The extent of the error which arises from not distinguishing between the sexes may be seen in Table III. constructed by Professor De Morgan from the statistical returns of the whole of Belgium for three successive years, as given by M. Quetelet and Smits, in the Recherches sur la Reproduction, &c. Brussels, 1832. This Table is calculated to shew the "expectation of life," that is, the average number of years remaining to any individual, at intervals of five years, from the age of 0 to 100. It distinguishes not only between male and female, but between town life and rural life; and the middle column gives the general average for the whole kingdom, male and female, town and country.

DISCUSSION AND INTERPRETATION OF
ANOMALOUS RESULTS.

458. Negative Results. It often happens, that the result of our operations for the solution of a proposed question or problem appears in a negative form, although, strictly speaking, there can be no such thing existent as an essentially negative quantity. But it will always be found, when such a result occurs, that there is something in the nature of the question which will either dispose of, or supply a meaning to, the negative result. Thus, to take a simple example; suppose a man wishes to ascertain the amount of his property-he puts down what he has, together with what is due to him, as positive, and all his debts with a negative sign. If then he finds that by taking the sum of both positive and negative quantities, the result is negative, its meaning will be sufficiently obvious, viz. that his property is so much less than nothing, that is, he is so much in debt. See Scholium, p. 44; and Art. 214.

Also, see Art. 282. Ex. 4. In this and like cases it is true that two solutions may be found for the equation, that is, two values of n; but when either of those values is fractional or negative, it is clearly inadmissible as a solution of the question proposed.

It may be observed also here generally, that when in solving a problem, expressed algebraically, we find it necessary, as in the above Example, to extract the square root of a quantity, the double sign ±, (that is,+ or -), need not to be prefixed to the root, at least for the object before us, if we have sufficient data beforehand for determining which sign the problem requires. Is it to be wondered at, that we produce an anomalous and unintelligible result, if we wilfully make a quantity negative which we know to be positive, or vice versâ?

Oftentimes, however, the negative solution, whether it results from carelessness or necessity, will satisfy another problem cognate with the proposed one; which may be determined by substituting the negative quantity for the positive in that step of the process which most clearly expresses the conditions of the question; and then interpreting the resulting equation with the assistance of the given problem. This is done in

Art. 214.

n

Ex. The sum of n terms of the series a, a + b, a + 2b, &c., obtained on the supposition that ̧n is a positive integer, is (2a + n − 1.b); find the series of which this is the sum, supposing n a negative integer.

=

If n be negative, the expression for the sum becomes (− 2a+n+1.b)~; and to find the series, let n = 1, then the sum = b-a the first term; again, let n=2, then the sum of two terms 36-2a; .. the second term = 3b − 2 a − (b − a) = 2b — a ; and .. the common difference b, as in the original series. Hence the required series is

b-a, 2b-a, 3b-a, 4b-a, &c.

=

459. Interminable Quotients. Strictly speaking a Quotient can only exist when after the division by which it is determined there is no Remainder; but the term is applied to those cases also where a remainder is left which cannot be got rid of. Thus we say generally, that the quotient of 11-x is 1 + x + x2 + x3 + ; whereas the true quotient is 1 + x + x2 + x3 +

1

Thus, whatever be the value of x,

x

1

1 X

-x

=

N. B. By taking the Remainder into account no unintelligible result can arise from substituting any particular value for x.

COR. 1. If x<

1, then the Remainder may be neglected, if a sufficient

number of terms of the series are taken. (Art. 290. Cor. 2.)

1 + 1 + 1 + 1 + &c. in inf. = an infinitely great num

If a be negative, we have

in which if we put 1 for x, we get

acccording as an even or an odd number of terms is taken; both of which

results are obviously impossible.

Now, taking the Remainder into account, we have

1

2.

(1 − x)**

without which fractional Remainder no arithmetical equality subsists between the series and And it may be observed generally, that no equality subsists, for purposes of calculation, betwixt any infinite series without the "Remainder," and the primitive quantity from which it was derived, unless the series is convergent, so that we can make the Remainder after r terms, by increasing r, as small as we please.

a

460. To explain generally the results which assume the forms a × 0, , ao, 0o,

a 0

0' 0°

(1). Since a xb signifies a taken b times, it is clear that, if 0 is to follow the same rule as other multipliers, a ×0 signifies a taken 0 times, and is therefore equal to 0.

(2). Since

b

a

0

a

expresses the number of times a is contained in b, will signify the number of times a is contained in 0, that is, 0 times, or

0

a

= 0.

(3). By the general Rule of Indices, a” × a” = a”+”, and a” ÷ a" = a*-*. Now in the first of these let m = 0, then ao.a" = a", if the same rule holds when one of the indices is 0; therefore ao, as far as regards the rule for multiplication of powers, is equivalent to 1, or ao = 1.

when the powers are equal; therefore it also accords with the general rule for division that ao:

=

1.

Hence ao, 6o, co, &c. are separately equal to 1, if 0 be admitted as an index subject to the same rule as other indices.

(4). Since it has been already explained, that any quantity raised to a power represented by O may be safely expressed by 1, it follows that (a - b) = 1, whatever a and b may be. If then ab, we have 0o= 1.

a

(5). When an algebraical quantity is made to assume the form, it is said to be infinitely great, and its value is expressed by the symbol. All that is meant is, that if the denominator be made less than any assignable or appreciable quantity the fraction becomes greater than any assignable quantity. This is easily shewn by taking any fraction, which 10a: for if the denominator be successively diminished one-tenth, we obtain the series of quantities 100 a, 1000a, 10000 a, &c., proving that as the denominator of the fraction is diminished, the value of the fraction is increased, and without limit.

a

as

0

(6). Suppose that an expression involving x assumes the form when some particular value (a) is substituted for x, then it is clear that the

expression is capable of being reduced to the form P(-a)TM

where p

¶ (x − a)" and q have no factor x-a in them; and by dividing numerator and denominator by their highest common factor, the value of the fraction may be found when x = a. (See Art. 381.)

Thus it appears that a quantity which assumes the form

determinate value. And, conversely, since p is equal to p.
0
a, p=ō'

ever be the values of x and a; if x =

may have a

x - a

x - a

what

that is, any quantity may be

made to assume the form. But this is, in fact, multiplying and dividing by O, on the supposition that the rules which apply to finite quantities apply also to 0 as a multiplier.

It may be said, generally, that to speak of absolute nothing as the subject of mathematical calculation is absurd ; and that it can only become so when it is taken to represent some finite quantity in that state when by indefinite diminution it has become less than any appreciable quantity. The mathematical symbol 0 has, then, always reference to some other quantity from which it is derived; and it is this primitive quantity which must be the subject of our investigations when by becoming 0 it produces an anomalous result that requires to be explained.

That mistakes will constantly arise from considering 0 as an absolute quantity is easily seen: Thus, it has been shewn that ao = 1 = 6o, therefore we might say, if 0 is a quantity, that a = b; or, since 2o = 4o, that 2 = 4, both of which conclusions are manifestly wrong.

This amounts to saying that, because a × 0=0, and 6 × 0=0, therefore a = b, which is obviously incorrect.

461. Given ax + b = cx+d, and .. X = (1) when a = c; and (2) when a = c, b = d.

d-b
0

d-b

to explain the result

a -C

(1) When a = c, x = =8. c. In this case the proposed equation becomes ax + b = ax+d, which can only hold true on the supposition of a being such, that ax is not affected to any appreciable amount whichsoever of the two different quantities b or d be added to it, that is, a must be immeasurably great, agreeing with the result already found.

equation becomes ax+b=ax+b, an identity which is clearly satisfied by

any value whatever of x; and this is the meaning of

462. Given ax+by = c
and a'x + b'y = c']'

explain the results when a

0

0

in this case.

ma, b' = mb, c' = mc.

or a'bab', ac' a'c, b'c = bc'; .. x =

=

and

=

с

0

y

From the original equations we have ax + by = C=

x + y

m m

m

= ax + by, an identity which is satisfied by any values whatever of x and y; agreeing with the results before obtained.

463. Given ax2 + bx + c = 0, and .. x =

this result when a = 0.

When a = 0, one value of a becomes O, the other

or co. From

which is the value

b'

464. A and B are travelling along the same road, and in the same direction, at a uniform pace of a miles and 6 miles per hour respectively. Given that at a known time B is d miles before A, find the time when they are together; and explain the result (1) when a = b, and (2) when а < <b.

Let x be the number of hours from the known time to the time when they are together. Then in that time A travels ax miles, and B travels bæ miles, and by the supposition