# Properties of Affine Spaces

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## Theorem

Let $\mathcal E$ be an affine space with difference space $V$.

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

Then the following hold for all $p,q,r \in \mathcal E$ 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:

\(\displaystyle \left({p - p}\right) + \left({q - p}\right)\) | \(=\) | \(\displaystyle \left({p + \left({q - p}\right)}\right) - p\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle q - p\) |

From Zero Element is Unique:

- $p - p = 0$

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

Using $(1)$ we see that:

\(\displaystyle p + 0\) | \(=\) | \(\displaystyle p + \left({p - p}\right)\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle p\) |

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

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:

\(\displaystyle p + u\) | \(=\) | \(\displaystyle p + v\) | |||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle \left({p + u}\right) - p\) | \(=\) | \(\displaystyle \left({p + v}\right) - p\) | ||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle \left({p - p}\right) + u\) | \(=\) | \(\displaystyle \left({p - p}\right) + v\) | ||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle u\) | \(=\) | \(\displaystyle v\) | by $(1)$ |

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

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

\(\displaystyle q - p\) | \(=\) | \(\displaystyle r - p\) | |||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle p + \left({q - p}\right)\) | \(=\) | \(\displaystyle p + \left({r - p}\right)\) | ||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle q\) | \(=\) | \(\displaystyle r\) | By the hypothesis $q - p = r - p$ |

$\blacksquare$