Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

:

[ocr errors]

= 3.

2 ACCB, or 89 = 52 + 32 +2 X 5 X 3 = 25 + 9 + 30 = 64, and 82 = 64.

The propositions of the fifth book may also be illustrated by numbers. If A = 12, B = 4, C=18, D= 6, then A:B=C:D A

-C or 12:45 18:6. Also A +B:B=C+D:D or 12 + 4:4=18+6:6 or 16:4= 24:6.

A C 12 18
Also
B =ū, or

D 4 6 II. The following are examples of products and quotients, and simple cases of fractions

:3, and n = 5, then, 1. mn =

3 X 5 = 15. 2. mn2 =3 X 52 = 3 X 25 = = 75.

mn2 3.

=mn, for mn multiplied by n, and di

3 X 52 3x5x5 vided by n, is just mn; or —

=3X5=15 5

5

If m =

mnn

n

12

[ocr errors]

= mn.

mn

mn

[ocr errors]
[ocr errors]
[ocr errors]

n2

nn

n

5. mnpq

[ocr errors][ocr errors]

3X5 3X5 3 4.

or n

52 5 X 5 5 If also p =

:2, and

9

= 4, then, = 3X 5 X 2 X 4= 120; or mpnq=3X 2X 5 X 4 = 120. mnpa

3X5 X 2X4 6. = mnp, or

=3X5 X 2 = 30 = 2

4 mnp. тпpg mping

3X 2X5 X4 7. = mp, or

=3 X 2 = ng ng

5X4

[ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small]

52 X 42

ng'

6 20

mnpq naqa mp ng

The rules for numerical vulgar fractions apply to the simple algebraical fractions treated of here.

The three following articles are propositions respecting common fractions :

b III. Let m = m; a, b, and m, being integers, then

, 1

a

[ocr errors]
[merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][ocr errors]

三4m

m

[ocr errors]

a

та

a

ma

ma IV. If a, b, and m, be integers, then

mbi

6 Let a = 6,6 = 2, and m = 3, then

ū=

= 3; and 3X6 18 mb = 3X2 6

= 3; or

ī mbi V. Let a, b, c, and d, be integers, then 7 a=id , ,

6 8 If a= 6,6 = 3,c=8, and d=5, then a

ūd=3*5

X 3 1

6x8 48 1 = 2x1

3
5

and
5 ;
= 3x5 15

6 d

a

с

ac

.

a

[ocr errors]
[ocr errors]

ac

a

[blocks in formation]
[merged small][ocr errors]

id VI. The propositions in the last three articles are also true when a, b, c, d, and m, are fractional terminate numbers. а,

g

k For example, let a=

b = and m = 7

the numh

T.

ma bers e, f, g, h, k, and l, being integers ; then

b mbi
:k ke

k 9 kg
For ma =
and mb =

therefore
ījTT
!"

ī Ih'

[ocr errors]

e

f

ma

a

[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]

M =

[ocr errors]

ma

a

721 Therefore

a

ke lh
eh

h eh mb If kg fg

And =

; therefore

f g fg ma mbi

2 7 9 As a numerical illustration, let a = 6

3 5 II 9 2 6

9 7 63
Then ma =
11 X3

; mb= Х
II

therefore 55

; 6 55 10

2 5 10

Also mb =ī X 63 =21:

6

3 X7-21 ma 6 mb

In a similar manner, the propositions in Art. III. and V. may be proved.

The same propositions are true when a, b, c, d, and m, are interminate numbers.

Their demonstrations are given in the four following articles. The accented letter in every case denotes an interminate number greater than that denoted by the same letter unaccented, and the same letter doubly accented denotes a terminate number intermediate between the other two. Thus a' z a, a" <a', and a" za; where a' is the intermediate terminate number. When a contradictory conclusion is arrived at on the hypothesis that a' 7 a, it is of course false, and it

may

in each case be similarly proved that the hypothesis of a' za is also false ; and hence if the first hypothesis be proved false in any case,

it
may

be concluded that a' must = a. These remarks will prevent unnecessary repetitions. VII. Let a, b, and m, be one or all of them interminate, 6 1

6 1

The values of and each supposed to be expressed by a decimal fraction, and the proposition asserts the equality of these decimals. 1. When a, and therefore m, is interminate. 3 1 6 1

1 1 Let

m (VI.); but m7
)

for

and is

= m, then

are m

a

m

a

[ocr errors]

1

[blocks in formation]
[ocr errors]

=m"

a

7>

6 b m" <m'; therefore

or a" za. But a"

a" (VI.), and m" 7 m, therefore

or a' > a, and it 7

6 1 was also shown to be less, therefore

2. When 6, and therefore m, is interminate.
The proof is similar to the last.
3. When a, b, and m, are interminate.
6 1

3 1
Let
and also let

Then (1st case) ;

a

m

a

m

a

n

a

q=n; but

a
7
=m, therefore n < m, for 7" 7 b. But 6'

1 1 > ", and therefore

or n 7 m, and was also m

6 shown to be less ; hence

[ocr errors]
[ocr errors]

a

m

a

c

a

Letī=

7 d COR.-If -If==m, then

7 VIII. When a, b, and m, are, one or all, interminate,

mat

mbi 1. When a is interminate. ma' a'' ma'

ma' ma Then mb' 7 mb

(VI.); but

mb mb' a" o <t

za;

therea ma fore

7 mb
2. When 6 is interminate.
This is proved like the last case.
3. When a and b are interminate.
ma
a" ma"

ma"
Then
b

mb
(2d case); but

mb

a

therefore , and hence a" <a; but a”

[ocr errors][merged small][merged small]

a

ma'

a' therefore

o <ão

or a' <a; but a" z a; hence

mo'

[blocks in formation]

a

a

ūmo
4. When m is interminate.
m'a

ma

(VI.); and m'a < m'a, 7

'm"

a m'a and m'b 7 mb, therefore < but these are.equal,

ml

Let y = mb

Then 7 =

[ocr errors]

m'a

mbi

a

a

= ĝ, therefore

ma for they are each =

ū mbi
5. When a and m are interminate.
ma'

a"
ma"

ma' Let

Then
mo

mb
(4th case); but

mb ma

a" therefore <

ū

and a" Za; but a' Za; therefore

a

[ocr errors]

a

IN

N

[blocks in formation]

7 mb

6. When 6 and m are interminate.
The proof is similar to the last.
7. When a, b, and m, are interminate.
ma'
a" ma'

ma"
Let

(6th case); but 6 mbo mb

mb ma' a'

and hence a' ca; but a'
2
7

a; there õi a ma fore

7 mbo

[blocks in formation]

IX. Let a, b, c, and d, be one or all interminate,

d

ac

=id
1. When one of them as a is interminate.

a'c

a" ac d=bd ī'ā= od

.

[merged small][merged small][ocr errors][merged small][merged small]
« ΠροηγούμενηΣυνέχεια »