VII

We came across equations of the form $x^2=-3$, to which no solutions could be

assigned in terms of positive and negative real numbers. We then found that all our difficulties would vanish if we could interpret the equation $x^2=-1$, i.e., if we could so define $\sqrt{(-1)}$ that $\sqrt{(-1)}\times \sqrt{(-1)}=-1$.

Now let us consider the three special

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 $(a,0)\times (x,y)=(ax,ay).$

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 $(0,b)\times (x,y)=(-by,bx).$