was mentioned in VI., on the laws of motion, as applying to velocities and forces. It will be remembered that, if $O P$ and $O Q$ represent two velocities, a particle is said to be moving with a velocity equal to the two velocities added together if it be moving with the velocity $O R$ . In other words $O R$ is said to be the resultant of the two velocities $O P$ and $O Q$ . Again forces acting at a point of a body can be represented by lines just as velocities can be; and the same parallelogram law holds, namely, that the resultant of the two forces $O P$ and $O Q$ is the force represented by the diagonal $O R$ . It follows that we can look on an ordered couple as representing a velocity or a force, and the rule which we have just given for the addition of ordered couples then represents the fundamental laws of mechanics for the addition of forces and

velocities. One of the most fascinating characteristics of mathematics is the surprising way in which the ideas and results of different parts of the subject dovetail into each other. During the discussions of this and the previous chapter we have been guided merely by the most abstract of pure mathematical considerations; and yet at the end of them we have been led back to the most fundamental of all the laws of nature, laws which have to be in the mind of every engineer as he designs an engine, and of every naval architect as he calculates the stability of a ship. It is no paradox to say that in our most theoretical moods we may be nearest to our most practical applications.

[Imaginary Numbers] VIIIImaginary Numbers (Continued)

The definition of the multiplication of ordered couples is guided by exactly the same considerations as is that of their addition. The interpretation of multiplication must be such that

() the result is another ordered couple,

() the operation is commutative, so that $(x, y) \times (x^{'}, y^{'}) = (x^{'}, y^{'}) \times (x, y),$

() the operation is associative, so that ${(x, y) \times (x^{'}, y^{'})} \times (u, v) = (x, y) \times {(x^{'}, y^{'}) \times (u, v)},$

() must make the result of division unique [with an exception for the case of the zero couple $(0, 0)$ ], so that when we seek to determine the unknown couple $(x, y)$ so as to satisfy the equation $(x, y) \times (a, b) = (c, d),$ there is one and only one answer, which we can represent by $(x, y) = (c, d) \div (a, b), or by (x, y) = \frac{(c, d)}{(a, b)} .$

() Furthermore the law involving both addition and multiplication, called the distributive law, must be satisfied, namely

All these conditions (), (), (), (), () can be satisfied by an interpretation which, though it looks complicated at first, is capable of a simple geometrical interpretation.

By definition we put $(x, y) \times (x^{'}, y^{'}) = {(x x^{'} - y y^{'}), (x y^{'} + x^{'} y)} . \quad\ensuremath{(A)}$

This is the definition of the meaning of the symbol $\times$ when it is written between two ordered couples. It follows evidently from this definition that the result of multiplication is another ordered couple, and that the value of the right-hand side of equation (A) is not altered by simultaneously interchanging $x$ with $x^{'}$ , and $y$ with $y^{'}$ . Hence conditions () and () are evidently satisfied. The proof of the satisfaction of (), (), () is equally easy when we have given the geometrical interpretation, which we will proceed to do in a moment. But before doing this it will be interesting to pause and see whether we have attained the object for which all this elaboration was initiated.