BOOK V.

RATIO AND PROPORTION.

Multiples.

451. NOTATION. In Book V., capital letters denote magnitudes.

Magnitudes which are or may be of different kinds are denoted by letters taken from different alphabets.

A+ A is abbreviated into 2A.

A+ A+ A = 3A.

The small Italic letters m, n, p, q, denote whole numbers.

452. A greater magnitude is said to be a Multiple of a lesser magnitude when the greater is the sum of a number of parts. each equal to the less; that is, when the greater contains the less an exact number of times.

169

453. A lesser magnitude is a Submultiple, or Aliquot Part, of a greater magnitude when the less is contained an exact number of times in the greater.

454. When each of two magnitudes is a multiple of, or exactly contains, a third magnitude, they are said to be Commensurable.

455. If there is no magnitude which each of two given magnitudes will contain an exact number of times, they are called Incommensurable.

456. REMARK. It is important the student should know, that of two magnitudes of the same kind taken at hazard, or one being given, and the other deduced by a geometrical construction, it is very much more likely that the two should be incommensurable than that they should be commensurable.

To treat continuous magnitudes as commensurable would be to omit the normal, and give only the exceptional case. This makes the arithmetical treatment of ratio and proportion radically incomplete and inadequate for geometry.

PROBLEM I.

457. To find the greatest common submultiple or greatest common divisor of two given magnitudes, if any exists.

Let AB and CD be the two magnitudes.

From AB, the greater, cut off as many parts as possible, each equal to CD, the less. If there be a remainder FB, set it off in like manner as often as possible upon CD. Should there be a second remainder HD, set it off in like manner upon the first remainder, and so on.

The process will terminate only if a remainder is obtained which is an aliquot part of the preceding one; and, should it so terminate, the two given magnitudes will be commensurable, and have the last remainder for their greatest common divisor.

For suppose HD the last remainder.

Then HD is an aliquot part of FB, and so of CH, and therefore of CD, and therefore of AF. Thus being a submultiple of AF and FB, it is contained exactly in AB. And, moreover, it is the greatest common divisor of AB and CD.

For since every divisor of CD and AB must divide AF, it must divide FB or CH, and therefore also HD.

Hence the common divisor cannot be greater than HD.

458. INVERSE OF 457. If two magnitudes be commensurable, the above process will terminate.

For now, by hypothesis, we have a greatest common divisor G.

But G is contained exactly in every remainder.

For G, being a submultiple of CD, is also an aliquot part of AF, a multiple of CD; and therefore, to be a submultiple of AB, it must be an aliquot part of FB the first remainder. Similarly, CD and the first remainder FB being divisible by G, the second remainder HD must be so, and in the same way the third and every subsequent remainder.

But the alternate remainders decrease by more than half, and so the process must terminate at G; for otherwise a remainder would be reached which, being less than G, could not be divisible by G.

459. CONTRANOMINAL OF 457. The above process applied to incommensurable magnitudes is interminable.

460. OBVERSE OF 457. If the above process be interminable, the magnitudes are incommensurable.

On this depends the demonstration, given in 461, of a remarkable theorem proved in the tenth book of Euclid's "Elements."

THEOREM I.

461. The side and diagonal of a square are incommensurable.

A

A

HYPOTHESIS. Let ABCD be a square; AB, a side; AC, a diagonal.

CONCLUSION. Then will AB and AC be incommensurable.

PROOF. AC > AB, but < 2AB,

Therefore a first remainder EC is obtained by setting off on AC a part AE = AB.

Erect EF perpendicular to AC and meeting CD in F. Join AF.

Δ ADF = Δ ΑΕΡ,

(179. Right triangles are congruent when the hypothenuse and one side are equal respectively in each.)

:. DF = FE.

Again, rt. A CEF is isosceles, because one of the complemental angles, ECF, is half a rt. 4,

. CE EF = FD.

Hence a common divisor of EC and DC would be also a common divisor of EC and FC.

But EC and FC are again the side and diagonal of a square, therefore the process is interminable.