It follows from this, that we may regard, as but one single question, those, the enunciations of which are connected together in such a manner, that the solutions, which satisfy one of the enunciations, will, by a mere change of sign, satisfy the other also.

62. Since negative quantities resolve in a certain sense the problems, which give rise to them, it is proper to inquire a little more particularly into the use of these quantities, and to settle once for all the manner of performing operations in which they are concerned.

We have already made use of the rule for the signs, which had been previously determined for each of the fundamental operations; but the rules have not been demonstrated with reference to insulated quantities. In the case of subtraction, for example, we supposed that there was to be taken from a, the expression b— c, in which the negative quantity c was preceded by a positive quantity b. Strictly speaking, the reasoning does not depend upon the value of b; it would still apply when b = 0, which reduces the expression b c to— c. But the theory of negative quantities being at the same time one of the most important and most difficult in algebra, it should be established upon a sure basis. To effect this, it is necessary to go back to the origin of negative quantities.

The greatest subtraction, that can be made from a quantity, is to take away the quantity itself, and in this case we have zero for a remainder; thus aa0. But when the quantity to be subtracted exceeds that from which it is to be taken, we cannot subtract it entirely; we can only make a reduction of the quantity to be subtracted, equal to the quantity from which it was to be taken. When, for example, it is required to subtract 5 from 3, or when we have the quantity 3-5; to take, in the first place, 3 from 5, we decompose 5 into two parts 3 and 2, the successive subtraction of which will amount to that of 5, and thus, instead of 35, we have the equivalent expression 3—3—2, which is reduced to 2. The sign, which precedes 2, shows what is necessary to complete the subtraction; so that, if we had added 2 to the first of the quantities, we should have had 3+2-5, or zero. We express then, with the help of algebraic signs, the idea that is to be attached to a negative quantity a, by forming the equation aa 0, or by regarding the symbols a—a bb, &c., as equivalent to zero,

This being supposed, it will be understood, that if we add to any quantity whatever the symbol bb, which in reality is only zero, we do not change the value of this quantity, and that, consequently, the expression a + b — b, 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 + b and 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 α b, 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 a ba b, because the multiplicand being equal to zero, the product must be zero; and the first term being a b, the second must necessarily be ab to destroy the first.

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

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 -a b, to destroy the first + ab. Whence+a, multiplied by b, must give give — Lastly, if we multiply a by b b, the first term of the product being, according to what has just been proved, — a b, 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 a b.

a b.

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 insulated, 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 the distance AR passed over by the first, the distance BR passed over by the second, and as

Y

I have the equation,

AR + BR = AB,

x + y = a.

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 , a time denoted by

x

b

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

C

(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,

x y

=

b с

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

As the sign does not enter into the values of x and y, it is evident that whatever numbers are put for the letters abc, 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.

Α

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

expressing only the equality of the times employed by the couriers in passing over the distances AR and BR, 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 AB.

If we now suppose b smaller than c, and take, for example, b = 10, c = 20,

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