# Zero Vector has no Direction

## Theorem

A zero vector has no direction.

## Proof

Let $\mathbf 0$ denote a zero vector.

Aiming for a contradiction, suppose $\mathbf 0$ has a direction.

Then $\mathbf 0$ can be represented as an arrow in a real vector space $\R^n$ with a Cartesian frame.

Let $\mathbf 0$ be so embedded.

Thus it consists of a line segment between two points with an initial point $A$ and a terminal point $B$.

The initial point and a terminal point are distinct from each other.

Let these points be identified as:

\(\ds A\) | \(=\) | \(\ds \tuple {a_1, a_2, \ldots, a_n}\) | ||||||||||||

\(\ds B\) | \(=\) | \(\ds \tuple {b_1, b_2, \ldots, b_n}\) |

Hence we have that the length of $\mathbf 0$ is defined as:

- $\norm {\mathbf 0} = \ds \sqrt {\sum_{i \mathop = 1}^n \paren {a_i - b_i}^2} > 0$

which means that at least one of $a_i - b_i$ is non-zero.

But this contradicts the definition of $\mathbf 0$ being the zero vector.

It follows by Proof by Contradiction that our assumption that $\mathbf 0$ has a direction must be false.

Hence the result.

$\blacksquare$

## Sources

- 1957: D.E. Rutherford:
*Vector Methods*(9th ed.) ... (previous) ... (next): Chapter $\text I$: Vector Algebra: $\S 1$. - 2014: Christopher Clapham and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(5th ed.) ... (previous) ... (next): Entry:**null vector**