1st. Every negative value found for the unknown quantity in a problem of the first degree, will, when taken with its proper sign, verify the equation from which it was derived.

2d. That this negative value, taken with its proper sign, will also satisfy the enunciation of the problem, understood in its algebraic

sense.

3d. The negative result shows that the enunciation is impossible, regarded in its arithmetical sense. The language of Algebra detects the error of the arithmetical enunciation, and indicates the general relation of the quantities.

4th. The negative result, considered without reference to its sign, may be regarded as the answer to a problem of which the enunciation only differs from the one proposed in this: that certain quantities which were additive have become subtractive, and reciprocally.

106. As a further illustration of the change which an algebraic sign may produce in the enunciation of a problem, let us resume that of the laborer (page 76).

Under the supposition that the laborer receives a sum c, we have the equations

If at the end of the time, the laborer, instead of receiving a sum c, owed for his board a sum equal to c, then, by would be greater than ax, and under this supposition, we should have the equations

Now, it is plain that we can obtain immediately the values of x and y, in the last equations, by merely changing the sign of c in each of the values found from the equations above; this gives

X=

bn
a+b'

- C

The results for both enunciations, may be comprehended in the same formulas, by writing

The double sign ±, is read plus or minus, and nus or plus. The upper signs correspond to the

, is read, micase in which

the laborer received, and the lower signs, to the case in which he owed a sum c. These formulas also comprehend the case in which, in a settlement between the laborer and his employer, their accounts balance. This supposes c = 0, which gives

Discussion of Problems. Explanation of the terms
Nothing and Infinity.

107. When a problem has been resolved generally, that is, by means of letters and signs, it is often required to determine what the values of the unknown quantities become, when particular suppositions are made upon the quantities which are given. The determination of these values, and the interpretation of the peculiar results obtained, form what is called the discussion of the problem.

The discussion of the following question presents nearly all the circumstances which are met with in problems of the first degree.

108. Two couriers are travelling along the same right line and in the same direction from R' toward R. The number of miles travelled by one of them per hour is expressed by m, and the number of miles travelled by the other per hour, is expressed by n. Now, at a given time, say 12 o'clock, the distance between them is equal to a number of miles expressed by a: required the time when they will be together.

At 12 o'clock suppose the forward courier to be at B, the other at A, and R to be the point at which they will be together. Then, AB = 0. their distens apart at 12 o'clock.

Let

t=

and

x=

the number of hours which must elapse, before they come together;

the distance, which is to be passed over by the forward courier.

Then, since the rate per hour, multiplied by the number of

hours, will give the

Now, so long as m>n, t will be positive, and the problem will be solved in the arithmetical sense of the enunciation. For, if m>n, the courier from A will travel faster than the courier from B, and will therefore be continually gaining on him the interval which separates them will diminish more and more, until it becomes 0, and then the couriers will be found upon the same point of the line.

In this case, the time t, which elapses, must be added to 12 o'clock, to obtain the time when they are together.

But, if we suppose m<n, then, m -n will be negative, and the value of t will be negative. How is this result to be inter

preted?

It is easily explained from the nature of the question, which considered in its most general sense, demands the time when the couriers are together.

Now, under the second supposition, the courier which is in advance, travels the fastest, and therefore will continue to separate himself from the other courier. At 12 o'clock the distance between them was equal to a: after 12 o'clock it is greater than a; and as the rate of travel has not been changed, it follows that previous to 12 o'clock the distance must have been less than a. At a certain hour, therefore, before 12, the distance between them must have been equal to nothing, or the couriers were together at some point R'. The precise hour is found by subtracting the value of t from 12 o'clock.

This example, therefore, conforms to the general principle, that, if the conditions of a problem are such as to render the unknown quantity essentially negative, it will appear in the result with the minus sign, whenever it has been regarded as positive in the enunciation.

If we wish to find the distances AR and BR, passed over by the two couriers before coming together, we may take the equation

t =

a

m-n

and multiply both members by the rates of travel respectively: this will give

from which we see, that the two distances AR and BR, will both be positive when estimated toward the right, and that AR' and BR will both be negative when estimated in the contrary direction.

109. To explain the terms nothing and infinity, let us consider the equation

t=

a

m -N

If the couriers travel at different rates, m n will be a finite quantity, and its sign will depend on the relative values of m and n. Designate this quantity by A.

Now, if we suppose a = 0, we shall have

an equation which can only be satisfied by making t=0. To interpret this result, let us go back to the enunciation of the problem. If a = 0, the couriers are together at 12 o'clock; and since they travel at different rates, they can never be again together: hence, t can have no other value than 0. Therefore, we conclude that, the quotient of 0 divided by a finite quantity, is 0. 110. Let us resume the equation

t =

a

m -n

If in this equation we make m=n, then m

the value of t will reduce to

an equation which cannot be satisfied for any finite value of t.

In order to interpret this new result, let us go back to the enunciation of the question. We see at once, that it is absolutely impossible to satisfy the enunciation for any finite value for t; for, whatever time we allow to the two couriers, they can never come together, since being once separated by an interval a, and travelling equally fast, this interval will always be preserved.

a

Hence, the result, may be regarded as a sign of impossi

0

bility for any finite value of t.

Nevertheless, algebraists consider the result,

as forming a species of value, to which they have given the name of infinite value, for this reason:

When the difference mn, without being absolutely nothing, is supposed to be very small, the result

In short, if the difference between the rates is not zero, the couriers will come together at some point of the line, and the time will become greater and greater, as this difference is diminished.

Hence, from analogy, if the difference between the rates is less than any assignable number, the time expressed by

will be greater than any assignable or finite number. Therefore for brevity, we say, when m― n = 0, the result,