The following Table contains a summary of the whole doctrine. Periods. Quadrill.; Trillions; Billions; Millions; Units. Half-per. th. un. th. un. th. un. th. un. th. un. pola a to play Figures. 123,456; 789,098; 765,432; 101,234; 567,890. NUMERATION is the reading of any number in words that is proposed or set down in figures; which will be easily done by help of the following rule, deduced from the foregoing tablets and observations--viz. Divide the figures in the proposed number, as in the summary above, into periods and half-periods; then begin at the left-hand side, and read the figures with the names set to them in the two foregoing tables. EXAMPLES 13405670 47050023 309025600 4723507689 274856390000 7523000 6578600307024 Notation is the setting down in figures any number proposed in words; which is done by setting down the figures instead of the words or names belonging to them in the summary above; supplying the vacant places with ciphers where any words do not occur. EXAMPLES. Set down in figures the following numbers; Fifty-seven. Two hundred eighty-six. Nine thousand two hundred and ten. Twenty-seven thousand five hundred and ninety-four. Six hundred and forty thousand, four hundred and eighty-one. Three millions, two hundred sixty thousand, one hundred and six. Foar = II Four hundred and eight millions, two hundred and fifty-five thousand, one hundred and ninety-two. Twenty-seven thousand and eight millions, ninety-six thou sand two hundred and four. Two hundred thousand and five hundred and fifty millions, one hundred and ten thousand, and sixteen. Twenty-one billions, eight hundred and ten millions, sixtyfour thousand, one hundred and fifty. OF THE ROMAN NOTATION. 1 = I As often as any character is re- peated, so many times is its value repeated. diminishes its value. A less character after a greater increases its value. comes 10 times as many. 1000 = M or CIO For every C and ?, placed one 2000 = MM at each end, it becomes 10 times as much. 5000 = V or 150 A bar over any number in6000 = VI creases it 1000 fold. 10000 == X or CCIII 50000 = L or 1 •• 60000 = LX 100000 Ĉ CCCIO 1000000 = M or CCCCIII 2000000 = MM &c. &c. = VI or EXPLANATION OF CERTAIN CHARACTERS. There are various characters or marks used in Arithmetic, and Algebra, to denote several of the operations and propositions; the chief of which are as follow : X or : + signifies plus, or addition. minus, or subtraction. known which is the greater. Thus, &c. OF ADDITION. Addition is the collecting or putting of several numbers together, in order to find their sum, or the total amount of the whole. This is done as follows: Set or place the numbers under each other, so that each , figure may stand exactly under the figures of the same value, that that is, units under units, tens under tens, hundreds under hundreds, &c. and draw a line under the lowest number, to separate the given numbers from their sum, when it is found. -Then add up the figures in the column or row of units, and find how many tens are contained in that sum.-Set down exactly below, what remains more than those tens, or if nothing remains, a cipher, and carry as many ones to the next row as there are tens.-Next add up the second row, together with the number carried, in the same manner as the first. And thus proceed till the whole is finished, setting down the total amount of the last row. TO PROVE ADDITION. First Method.—Begin at the top, and add together all the rows of numbers downwards; in the same manner as they were before added upwards; then if the two sums agree, it may be presumed the work is right.—This method of proof is only doing the same work twice over, a little varied. Second Method.-Draw a line below the uppermost number, and suppose it cut off.-Then add all the rest of the nunibers together in the usual way, and set their sum under the number to be proved. Lastly, add this last found number and the uppermost line together; then if their sum be the same as that found by the first addition, it may be presumed the work is right. This method of proof is founded on the plain axiom, that “ The whole is equal to all its parts taken together." EXAMPLE I. 5 Third Method.-Add the figures in the uppermost line together, and find how many nines are contained in their sum.-Reject those nines, and 3497 set down the remainder towards the 6512 right hand directly even with the 8295 figures in the line, as in the annexed example. Do the same with each 18304 of the proposed lines of numbers, setting all these excesses of nines in a column on the right-hand, as here 5, 5, 6. Then, if the excess of 9's in this sum, found as before, be equal to the excess of 9's in the total súm 18304, the work is probably right.Thus, the sum of the right-hand column, 5, 5, 6, is 16, the excess of which above 9 is 7. Also the sum of the figures in the Excess of nines. the sum total 18304, is 16, the excess of which above 3 is also 7, the same as the former*. * This method of proof depends on a property of the number 9, which, except the number 3, belongs to no other digit whatever ; namely, that " any number divided by 9, will leave the same remainder as the sum of its figures or digits divided by 9:" which may be demonstrated in this manner. Demonstration. Let there be any number proposed, as 4658. This, separated into its several parts, becomes 4000 + 600 + 50 + 8. But 4000 = 4 x 1000 =4 X (999 + 1) = 4 x 999 + 4. In like manner 000 = 6 x 99 +'6; and 50 = 5 x 9 + 5. Therefore the given number 4658 = 4 x 999 +4 + 6 x 99 +6+ 5 X 9 + 5 + 8 = 4 x 999 + 6 x 99 + 5 x 9 +4 +6 + 5 +8; and 4058 • 9 = (4 x 999 + 6 x 99 + 5 x 9 +4 +6 + 5 + 8) • 9. But 4 X 999 + 6 x 99 + 5 x g is evidently divisible by 9, without a remainder ; therefore if the given numbør 4658 be divided by 9, it will leave the same remainder as 4 + 6 + 5 + 8 divided by 9. And the same, it is evident, will hold for any other number whatever. In like manner, the same property may be shown to belong to the number 3 ; but the preference is usually given to the number 9, on account of its being more convenient in practice. Now, from the demonstration above given, the reason of the rule itself is evident; for the excess of y's in two or more numbers being taken separately, and the excess of g's taken also out of the sum of the former excesses, it is plain that this last excess must be equal to the excess of 9's contained in the total sum of all these numbers; all the parts taken together being equal to the whole. -This rule was first given by Dr. Wallis in his Arithmetic, published in the year 1657. Ex |