An Introduction to Mathematics · Alfred North Whitehead
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
(α) the result is another ordered couple,
(β) the operation is commutative, so that (x, y) × (x', y') = (x', y') × (x, y),
(γ) the operation is associative, so that {(x, y) × (x', y')} × (u, v) = (x, y) × {(x', y') × (u, v)},
(δ) 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) × (a, b) = (c, d), there is one and only one answer, which we can represent by (x, y) = (c, d) ÷ (a, b),or by (x, y) = (c, d)/(a, b).
(ε) Furthermore the law involving both addition and multiplication, called the distributive law, must be satisfied, namely (x,y) × {(a, b) + (c, d)} ; = {(x, y) × (a, b)} + {(x, y) × (c, d)}.
All these conditions (α), (β), (γ), (δ), (ε) 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) × (x', y') = {(xx' - yy'), (xy' + x'y)}. (A)
This is the definition of the meaning of the symbol × 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 (α) and (β) are evidently satisfied. The proof of the satisfaction of (γ), (δ), (ε) 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.
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)) × sqrt((-1)) = -1.
Now let us consider the three special
ordered couples (For 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) × (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 × 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 × 1 = x , for all values of x. Now by our law of multiplication (x, y) × (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)) × sqrt((-1)) = -1. Now by the law of multiplication for ordered couples (0, 1) × (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 ±sqrt((-1)). Consider (0, -1) ; here again remembering that (-1)^2 = 1 , we find, (0, -1) × (0, -1) = (-1, 0).
Thus (0, -1) is the other square root of sqrt((-1))-1. Accordingly the ordered couples (0, 1) and (0, -1) are the interpretations of ±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) × (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) × (x, y) = (-by, bx).
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) × (0, b) =(0, ab).
Fourthly, we multiply the two "real" couples (a, 0) and (a', 0) and obtain (a, 0) × (a', 0) =(aa', 0).
Fifthly, we multiply the two "imaginary couples" (0, b) and (0, bb') and obtain (0, b) × (0, b') = (-bb', 0).
We now turn to the geometrical interpretation, beginning first with some special cases.
Take the couples (1, 3) and (2, 0) and consider the equation (2, 0) × (1, 3) = (2, 6).
In the diagram (fig. 11) the vector OP represents (1, 3) , and the vector ON represents (2, 0) , and the vector OQ represents (2, 6). Thus the product (2, 0) × (1, 3) is found geometrically by taking the length of the vector OQ to be the product of the lengths of the vectors OP and ON , and (in this case) by producing OP to Q to be of the required length. Again, consider the product (0, 2) × (1, 3) , we have (0, 2) × (1, 3) = (-6, 2).
The vector ON_1 , corresponds to (0, 2) and the vector OR to (-6,2). Thus OR which
represents the new product is at right angles to OQ and of the same length. Notice that we have the same law regulating the length of OQ as in the previous case, namely, that its length is the product of the lengths of the two vectors which are multiplied together; but now that we have ON_1 along the "ordinate" axis OY , instead of ON along the "abscissa" axis OX , the direction of OP has been turned through a right-angle.
Hitherto in these examples of multiplication we have looked on the vector OP as modified by the vectors ON and ON_1. We shall get a clue to the general law for the direction by inverting the way of thought, and by thinking of the vectors ON and ON_1 as modified by the vector OP. The law for the length remains unaffected; the resultant length is the length of the product of the two vectors. The new direction for the enlarged ON (i.e. OQ ) is found by rotating it in the (anti-clockwise) direction of rotation from OX towards OY through an angle equal to the angle XOP : it is an accident of this particular case that this rotation makes OQ lie along the line OP. Again consider the product of ON_1 and OP ; the new direction for the enlarged ON_1 (i.e. OR ) is found by rotating ON in the anti-clockwise direction of rotation through an angle equal to the angle XOP , namely, the angle N_1OR is equal to the angle XOP.
The general rule for the geometrical representation of multiplication can now be enunciated thus:
The product of the two vectors OP and OQ is a vector OR , whose length is the product of the lengths of OP and OQ and whose direction OR is such that the angle XOR is equal to the sum of the angles XOP and XOQ.
Hence we can conceive the vector OP as making the vector OQ rotate through an angle XOP (i.e. the angle QOR = the angle XOP ), or the vector OQ as making the vector OP rotate through the angle XOQ (i.e. the angle POR = the angle XOQ ).
We do not prove this general law, as we
should thereby be led into more technical processes of mathematics than falls within the design of this book. But now we can immediately see that the associative law [numbered (γ) above] for multiplication is satisfied. Consider first the length of the resultant vector; this is got by the ordinary process of multiplication for real numbers; and thus the associative law holds for it.
Again, the direction of the resultant vector is got by the mere addition of angles, and the associative law holds for this process also.
So much for multiplication. We have now rapidly indicated, by considering addition and multiplication, how an algebra or "calculus" of vectors in one plane can be constructed, which is such that any two vectors in the plane can be added, or subtracted, and can be multiplied, or divided one by the other.
We have not considered the technical details of all these processes because it would lead us too far into mathematical details; but we have shown the general mode of procedure. When we are interpreting our algebraic symbols in this way, we are said to be employing "imaginary quantities" or "complex
quantities." These terms are mere details, and we have far too much to think about to stop to enquire whether they are or are not very happily chosen.
The nett result of our investigations is that
any equations like x + 3 = 2 or (x + 3)^2 = -2 can now always be interpreted into terms of vectors, and solutions found for them. In seeking for such interpretations it is well to note that 3 becomes (3, 0) , and -2 becomes (-2, 0) , and x becomes the "unknown" couple (u, v) : so the two equations become respectively (u, v) + (3, 0) = (2, 0) , and {(u, v) + (3, 0)}^2 = (-2, 0).
We have now completely solved the initial difficulties which caught our eye as soon as we considered even the elements of algebra. The science as it emerges from the solution is much more complex in ideas than that with which we started. We have, in fact, created a new and entirely different science, which will serve all the purposes for which the old science was invented and many more in addition. But, before we can congratulate ourselves on this result to our labours, we must allay a suspicion which ought by this time to have arisen in the mind of the student. The question which the reader ought to be asking himself is: Where is all this invention of new interpretations going to end? It is true that we have succeeded in interpreting algebra so as always to be able to solve a quadratic equation like x^2 - 2x + 4 = 0 ; but there are an endless number of other equations, for example, x^3 - 2x + 4 = 0 , x^4 + x^3 + 2 = 0 , and so on without limit. Have we got to make a
new science whenever a new equation appears?
Now, if this were the case, the whole of our preceding investigations, though to some minds they might be amusing, would in truth be of very trifling importance. But the great fact, which has made modern analysis possible, is that, by the aid of this calculus of vectors, every formula which arises can receive its proper interpretation; and the "unknown" quantity in every equation can be shown to indicate some vector. Thus the science is now complete in itself as far as its fundamental ideas are concerned. It was receiving its final form about the same time as when the steam engine was being perfected, and will remain a great and powerful weapon for the achievement of the victory of thought over things when curious specimens of that machine repose in museums in company with the helmets and breastplates of a slightly earlier epoch.
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End of part 8 of 19
Continue readingChapter IX: Coordinate Geometry
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