### Roots and Coefficients

If xyz = 1 and x+y+z =1/x + 1/y + 1/z show that at least one of these numbers must be 1. Now for the complexity! When are the other numbers real and when are they complex?

### Target Six

Show that x = 1 is a solution of the equation x^(3/2) - 8x^(-3/2) = 7 and find all other solutions.

### 8 Methods for Three by One

This problem in geometry has been solved in no less than EIGHT ways by a pair of students. How would you solve it? How many of their solutions can you follow? How are they the same or different? Which do you like best?

# Two and Four Dimensional Numbers

##### Age 16 to 18Challenge Level

Do you think of complex numbers as very abstract and strange, and the idea of 4-dimensional numbers as even stranger? If you can add and multiply two by two matrices then you will see here how two by two matrices provide a model for complex numbers with a two by two matrix corresponding to i (the square root of -1).

Two by two matrices also provide a model for the 4-dimensional numbers called quaternions and for the many different square roots of -1 that occur in this number system. Read on...

Here is another excellent solution from Andrei (Bucharest, Romania).
(1) (a) First we shall add and multiply the two matrices, obtaining: $$\pmatrix {x & -y \cr y & x} + \pmatrix {u & -v \cr v & u} = \pmatrix {x+u & -y-v)\cr y+v & x+u}$$ and $$\pmatrix {x & -y \cr y & x} \pmatrix {u & -v \cr v & u} = \pmatrix {xu-yv & -xv-yu\cr xv+yu & xu-yv}.$$ [Note the similarities here to the addition and multiplication of complex numbers.]

(b) By simple calculations we observe that $\pmatrix {0 & 0 \cr 0 & 0}$ is the identity for addition and $\pmatrix {1 & 0 \cr 0 & 1}$ is the identity for multiplication. The inverses for addition and multiplication are obtained from the conditions that

i) addition of the given matrix with its inverse gives the identity for addition, so the inverse of $\pmatrix {x & -y \cr y & x}$ for addition is: $\pmatrix {-x & y \cr -y & -x}$ and

ii) multiplying the given matrix with its inverse gives the identity for multiplication; the inverse is ${1\over (x^2+y^2)}\pmatrix {x & y \cr -y & x}.$ We see that both identity matrices and both inverses are from the set C*.

(c) Here, we shall consider R* as the set of matrices of the form $\pmatrix {x & 0 \cr 0 & x}$. These matrices could be written as $x\pmatrix {1 & 0 \cr 0 & 1} = xI_2$. Evidently $\pmatrix {x & 0 \cr 0 & x} \pm \pmatrix {y & 0 \cr 0 & y} = \pmatrix {x\pm y & 0 \cr 0 & x\pm y} = (x\pm y)I_2$.

The identity for addition is $\pmatrix {0 & 0 \cr 0 & 0}$ and the inverse of $\pmatrix {x & 0 \cr 0 & x}$ is $\pmatrix {-x & 0 \cr 0 & -x}$, which are both from R*.
For multiplication $\pmatrix {x & 0 \cr 0 & x}\pmatrix {y & 0 \cr 0 & y} =x\pmatrix {1 & 0 \cr 0 & 1}\cdot y\pmatrix {1 & 0 \cr 0 & 1} = \pmatrix {xy & 0 \cr 0 & xy}$.

The multiplicative identity is $\pmatrix {1 & 0 \cr 0 & 1}$ and the multiplicative inverse of $\pmatrix {x & 0 \cr 0 & x}$ is $\pmatrix {{1\over x} & 0 \cr 0 & {1\over x}}$.

The distributive law of addition and multiplication is the same as that of real numbers: $$\pmatrix {x & 0 \cr 0 & x}\cdot \left(\pmatrix {y & 0 \cr 0 & y}+ \pmatrix {z & 0 \cr 0 & z}\right)= \pmatrix {x & 0 \cr 0 & x}\cdot \pmatrix {y & 0 \cr 0 & y} + \pmatrix {x & 0 \cr 0 & x}\cdot \pmatrix {z & 0 \cr 0 & z} = \pmatrix {x(y+z) & 0 \cr 0 & x(y+z)}.$$ This proves that the arithmetic of R* is the same as the arithmetic of real numbers.

(d) Considering 2 complex numbers, $x + iy$ and $u + iv$:
$(x + iy) + (u + iv) = (x + u) + i(y+v)$
$(x + iy) (u + iv) = (xu - yv) + i(xv + yu).$

By simple computation we can show that multiplication in the set of two by two matrices C* is commutative and that the distributive law holds in C*. This is so because the laws apply to every operation on the components. We observe that addition and multiplication in C* are the same as addition and multiplication of complex numbers.

(e) Note that $\pmatrix {0 & -1 \cr 1 & 0}^2 = \pmatrix {-1 & 0 \cr 0 & -1}$ so we see that C* contains a model for $\sqrt -1$, the complex number $i$.

The matrix $\pmatrix {x & -y \cr y & x}$ corresponds to the complex number $x + iy$. Multiplying this number by $i$, we obtain $-y + ix$, i.e. the matrix $\pmatrix {-y & -x \cr x & -y}$. We shall consider that matrix $\pmatrix {a & b \cr c & d}$ corresponds to $i$: So, $\pmatrix {x & -y \cr y & x}\pmatrix {a & b \cr c & d} =\pmatrix {-y & -x \cr x & -y}$. The only solution for this equation is $\pmatrix {0 & -1 \cr 1 & 0}$.

A different approach is the following: we associate the point $(x,y)$ in the complex plane with the complex number $(x + iy)$ and also with the matrix $\pmatrix {x & -y \cr y & x}$. The geometrical significance of the multiplication by $i$ of a complex number is the counterclockwise rotation of the point by $\pi /2$ so $(x,y) \to (-y,x)$. In matrix notation this corresponds to $$\pmatrix {0 & -1 \cr 1 & 0}\pmatrix {x & -y \cr y & x} =\pmatrix {-y & -x \cr x & -y}.$$ (2)(a)Working out the squares of the matrices $B= \pmatrix {i & 0\cr 0 & -i}$ , $C= \pmatrix {0 & 1\cr -1 & 0}$ and $D=\pmatrix {0 & i\cr i & 0}$ we get $B^2=C^2=D^2=\pmatrix{-1 & 0\cr 0 & -1}$. So all these three matrices are square roots of -1.

(b) Similarly $BC=D=-CB$, (c)$CD=B=-DC$ (d) $DB=C=-BD$ so these matrices are models of the quaternions $i, j$ and $k$. Then $$a\pmatrix {1 & 0\cr 0 & 1}+b\pmatrix {i & 0\cr 0 & -i} +c\pmatrix {0 & 1\cr -1 & 0}+d\pmatrix {0 & i\cr i & 0}$$ provides a model for the quaternion number system. Simple calculations show that all the field axioms hold except t

(3) Now, we shall calculate $i, i j, i j k, i j k i,...$ and we shall try to find a pattern. Using the relationships already established: $ij = k$, $ijk = k^2 = -1$, $ijki = -i$, $ijkij = -ij = -k$, $ijkijk = -kj = i$.

This means that the sequence is periodic with period 6, namely $i, k, -1, -i, -k, 1, i, k ...$.