$\Box$

### $(3): \quad p + u = p + v \iff u = v$

Let $u = v$.

By definition a mapping has a unique image point on a given element.

It follows that:

$p + u = p + v$

Let $p + u = p + v$.

We have:

 $\ds p + u$ $=$ $\ds p + v$ $\ds \leadsto \ \$ $\ds \paren {p + u} - p$ $=$ $\ds \paren {p + v} - p$ $\ds \leadsto \ \$ $\ds \paren {p - p} + u$ $=$ $\ds \paren {p - p} + v$ $\ds \leadsto \ \$ $\ds u$ $=$ $\ds v$ by $(1)$

$\Box$

### $(4): \quad q - p = r - p \iff q = r$

Let $q = r$.

By definition a mapping has a unique image point on a given element.

It follows that:

$q - p = r - p$

Let $q - p = r - p \in V$.

Then:

 $\ds q - p$ $=$ $\ds r - p$ $\ds \leadsto \ \$ $\ds p + \paren {q - p}$ $=$ $\ds p + \paren {r - p}$ $\ds \leadsto \ \$ $\ds q$ $=$ $\ds r$ by hypothesis $q - p = r - p$

$\blacksquare$