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): null vector