EXAMPLES. Write down in figures the following numbers : Eighty one. Two hundred and eleven. One thousand and thirty nine. A million and a half. A hundred and four-score and five thousand. Eleven thousand million, eleven hundred thousand and eleven. Thirteen billion, six hundred thousand million, four thousand and one. SIMPLE ADDITION. Simple Addition teacheth to collect several numbers of the same denomination into one total. RULE.* 1. Place the numbers under each other, so that units may stand under units, tens under tens, &c. and draw a line under them. 2. Add 1000=M or CIO For every C and O, put one at each end, 2000=MM it becomes ten times as much. 5000=100: or A line over any number increases * This rule, as well as the method of proof, is founded on the known axiom, "the whole is equal to the sum of all its parts." All that requires explaining is the method of placing the num2. Add up the figures in the row of units, and find how many tens are contained in their sum. bers, 3. Set down the remainder, and carry as many units to the next row, as there are tens; with which proceed as before; and so on till the whole is finished. Method bers, and carrying for the tens; both which are evident from the nature of notation: for any other disposition of the numbers would entirely alter their value; and carrying one for every ten, from an inferior line to a superior, is evidently right, since an unit in the latter case is of the same value as ten in the former. Beside the method here given, there is another very ingenious one of proving addition by casting out the nines, thus : RULE 1. Add the figures in the uppermost line together, and find how many nines are contained in their sum. 2. Reject the nines, and set down the remainder directly even with the figures in the line. 3. Do the same with each of the given numbers, and set all these excesses of nine together in a row, and find their sum, then if the excess of nines in this sum, found as before, is equal to the excess of nines in the total sum, the question is right. This method depends upon a property of the number 9, which belongs to no other digit whatever, except 3; viz. 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 thus demonstrated. DEMON. Method of PROOF. 1. Draw a line below the uppermost number, and suppose it cut off. 2. Add all the rest together, and set their sum under the number to be proved. 3. Add DEMON. Let there be any number, as 3467; this separated into its several parts becomes 3000+400+60+7; but 3000=3 1 X1000=3×999+1=3×999+3. In like manner 400=4× 99+4, and 60=6×9+6. Therefore 3467=3×999+3+4× 99+4+6×9+6+7=3×999+4×99+6×9+3+4+6+7 And 3467-3×999+4×99+6×9 + 3+4+6+7. 9 But 3X 999+4×99+6×9 is evidently divisible by 9; therefore 3467 divided by 9 will leave the same remainder as 3+4+6+7 divided by 9; and the same will hold for any other number what. ever. Q. E. D. The same may be demonstrated universally thus : DEMON. Let N= any number whatever, a, b, c, &c. the digits of which it is composed, and n= as many cyphers as a, the highest digit, is places from unity. Then N=a with n, o's +6 with 1 n-1, o's + with n-2, o's, &c. by the nature of notation ; =axn-1, 9's+a+bxn-2, 9's+b+cxn-3, 9's+c, &c. =axn-1,9's+bxn-2,9's+cxn-3, 9's, &c. +a+b+c, &c. but an-1, 9's+bxn--2, 9's+cxn-3, 9's, &c. is plainly divisible by 9; therefore N divided by 9 will leave the same remainder, as a+b+c, &c. divided by 9. Q. E. D. In the very same manner, this property may be shown to belong to the number three; but the preference is usually given to the number 9, on account of its being more convenient in practice. Now from the demonstration here given, the reason of the rule itself is evident; for the excess of nines in two or more numbers being taken separately, and the excess of nines taken also out of the 3. Add this last found number and the uppermost line together, and if their sum be the same as that found by the first addition, the sum is right. 4. Add 8635, 2194, 7421, 5063, 2196, and 1245 to gether. Ans. 26754 5. Add the sum of the former excesses, it is plain this last excess must be equal to the excess of nines contained in the total sum of all these numbers; the parts being equal to the whole. This rule was first given by Dr. WALLIS, in his Arithmetic, published A. D. 1657, and is a very simple easy method; though it is liable to this inconvenience, that a wrong operation may sometimes appear to be right; for if we change the places of any two figures in the sum, it will still be the same, but then a true sum will always appear to be true by this proof; and to make a false one appear true, there must be at least two errors, and these opposite to each other; and if there be more than two errors, they must balance among themselves : but the chance against this particular circumstance is so great, that we may pretty safely trust to this proof. 5. Add 246034, 298765, 47321, 58653, 64218, 5376, 19821, and 340 together. Ans. 730528. 6. Add 562163, 21964, 56321, 18536, 4340, 279, and Ans. 663686. 83 together. 7. How many shillings are there in a crown, a guinea, Ans. 89. a moidore, and a six and thirty? 8. How many days are there in the twelve calendar Ans. 365. months ? 9. How many days are there from the 19th day of April, 1774, to the 27th day of November, 1775, both days exclusive? Ans. 586. SIMPLE SUBTRACTION. Simple Subtraction teacheth to take a less number from a greater of the same denomination, and thereby. shews the difference or remainder. The less number, or that which is to be subtracted, is called the subtrahend ; the other, the minuend; and the number that is found by the operation, the remainder of difference. RULE.* 1. Place the less number under the greater, so that units may stand under units, tens under tens, &c. and draw a line under them. 2. Begin * DEMON. I. When all the figures of the less number are less than their correspondent figures in the greater, the difference of the figures in the several like places must altogether make the true difference sought; because as the sum of the parts is equal to the whole, so must the sum of the differences of all the similar parts be equal to the difference of the wholes 2. When |