### A Knight's Journey

This article looks at knight's moves on a chess board and introduces you to the idea of vectors and vector addition.

### An Introduction to Complex Numbers

A short introduction to complex numbers written primarily for students aged 14 to 19.

### An Introduction to Vectors

The article provides a summary of the elementary ideas about vectors usually met in school mathematics, describes what vectors are and how to add, subtract and multiply them by scalars and indicates why they are useful.

# Vector Walk

##### Stage: 5 Challenge Level:

$\mathbf{P} = n\begin{pmatrix}2\\1\end{pmatrix} + m\begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix}2n\\n+m\end{pmatrix}$

The end destination tells you how many of each vector were used, but does not include cancelled steps: from the $x$ coordinate we can deduce how many net $b_1$ steps there were; this then allows us to determine how many net $b_2$ steps there must have been.

The vectors can together reach any point with integer $y$ coordinate and even $x$ coordinate: Thus, the set of possible destinations are the points, and only the points, with coordiantes $\begin{pmatrix}2p\\q\end{pmatrix}$ for any integers $p, q$.

Two other base vectors which give rise to exactly the same set of points are
$$\begin{pmatrix}2\\0\end{pmatrix} \,,\begin{pmatrix}0\\1\end{pmatrix}$$

Imagine now that we have two other more general base vectors
$$\mathbf{a} = \begin{pmatrix} x_a \\ y_a\end{pmatrix} \quad \mathbf{b} = \begin{pmatrix} x_b \\y_b\end{pmatrix}$$
If these are to reach exactly the same set of points we must have, for any integers $N$ and $M$ that
$$N\begin{pmatrix} x_a \\ y_a\end{pmatrix} +M\begin{pmatrix} x_b\\ y_b\end{pmatrix}\mathbf{b} = \begin{pmatrix} 2p \\ q\end{pmatrix}$$
Clearly to reach these, and only these points, I would need both $a$ and $b$ to have even $x$ component. Furthermore, if I can reach the points $(2, 0)$ and $(0, 1)$ with combinations of my base vectors then I would be able to reach all of the points $(2n, m)$ by adding together $n$ and $m$ lots of these combinations.

Algebra quickly becomes very complicated, as there are too many variable to give me anything to 'solve'.

So, look at some more particular cases. Look at a base vector with an $x$-component of $4$.
$$\begin{pmatrix} 4 \\ 0\end{pmatrix}$$
Clearly, to be able to reach the point $(2, 0)$ the other base vector would need to be $\begin{pmatrix} \pm 2 \\ 0\end{pmatrix}$, but this would not allow us to reach the point $(0, 1)$. Thus, any other possible pairs of base vectors would have to have non-zero $x$ and $y$ components.
A suitable pair would be
$$\mathbf{a}= \begin{pmatrix} 4 \\ 1\end{pmatrix}\,, \mathbf{b} =\begin{pmatrix}2 \\ 1\end{pmatrix}$$
To see why, note that the combinations $\mathbf{a}-\mathbf{b}$ and $2\mathbf{b}-\mathbf{a}$ take us to the points $(2, 0)$ and $(0, 1)$.
Another suitable pair would be
$$\mathbf{a}= \begin{pmatrix}6 \\ 2\end{pmatrix}\,, \mathbf{b} =\begin{pmatrix}4 \\ 1\end{pmatrix}$$ To see why, note that the combinations $2\mathbf{b}-\mathbf{a}$ and $2\mathbf{a}-3\mathbf{a}$ take us to the points $(2, 0)$ and $(0, 1)$.

For base vectors which yield none of the points in the original lattice, we would need irrational components, because if I had a base vector with rational components $\frac{n}{m}$ and $\frac{p}{q}$ then taking $2mq$ lots of this vector would take us to the point $(2qn, 2pm)$, which is in our original lattice of points.