ordered couplesFor the future we follow the custom of omitting the $+$ sign wherever possible, thus $(1, 0)$ stands for $(+ 1, 0)$ and $(0, 1)$ for $(0, + 1)$ . $(0, 0)$ , $(1, 0)$ , and $(0, 1)$ .

We have already proved that $(x, y) + (0, 0) = (x, y) .$

Furthermore we now have $(x, y) \times (0, 0) = (0, 0) .$

Hence both for addition and for multiplication the couple $(0, 0)$ plays the part of zero in elementary arithmetic and algebra; compare the above equations with $x + 0 = x$ , and $x \times 0 = 0$ .

Again consider $(1, 0)$ : this plays the part of $1$ in elementary arithmetic and algebra. In these elementary sciences the special characteristic of $1$ is that $x \times 1 = x$ , for all values of $x$ . Now by our law of multiplication $(x, y) \times (1, 0) = {(x - 0), (y + 0)} = (x, y) .$

Thus $(1, 0)$ is the unit couple.

Finally consider $(0, 1)$ : this will interpret for us the symbol $\sqrt{(- 1)}$ . The symbol must therefore possess the characteristic property that $\sqrt{(- 1)} \times \sqrt{(- 1)} = - 1$ . Now by the law of multiplication for ordered couples $(0, 1) \times (0, 1) = {(0 - 1), (0 + 0)} = (- 1, 0) .$

But $(1, 0)$ is the unit couple, and $(- 1, 0)$ is the negative unit couple; so that $(0, 1)$ has the desired property. There are, however, two roots of $- 1$ to be provided for, namely $\pm \sqrt{(- 1)}$ . Consider $(0, - 1)$ ; here again remembering that $(- 1)^{2} = 1$ , we find, $(0, - 1) \times (0, - 1) = (- 1, 0)$ .

Thus $(0, - 1)$ is the other square root of $- 1$ . Accordingly the ordered couples $(0, 1)$ and $(0, - 1)$ are the interpretations of $\pm \sqrt{(- 1)}$ in terms of ordered couples. But which corresponds to which? Does $(0, 1)$ correspond to $+ \sqrt{(- 1)}$ and $(0, - 1)$ to $- \sqrt{(- 1)}$ , or $(0, 1)$ to $- \sqrt{(- 1)}$ , and $(0, - 1)$ to $+ \sqrt{(- 1)}$ ? The answer is that it is perfectly indifferent which symbolism we adopt.

The ordered couples can be divided into three types, (i) the "complex imaginary" type $(x, y)$ , in which neither $x$ nor $y$ is zero; (ii) the "real" type $(x, 0)$ ; (iii) the "pure imaginary" type $(0, y)$ . Let us consider the relations of these types to each other. First multiply together the "complex imaginary"

couple $(x, y)$ and the "real" couple $(a, 0)$ , we find $(a, 0) \times (x, y) = (a x, a y) .$

Thus the effect is merely to multiply each term of the couple $(x, y)$ by the positive or negative real number $a$ .

Secondly, multiply together the "complex imaginary" couple $(x, y)$ and the "pure imaginary" couple $(0, b)$ , we find $(0, b) \times (x, y) = (- b y, b x) .$

Here the effect is more complicated, and is best comprehended in the geometrical interpretation to which we proceed after noting three yet more special cases.

Thirdly, we multiply the "real" couple $(a, 0)$ by the imaginary $(0, b)$ and obtain $(a, 0) \times (0, b) = (0, a b) .$

Fourthly, we multiply the two "real" couples $(a, 0)$ and $(a^{'}, 0)$ and obtain $(a, 0) \times (a^{'}, 0) = (a a^{'}, 0) .$

Fifthly, we multiply the two "imaginary couples" $(0, b)$ and $(0, b^{'})$ and obtain $(0, b) \times (0, b^{'}) = (- b b^{'}, 0) .$

We now turn to the geometrical interpretation, beginning first with some special cases.