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Cor. 1. In the same manner it may be shown, that the cube root of an integer, that is not a complete cube, can never be represented by any rational fraction; and generally, if a be an integer, and

a cannot be represented by an integer, it is also impossible to represent it by any rational fraction.

Cor. 2. The product of the square root of two numbers prime to each other cannot be expressed by any rational fraction. For if P and q were two

m

numbers prime to each other, and √p× √q= n

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by the above proposition, because the product pq is not a square (cor. 2, prop. xvii.).

Cor. 3. In the same manner it may be shown, that the product of the cube roots of two numbers prime to each other, and not both cubes, cannot be represented by any rational fraction; and generally, if p and q be any two integer numbers, prime to each other, then the product /pq, cannot become equal to any rational quantity, unless p and q be each complete nth powers, and in this case the product is an integer.

Cor. 4. All that has been demonstrated in the above two corollaries, of two quantities prime to each other, is equally true of number of quantities under the same restrictions.

any

PROP. XIX.

19. Neither the sum, nor difference, of the square roots of two quantities, prime to each other, can be represented by any rational quantity; nor by the

square root of any rational quantity; unless each of those numbers be a complete square.

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c representing any rational quantity, or the square root of any rational quantity; then, by squaring, we

have

( √p± √q)2 = c2=p+q±2 √pq;

and since c is either rational, or the square root of a rational quantity, c2 is in either case rational; and, consequently, also

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is also a rational fraction, which is impossible (cor. 2, prop. xviii.); and, consequently,

√p± √q=c, or = √c,

--

is also impossible. — Q. E. D.

Cor. It may likewise be demonstrated, by means of this proposition, p and q, being prime as before, and not both squares, that neither the sum nor difference of their roots can be represented by the sum or difference of the roots of any other two integral quantities whatever, that are prime to p that is,

and q;

is impossible.

√p± √q= √rì vs

For, by squaring, we have

p+q± 2 √ pq=r+s±2 √rs, or

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Now either rs is rational, or it is not; if it be rational, so also must be

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which it cannot be (cor. 2, prop. xviii.); therefore, Ars cannot be rational.

Again, if rs be not rational, then we should have

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a rational quantity; which is also impossible by the above proposition, because pq and rs are prime to each other, and, consequently,

√p± √q= vr± √s

is impossible, under the specified limitations.-Q.E.D.

PROP. XX.

20. If in any equation whatever, higher than the first degree, the coefficient of the first term be unity, and those of the other terms integral numbers, then no one of the roots of such an equation can be equal to a rational fraction.

For let xax" - ' + bx" ̃2±, &c., ±r=0, represent a general form of equation; and, if it be possible, let x=2, the fraction being supposed already in

Ρ

q

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its lowest terms; then, by substituting for x, the

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±, &c., ±r=0; or,

p❞ ap"-1 bp"-2

+

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q"

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-3

p"±aqp"-'±bq2p"-2±cq'p"-3±, &c.,

-1

q"

pr±q(ap”-1±bqp"-2±cq2p"-3±, &c.)

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= Fr; or,

= Fr.

Now as this is equal to r, an integer, it follows that the numerator is divisible by q"; and, therefore, also by q; but since the part

q(ap1bqp"-2+ cq2p"-3±, &c.)
bqp”~2±cq2p"

is divisible by q, it is evident that p" must also be divisible by q, if the whole quantity be so (cor. 1, prop. vi.); but this is impossible, because p is prime to q, therefore the numerator is not divisible by q;

and, consequently, x=2 is impossible.

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Q. E. D.

32

CHAP. II.

On Divisors, and the Theory of Perfect, Amicable, and Polygonal Numbers.

PROP. I.

21. Any number N being reduced to the form N=a"b"c"d", &c., the number of its divisors will be expressed by the formula

(m + 1) × (n + 1) × (p+1), &c.

For it is evident, that N will be divisible by a, and every power of a, to a"; that is, by each of

the terms

1, a, a, a', &c., a".

Also by b, and every power of b, to b"; that is, by every term in the series

1, b, b3, b3, &c., br.

And, in the same manner, N is divisible by c, and every power of c, to c"; by d, and every power of d, to d, &c.; and also by every possible combination of the respective terms of the above series; that is, by every term of the continued product,

(1 + a + a2+, &c., aTM) × (1+ b + b2 +, &c., b") × (i+c+c2+, &c., c3) × (1 + d + d2 +, &c., d2). But the number of terms of this product, since no two of them can be the same, is truly expressed by the formula

(m + 1) × (n + 1) × (p + 1) × (g + 1), &c.;

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