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SECTION I.

PROPOSITIONS 1-8.

ON MAGNITUDES AND THEIR MULTIPLES.

Art. 1. Number.

In this book except where otherwise stated the word Number will be used as an abbreviation for Positive Whole Number.

Notation for Number.

A Number will always be denoted by a small letter.

Art. 2. Notation for Magnitude.

A Magnitude will be denoted throughout this book by a capital letter*.

Art. 3. Def. 1. MULTIPLE.

One magnitude is said to be a multiple of another magnitude, when the former contains the latter an exact number of times.

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It will be in agreement with the above nomenclature, when A is equal to B, to say that A is the first multiple of B; and to call the rth multiple of B and the rth multiple of C the same multiples of B and C.

Art. 4. It is necessary to prove certain propositions regarding magnitudes and multiples of magnitudes before entering upon the discussion of the relations between magnitudes.

* A point will also be denoted by a capital letter, but this will not lead to any difficulty.

H. E.

1

Art. 5. PROPOSITION I.* (Euc. v. 1.)

ENUNCIATION. To prove that r(A + B)=rA +rB.

Construct the following diagram.

Draw a rectangle.

Draw one line parallel to one pair of sides, dividing it into two compartments. Draw (1) lines parallel to the other pair of sides, dividing each of the two compartments into r compartments.

In or upon each one of the upper row of compartments place the magnitude A; and in or upon each one of the lower compartments place the magnitude B.

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Fig. 1.

Then, adding together the two magnitudes in any column the result is A + B, and as there are r columns, the sum of all the magnitudes on the whole rectangle is r (A+B).

Again, adding together the magnitudes in the upper row the result is rA, and the sum of the magnitudes in the lower row is rB.

Hence the sum of all the magnitudes is rA +rB.

But the sum of all the magnitudes is independent of the order in which they are added.

.. r (A + B)=rA +rB.

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Art. 6. EXAMPLE 1.

By repeated application of Proposition I. prove that

r (A + B + ... + K) = rA + rB + ... + rK.

Art. 7. PROPOSITION II.* (Euc. v. 2.)

ENUNCIATION. To prove that (a+b) R = aR+ bR.

Take any rectangle.

Draw (a + b − 1) straight lines parallel to one pair of sides, thus dividing it into (a+b) compartments.

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* See Note 1.

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Then the sum of all the magnitudes is (a + b) R.

Again, separating the magnitudes into two groups, consisting respectively of the magnitudes in the first a compartments, and the magnitudes in the remaining b compartments, the sum of the magnitudes in the first group is aR, and the sum of those in the second group is bR.

Hence the sum of all the magnitudes is aR + bR.

But this sum was shown above to be (a + b) R.

.. (a+b) R= aR+bR.

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3. If A and B are both multiples of G, prove that A + B is a multiple of G.

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Art. 10. PROPOSITION IV.* (Euc. v. 6.)

ENUNCIATION. If a > b, prove that (a - b) R = aR − bR.
Since a >b, and each is a positive integer,

.. (a - b) is a positive integer which may be called c.

.. a = b + c.

.. aR (b+c) R

=

=bR+cR.

.. cRaRbR

.. (ab) RaR-bR.

Art. 11. EXAMPLE 4.

[Prop. 2.

If A and B are multiples of G, then the difference of A and B is a multiple of G.

Art. 12. PROPOSITION V.†

ENUNCIATION. To prove that

r (sA) = rs (A) = sr (A) = s (rA).

Let a rectangle be drawn and divided into compartments standing in r columns and s rows.

Place the magnitude A in each compartment.

Then the sum of the magnitudes in any row is rA, and the sum of those in any column is sA.

Since there are s rows, the sum of all the magnitudes is s (rA).

Since there are r columns, the sum of all the magnitudes is r (sA).

If the number of the magnitudes be counted, it is rs, or it may also be expressed

as sr.

Hence the sum can be written in either of the forms rs (A) or sr (A).

But the sum of the magnitudes is the same in whatever way it is determined. :. r(sA)=rs(A) = sr (A) = s (rA).

Art. 13. EXAMPLE 5.

If A and B are multiples of G, then the sum and difference of rA and sВ are multiples of G.

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Both these results are contradictory to the hypothesis that rA > rB.

Hence A must be greater than B.

The second and third cases can be proved in like manner.

[Prop. 1.

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