or 2 x 2 x 5=20, the last denominator, and in each case continue to multiply by the remaining numbers. The product must in all cases be a multiple of the first respective product of 6, 8, or 20, that is to say, a multiple of each original denominator. As this operation and demonstration are available for all other fractions, and any number of them, we are justified to give the following rule for finding the least common denominator : Decompose each original denominator into its lowest factors, and put each of these latter down only as often as it occurs in the denominator which contains it most ; multiply the numbers thus found, and you have the denominator required. To bring afterwards each original fraction to this denominator we must compare its own denominator with the common one, or see how many times it is contained therein, and with the number thus obtained we multiply its numerator. M. Bring to the same denomination : and į ; 5, and so ; , and ši and ; , 24, 1962 g and joined and openi 15, ha, 1836. Add together + 'tig; + +i: +1+30 i 8 + 80%+ Subtract 11-18 11-15, 35-16, 29435–11814, 6342346-3291356 IX. To reduce Fractions to lower Terms. M. We have seen before that it does not affect the value of a fraction if we multiply both its terms by the same number, as =,=h We may, of course, restore the fractions thus obtained to their original form by again dividing each term by the same number, as =ş Hi= But you will also easily perceive that in general use we may divide the two terms of a fraction by the same number without altering its value. What does the denominator signify? How does it affect the fraction if I divide the denominator ; what effect has this upon the parts -A. The parts become so many times larger, and the fraction therefore increases. M. What does the numerator signify, and how does it affect the fraction if I divide the numerator -A. I get so many times less of the parts, and therefore the fraction decreases. M. If I first double the fraction by dividing its denominator, and then take one-half of the doubled fraction, what must be the final result?-A. I have as much as in the beginning. M. And it is the same if I first increase the fraction 3, 4, 5, &c. times, and then take ži , ļ, &c., off the product. You will more clearly I perceive the correctness of this statement if I show it on some lines. say, for instance, =, = = SASARL M. This rule is of importance, as it enables us in many cases to reduce fractions to lower terms, which of course renders our operations with them more easy. Suppose now you want to reduce the following fractions to lower terms: 5, 18, 1, 36, 370, 300, 1900, 16: X. To recognize certain Divisors or Measures of Numbers. M. What is the first question you ask yourself in all these examples ? -A. By what number can I divide both terms of that fraction ? M. In many cases, particularly with fractions the terms of which are low, this is easily found if one only knows the tables of multiplication. But it is very useful to know how to reduce fractions with large terms; and there are some outward signs in many numbers by which you may recognize whether a reduction be possible or not. As you already stated, the question always is : by which number can I divide the numerator and denominator without leaving a remainder. First for the number 2. Each large number consists of units, tens, hundreds, thousands, and so forth. 7384=4 units, 8 tens, 3 hundreds, 7 thousands. Now 10 and any number of tens are always divisible by 2 without a remainder; 100 and any multiple of 100, 1,000 and any multiple of 1,000, and so forth, leave no remainder. I have, therefore, only to consider the units. If they leave no remainder in the division by 2, the whole number cannot leave any. What are the units that can be exactly divided by 2 ? A. 2, 4, 6, 8, O. M. Therefore I say : A number may be divided by 2 without a remainder if the units are either 2, 4, 6, 8, or 0; in other words, if the last figure is an even number. Say and write many large numbers which you may divide by 2 without a remainder. Write fractions with large terms that may be reduced by 2.--The number 4. M. First observe that 100, 1,000, 10,000, &c., are exactly divisible by 4; we have, therefore, only to examine the tens and units, or the two last figures. If they exactly contain 4, the whole number will leave no remainder. M. Say and write down what may be the two last figures of a number that the whole may be easily divisible by 4. Begin with 00, 04, 08, 12, &c. Write large numbers and fractions as above. To see whether we can exactly divide by 8, we have to consider in the same way and for similar reasons the last three figures of the number; we might even go on with 16 and 32, but this would scarcely be of any practical use. M. Let us now find out when we may divide a number by 5 without leaving a rest ? Can you tell me which of its figures we only consider ? The pupil will easily discover that, as 10, 100, and all their multiples are divisible by 5, the last figure of a number is alone of importance in this respect, and that if it be either 5 or 0, the whole number leaves no remainder in the division. The cases for 10, 100, 1,000, &c., are quite as easily discovered and explained ; the pupil must only be induced to state the reasons for each rule, and then give many instances of large numbers and fractions as above. The numbers 9 and 3. M. We are now going to find an outward sign by which we ascertain whether a given number be divisible by 9. Divide 1, 10, 100, 1,000, 10,000, &c., by 9, and only state what remains f-A. In all these cases the remainder is 1. M. Now divide 2, 20, 200, 2,000, 20,000, by 9, and say what remains ?-A. In all these cases I get 2 for remainder. M. Divide 3, 30, 300, 3,000, &c.—4, 40, 400, 4,000, &c.-5, 50, 500, 5,000, &c., by 9. The pupil will find the respective remainders, 3, 4, 5, &c. M. Now let us consider a large number; for instance, 67,545. This number, as it is written, consists of 5 parts, i. e. 5 units, 4 tens, 5 hundreds, 7 thousands, and 6 tens of thousands. Divide each of these parts by 9, and state what are the remainders.—A. These will be 5, 4, 5, 7, and 6. M. Add them together.--A. The sum is 27. M. Then I say: The whole number having already been divided by 9, except these remainders, and these remainders being exactly divisible by 9, the whole number must be so; and if you try, you will find that it really is so. Let us now consider the number 478,386 ; name the single parts it consists of; state the remainder left in the division of each part by 9; add the remainders together, and draw your conclusion from it. After some practice the pupil will find that the single figures contained in a number represent the remainders left in the divisions by 9, and the rule is : If the sum of the single figures of which a number consists is divisible by 9, the whole number is so · Observationi. In order not to complicate the rule and operation it will be better to regard the figures 9 occurring in a number as remainders, as if we bad 9:9=0+9 left, 90:9=9+9 over, 900:-9= 99 + 9 over. The same observation applies to the number 3, for which the same considerations and demonstrations are to be made as for 9, considering 3, 4, 5, &c., as remainders of a division not completely finished, as if it were, for instance, 3+3=0+3 over, 4;3=0+4 over, 50+3=15+5 over, 500--3=165+5 over, 7,000+3=2,331 +7 over, &c. Now write many numbers which exactly contain 9, and fractions which may be reduced by 9; then the same with 3. XI. To find a Common Measure for two Numbers. M. The rules which we have found about dividing numbers without remainder by 2, 4, 8, 5, 10, 100, 9 and 3, are available for many cases ; but very often one term of the fraction only is divisible, and then we cannot reduce it to lower terms. There are, on the other hand, fractions the terms of which have a common divisor (or measure, as we often call it), such as 7, 13, 23, 37, but for these divisors or measures there exists no outward sign as for 2, 4, 8, 9, and the others. We have, however, a means of ascertaining whether the two terms of a fraction, or any two numbers given, may both be divided by the same number without leaving a remainder. As you would not be able to find it out yourselves I must show it to you, but will afterwards try to explain its correctness. The rule is : Divide the larger of the two numbers by the smaller one, this one by the remainder, the first remainder by the second, the second by the third, and so forth. The last divisor is the common divisor or measure for the original two numbers. For instance, I want to know whether the fraction 1299 can be reduced to lower terms, or whether 1209 and 1417 have a common measure. The operation stands thus (you had better write the divisor in this case on the left side) : 1209: 1417=1 1209 The last divisor being 13, I say this number is the common measure for 1209 and 1417. A trial will at once show that I am right, for 1209--13=93, 1417 •13=109. But you might, without trying first, infer from mere reasoning that it must come right; and as the same course of reasoning is available for all other cases, it shows the correctness of the rule given. The last divisor 13 is exactly contained in 39; 169 containing 4 x 39 +13, both of which numbers, 39 and 13, exactly contain 13, 169 must likewise exactly contain 13; 208=169 +39, both which parts are multiples of 13, as we have seen, and therefore 208 contains 13; 1209=5 X 208+169, and is for the same reason a multiple of 13; and lastly, 1417=1209+208, which two numbers exactly contain 13 without a remainder, and therefore 1417, as well as 1209, must exactly contain 13. Find now by the same rule the common measure for 3553 and 5321, 11,421 and 26,461. Reduce the fractions, 99610, 186, 1997 XII. Applications of the preceding Rules. M. You will often have occasion to apply the rules for reducing fractions to lower terms after addition, subtraction, and particularly after HUMILITY ESSENTIAL TO A TEACHER.-Nothing will impede a Master's progress more than vanity or pride. It will be at once detected, and stand in strange contrast with the lessons which it is his business to impart. It is a great blemish in any character, but in one devoted to teaching, it is most incongruous. The end of all our learning should be a conviction that our store of knowledge is but infinitesimal, compared with the vast treasures that are above us and around us. How often does pride shut a man's eyes and ears to lessons which he might learn, but which he will not even condescend to read, because they come from sources which he considers beneath his notice. A master, keenly alive to the interests of his school, will have his own mind open to receive instruction, let the source from which it comes be ever so lowly. Many a one who stands high in the estimation of those around him, does so, not so much from his great abilities as from the fact that he has used his natural gifts in a spirit of meekness and humility.–From an able Address on the Qualifications of the Church of England Schoolmaster, by the Rev. John Freeman, Rural Dean. |