This being supposed, it will be understood, that if we add to any quantity whatever the symbol b-b, which in reality is only zero, we do not change the value of this quantity, and that consequently the expression a+bb is nothing else but a different manner of writing the quantity a which is also evident from the consideration, that +b and b destroy each other. But having by this change of form introduced band-b into the same expression with a, we see that in order to subtract any one of these quantities, it is sufficient to efface it. If it were +b that we would subtract, we efface it, and there remains ab, which accords with the rule laid down in art. 2; if on the other hand it wereb, we efface this quantity and there would remain a + b, as might be inferred from art. 20.

With respect to multiplication it will be observed, that the product of aa by +b must be ab-ab, because the multiplicand being equal to zero, the product must be zero; and the first term being ab, the second must necessarily be ab to destroy the first.

We infer from this, thata, multiplied by +a, must give - ab.

By multiplying a by b-b, we have still ab-ab, because the multiplier being equal to zero, the product will also be equal to zero; it is therefore necessary that the second term should be - ab to destroy the first+ub.

Whence a multiplied by -b must give a b.

Lastly, if we multiplya by b-b, the first term of the product being, according to what has just been proved, ab, it is necessary that the second term should be +ab, as the product must be nothing when the multiplier is nothing.

Whencea, multiplied by -b gives+ab.

By collecting these results together we may deduce from them the same rules as those in art. 31 (A).

As the sign of the quotient, combined with that of the divisor according to the rules proper for multiplication, must produce the sign of the dividend, we infer from what has just been said, that the rule for the signs given in art. 42, corresponds with that which it is necessary to observe in fact, and that consequently, simple quantities, when they are insolated, are combined with respect to their signs, in the same manner, as when they make a part of polynomials.

63. According to these remarks we may always, when we meet with negative values, go back to the true enunciation of the question resolved, by seeking in what manner these values will satisfy the equations of the proposed problem; this will be confirmed by the following example, which relates to numbers of a different kind from those of the question in art. 56.

64. Two couriers set out to meet each other at the same time from two cities, the distance of which is given; we know how many miles (a) each travels per hour, and we inquire at what point of the route between the two cities they will meet.

To render the circumstances of the question more evident, I have subjoined a figure in which the points A and B represent the places of departure of the couriers.

A

R

B

I denote the things given, and those required in the usual way, by small letters.

a the distance in miles of the points of departure A and B,
b the number of miles per hour, which the courier from A
travels,

c the number of miles per hour which the courier from B
travels.

The letter R being placed at the point of meeting of the two couriers, I shall call a the distance AR passed over by the first, y the distance BR passed over by the second,

Considering that the distances x and y are passed over in the same time, we remark that the first courier, who travels a number b of miles in an hour, will employ, in passing over the distance x, a time denoted by T.

Also the second courier, who travels c miles in an hour, will y employ, in passing over the distance y, a time denoted by 2

; we

(a) In the original the distance is given in kilometres. It is here expressed by miles to avoid perplexing the learner.

The equations of the question therefore will be

x+y=a

XC y
=

b

Making the denominator b of the second to disappear, we have

putting this value in the place of x in the first equation, it be

Substituting this value of y in the expression for the value of

x, we obtain

x=

b+c

As the sign does not enter into the values of x and y, it is evident that whatever numbers are put for the letters a, b, c, we shall always find x and y with the sign+, and therefore the question proposed will be resolved in the precise sense of the enunciation. Indeed it is readily perceived, that in every case where two persons set off from different points and travel toward each other they must necessarily meet.

65. I will now suppose, that the two couriers proceed in the same direction, and that the one, who sets out from A, is pursuing the one who sets out from B, and who is travelling toward the same point C, placed beyond B, with respect to A.

A

B

R

C

It is evident that in this case, the courier, who starts from the point A, cannot come up with the courier who sets off from the point B, except he travels faster than this last, and the point of coming together R cannot be between A and B, but must be beyond B, with respect to A.

Having the same things given as before, and observing that when

AR-BR= A B,

expressing only the equality of the times employed by the couriers in passing over the distances AR and B R, undergoes no change.

The above equations, being resolved like the former ones, give

Here the values of x and y will not be positive, except when b is taken greater than c, that is to say, except the courier starting from the point A be supposed to travel faster than the other. If, for example, we make

from which it follows, that the point of their coming together is

distant from the point A twice A B.

If we now suppose b smaller than c, and take, for example

These values being affected with the sign, make it evident, that the question cannot be resolved in the sense in which it is onunciated; and indeed it is absurd to suppose that the courier

H

setting out from the point A, and proceeding only 10 miles in au hour, should ever be able to overtake the courier setting out from the point B and travelling 20 miles per hour, and who is in advance of the first.

66. Nevertheless, these same values resolve the question in a certain sense; for by substituting them in the equations

equations which are satisfied; since by making the reductions, the first member becomes equal to the second; and if we attend to the signs of the terms, which compose the first, we shall see how it is necessary to modify the enunciation of the question, in order to do away the absurdity.

Indeed, it is the distance a corresponding to x, and passed over by the first courier, which is in reality subtracted from the distance 2 a, corresponding to y and passed over by the second courier; it is then just as if we had changed y into x, and x into y, and had supposed that the courier starting from the point B had run after the other.

This change in the enunciation, produces also a change in the direction of the routes of the couriers; they are no longer travelling toward the point C, but in an opposite manner toward the point C', as represented in the figure below;

and their coming together takes place in R. The result from this is