18. In Multiplication and Division. On two straight lines, meeting at a convenient angle, construct OB, OB1, each equal to b, and OA,, OA, each equal to a. On OB, take 0Jj, the real unit, join JA,, JB, draw AM, B1N, parallel respectively to JB, JA,, intersecting OB (produced, if necessary) in M and N, and let OM=m, ON=n. By definition of a product (Art. 5),

and that determined by the parallels ƒ and g is

jban,

which by interchange of means (Prop. 11) becomes

M and N are hence one and the same point, and

b x a = a × b.

(Prop. 6.)

Inasmuch as the reciprocal of a line-magnitude is itself a line-magnitude, it follows from the above demonstration that

bx/a = axb;

= |a

or, observing the law of signs (Art. 13), this is

b/a = axb;
= /a

and if b carry with it its proper factor sign, the more explicit form of statement contained in this equation is

xb/a = a X b.

The formula for the commutative law in multiplication and division may therefore be written

× by a = xa xb,

and this is easily extended to products of three or more factors. In this formula, the same sign attaches to a or b on both sides of the equation.

19. Agenda: Theorems in Proportion.

(1). Prove that in a proportion the product of the means is equal to the product of the extremes.

(2). Prove that rectangles are to one another in the same ratio as the products of their bases by their altitudes. Show that the same relation holds for parallelograms and triangles.

(3). If a square whose side is I be taken as the unit of area, the area of any rectangle (or parallelogram) is equal to the product of its base by its altitude.

VI.

THE DISTRIBUTIVE LAW FOR REAL QUANTITIES. 20. With the Sign of Multiplication. On OH and OK (Fig. 19), take OC=c, OE = a + b, and construct the product (a + b) XcOG. Then if OJ be the real unit, laid off

a+b

K

on OK, EG and JC are parallel by construction. On OK take OA = a, and construct the product a X c = OF Then HAF and EG, being parallel to JC, are parallel to one another, equal to AE, that is, product b× c. Hence

(a + b) Xc=ax c + bxc=c × (a+b), which is the law of distribution in multiplication and addition.

The construction and proof are not essentially different when some or all the magnitudes involved are affected with the negative instead of the positive sign.

The second factor may also consist of two or more terms; for

(a + b) x (c + d) = a × (c + d) + b × (c + d),
by the distributive law,

= (cd) Xa+(c + d) × b,

by the commutative law,

and again, by the distributive and commutative laws, (c+a) Xa+(c+d) ×b=cXa+dXa+cxb+dXb

=aXc+axd+bXc+b× d.

Hence, replacing the sign + by the double sign ±, the completed formula for the distributive law in multiplication is

(±a+b)X(±c±d)=(±a)X(±c)+(±a)X(±d)

21.

+(±b)×(±c)+(±b)×(±ď).

With the Sign of Reciprocation. Since a, b, c, d, in the formula last written, are any real magnitudes whatever, they may be replaced by their reciprocals, | a, | b, /c, Id: For the same reason, in the formula

(±a±b)×(±c)=(±a)×(±c)+(±b)×(±c), X(c) may be replaced by (±c), giving the corresponding distributive law with the sign of reciprocation :

(±a±b)/(±c) = (±a) / (±c)+(±b) / (±c).

But while the sign X is distributive over the successive terms of a sum (Art. 20), that is:

X (±a±b±...) = X(±a) +× (±b)+X...,

the sign is not, as may be readily seen by constructing the product X / (a+b), and the sum of products c X / a +cX/b, and comparing the results, which will be found. to differ.

22. Agenda: Theorems in Proportion, Arithmetical Multiplication and Division.

(1). If

prove that

ABC:D :: E: F::.. .,

A: BA+ C + E + . . . : B + D + F + ...

(2). If prove that

ABCD,

: :

(Addendo.)

A + B : B :: C + D D * (Componendo.)

ABB: C~ D: D.

(Dividendo.)

and that

(3). If

prove that

(4). If

prove that

ABC: D,

A+BA~ B :: C + D : C ~ D.
ABPQ, B:C::Q: R,
C: DRS, and D: E: ST,

AEP: T.

(5). If A:B :: Q : R and B: C:: P: Q,

prove that

A: CPR.

(Ex æquali.)

(6). Show that corresponding to

cx (xa + xb) = X (X c × a + xc X b)

there is the analogous formula

(7).

result is 6.

c! (/a + /b) = /(/c/a+ / c / b).

Construct the product 3 X 2 and show that the

(8). Construct the quotient 1 2/3 and show that the result is 4.

(9). Construct 5/ 3 and 2 X 5/3.

VII.

EXPONENTIALS AND LOGARITHMS.

23. Definitions.* Suppose P, Q to be two points moving in a straight line, the former with a velocity, more strictly a speed, ** proportional to its distance from a fixed origin O, the latter with a constant speed. Let x denote

the variable distance of P, y that of Q from the origin, and let & be the speed of P when x= QJ1, the constant speed of Q. As it arrives at the positions J, P', P'' successively, P is moving at the rates:

λ, x'λ, x''λ, respectively,

the corresponding values of x and y are:

x=0J

OJ = 1, x'= 1 + JP', x''= 1 +JP", y = 0,

y' = OQ',

y"

OQ"

* Napier's definition of a logarithm. Napier: Mirifici logarithmorum canonis descriptio (Lond. 1620), Defs. 1-6, pp. 1-3, and the Construction of the Wonderful Canon of Logarithms (Macdonald's translation, 1889), Def. 26, p. 19. Also MacLaurin: Treatise of Fluxions, vol. i, chap. vi, p. 158, and Montucla: Histoire des Mathematiques, t. ii, pp. 16-17, 97.

**The speed of a moving point is the amount of its rate of change of position regardless of direction. Velocity takes account of change of direction as well as amount of motion. See Macgregor: Elementary Treatise on Kinematics and Dynamics, pp. 22, 23 and 55.