we may infer A2: B2 :: C2 : D2;

or any other property respecting four proportionals.

SUPPLEMENT I.

ON COMMENSURABLE AND INCOMMENSURABLE MAGNITUDES.

MAGNITUDES which are multiples of any other magnitude are said to be commensurable, and this other magnitude is called a common measure.

Magnitudes which have no common measure, that is, magnitudes which are not multiples of any other magnitude, are said to be incommensurable.

Numbers have always a common measure, for they can always be measured by a unit. When they can only be measured by a unit, they are called prime numbers: numbers are said to be prime to each other when no number except a unit will measure them. Thus, 43, 47, 53, 61, are prime numbers;

and 5, 8, 51, 77, are prime to each other, for no number except a unit is contained an exact number of times in all: yet 8 can be measured by 1, 2, and 4; 51 by 1, 17, and 3; and 77 by 1, 7, and 11.

PROP. I. THEO.

If a magnitude measure each of two others, it will also measure their sum and difference.

Let A measure B and C separately, it will also measure B + C and B — C, allowing B to be the greater of B and C; unless one of the magnitudes B or C be greater than the other, the difference cannot be measured by A, as it is 0.

Let A be contained in B, m times exactly, .. B = m A ;
and A to be contained in C, n times exactly, .. C = n A ;

it is evident that n must be less than m, for B is greater than C.

Therefore A is contained in B + C, m + n times, and in BC, mn

times; consequently A measures the sum and difference of B and C.

It will immediately follow, that if A measures B + C and C, it will also measure B, for B is the difference of B + C and C; or if A measures B + C and B, it will measure C for the same reason. And if A measures B - C and C, it will also measure B, for B is the sum of C and B - C; it is evident also, if A measures B - C and B, it will measure C, for C is the difference between B and BC.

PROP. II. PROB.

Two magnitudes of the same kind being given to find their greatest common measure, if it be possible.

Let A and B be the two magnitudes, and A less than B; it is required to find the greatest magnitude that will measure them both.

Let p be the greatest number of times that A is contained in B; take p A from B, and let the remainder be C, which must be less than A. Let g be the greatest number of times that C is contained in A; from A take 9 C, and let the remainder be D.

This operation is to be continued, always dividing the preceding divisor by the last remainder, till nothing remains; the remainder, which divides the preceding one exactly, is the greatest common

measure.

A) B (p
pA

C) A (4

q C

D) C (r

r D

E) D (8

s E

&c.

It must be observed, that A, B, C, D, E, &c., represent magnitudes of the same kind, and p, q, r, s, &c., whole numbers.

First, we shall prove that any magnitude which measures A and B, will also measure the remainders C, D, E, &c. Let M measure both A and B, it will measure any multiple of A; .. it will measure p A; and, as it measures B, it will also measure B pA or C (prop. i.); and, as it measures C, it will also measure 9 C; and M measures A, .. it will measure A q C or D; as it measures D, it will measure r D; and it has been shown to measure C, .. it will measure Cr D or E. Therefore any magnitude, M, which measures A and B, will measure all the remainder, C, D, E, &c., unto the last.

Now we have shown that every common measure of A and B, will measure the last remainder; and it is evident the greatest magnitude that will measure the last remainder is itself; therefore we have only to prove that the last remainder measures A and B, to show that the last remainder is the greatest common measure of A and B. This may be shown as follows::

Let E be the last remainder, or that E is contained in D an exact number of times; .. E will measure D, and it will also measure itself and r D; ... E will measurer D + E or C; and, as E will measure C, it will also measure C; and it has been shown to measure D, .. E will measure q C + D or A. Again, because E measures A, it will measure p A; and it has been shown that E measures C, .. E measures p A + C or B; .. the last remainder measures A and B. If there be no such thing as a last remainder, that is, when the above process does not terminate, the magnitudes have no common measure, or they are incommensurable.

PROP. III. THEO.

If two numbers be prime to each other, they are the least two numbers in that proportion.

B Let be a fraction where A is prime to B, and if it be possible let this A

fraction be equal to another,

B' 1 Α'

reduced to its lowest terms, by dividing by

the greatest common measure; and if it be possible let B be greater than B',

and A greater than A', at the same time,

B

=

A

B'
A''

A

or

=

B

Divide B by A, and B' by A', as in the last problem.

A'
B

the first quotients, m, m, are equal; and since

and because A is greater than A', C must be greater than C'.

D

D'

be greater than D'.

Again, because

and as

and because C is greater than C', D must

In the same manner it may be shown that E is greater

than E', &c.; .. a remainder in the first division must always be greater than the one similarly circumstanced in the latter; .. one of the remainders in the latter division will become unity sooner than in the former. Let E'

then E must be greater than 1; and since

D
E

D'

D

...

=

E

=

1;

D';

.. E, which is greater than 1, measures D, .. E must measure both A and B, according to the last proposition. But this is contrary to the hypothesis, for A and B were supposed prime to each other; consequently we may infer that when A is prime to B they are the least in the proportion, and when there are other numbers in the same proportion they must be multiples of these.

PROP. IV. THEO.

If two numbers be each prime to another number, their product will be prime to that number.

Let A and B be each prime to C;

then A x B will be prime to C:

for, if not, let A x B = m x P, and C = m x Q.

Since A and B are both prime to C, they are also prime to m × Q,

and.. to m.

Now, A x B = m × P; divide each by m B, and we have,

but as it has been shown that A is prime to m, therefore, by the last proposition, P must be a multiple of A, and B a multiple of m. But it has also been shown that B is prime to m, which is absurd. Therefore, if two numbers be prime to another number, their product will be prime to that number.

PROP. V. THEO.

If two numbers be prime to each other, one of them is prime to the square, cube, &c., of the other; and any power of one of them will be prime to any power of the other.

Let A be prime to B; then A2 is prime to B,

A3 to B,
&c., &c.

For A and A are prime to B, therefore, by the last proposition, A2 is prime to B; and as A and A2 are each prime to B, A3 must be prime to B (prop. iv.), and so on, it may be shown that A" is prime to B.