Imágenes de página

book is that which gives the prescribed subjects of examination required by each Government department. The extracts from the papers are useless, because they do not show for what departments they have been used. Lectures to Children on Scripture Doctrines. By Samuel Green, B.A. Pp. 115. (London : Sunday School Union, 1856.) Lectures to Children on the Bible. By Samuel Green, B.A. Pp. 117. (London : Sunday School Union, 1856.) These are good little books, conveying practical lessons of piety, and at the same time are interspersed with pretty anecdotes, which will aid in enforcing those duties which must be taught to every child.

SERIALS RECEIVED. Prazer's Magazine ; British Educator; Titan; Scottish Educational Journal; Educator*; L'Abeille ; Educational Record ; Penny Post; Gentleman's Magasine. We are glad to hail Mr. Sylvanus Urban evincing a degree of healthful vigour, which reminds us of his best days.] Lardner's Handbook of Natural Philosophy; Chambers's Educational Course ; Modern History; Scripture Geography; Hobson's Choice; Susan; Mary Thomas ; The British Soldier.

Intelligence. BRITISH ASSOCIATION.–At the recent Cheltenham Meeting, in the Section of Political Science, education was an important feature; not only were there able papers on Reformatories by Mr. Barwick Baker and Miss Carpenter,—there was an interesting and instructive paper by the Rev. C. Bromby on National Education. In the Mathematical Section Mr. Grove gave a remarkably clever paper on Correlative Forces; and Professor Whewell and Mr. Symons descanted on the Moon's Motion (disagreeing but slightly thereon); and Col. Sir T. Rawlinson at immense length on Cuneiform Inscriptions. There was a due quota of interesting geology, and a horrifying exposé of the effects of strychnine, very instructive in the art of murder made easy. Altogether it was a successful meeting.

Notices. REDUCED CHARGE FOR ADVERTISEMENTS. Orders and Advertisements must be sent ONLY to MESSRS. GROOMBRIDGE, 5, Paternoster Row; the latter, from strangers, must be accompanied by a remittance, according to the following scale :-If under 40 words, 3s. 6d.; for every additional ten words, 6d. ; a whole page, £2. 28. ; a half-page, or one column, £1. 58. Ten per cent. discount on all Advertisements inserted more than twice.

The JOURNAL will be sent, free of postage, for one year, on receipt of 6s. 6d. in advance.

TO CORRESPONDENTS. *** No other paper on Local Words can be found. We shall be glad of one.

Dr. Lardner is thanked for the offer of his Pamphlet, but it does not carry the argument any further than before. Some able treatises have also reached us, together with Herr von Hompach's voluminous volume, and Mr. Symons's paper at Cheltenham, both against the theory of rotation. The astronomers will eventually find the common sense view of this matter too strong for them, and will have to amend their description of what they admit to be the Moon's motion ; and which is demonstrably that of simple translation in a circular orbit, not having a particle of rotation, according to Barlow's definition of it, in his Standard Mathematical Dictionary, as follows:

Rotation; the motion of the different parts of a solid body about an axis, called the axis of rotation, being thus distinguished from the progressive motion of a body about some distant point or centre: thus the diurnal motion of the earth is a motion of rotation, but its annual motion one of revolution."


* “When a solid body turns round an axis, retaining its shape and dimensions unaltered, every particle is actually describing a circle round this axis, which axis passes through the centre of the circle, and is perpendicular to its plane."

We owe it to our readers to give them a truce on this subject, strongly recommending those who need further inforination to read the several able treatises on Kinematics published long ago by Mr. H. Perigal on this subject.



(Continued from page 382.) VII. Bringing several Fractions to the same Denomination. Adding and

subtracting them. M. You know already that fractions, which are to be added together or subtracted from one another, must have the same denominator. In the preceding examples only one fraction was to be changed, and brought under the denominator of the other. But in most cases both or all the fractions must be changed ; for instance, if we have +4. Which are the new denominators you can give to the fraction by multiplying the denominator 5.-A. Instead of 5 I may get the denominators 10, 15, 20, 25, 30, 35, &c.

M. Which are the denominators to which you can bring the second fraction 44-4. 14, 21, 28, 35, 42, &c.

M. Write down the two lists of denominators, and see which is common to both A. I find 35 in both lists.

M. Therefore as well as may be changed into 35ths. By which number must you multiply the denominator 5 in order to obtain 35 –

A. By 7. .: M. 7, which is the denominator of the second fraction. And by which number must you multiply 7 to obtain the same denominator 35 ? -A. By 5.

M. 5, which is the denominator of the first fraction. Now change both fractions and bring them to 35ths, but do not forget to multiply the numerator by the same number by which you must multiply the denominator.-A. = 1, {=}.

M. Now add them together.-A. +}=1:+=

M. Here I show you once more on lines and squares that you are right.



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

are to be repeated as in the preceding instance, and, after two or more such additions have been performed, the pupil will easily arrive at the practical rule, that—To find a common denominator for two fractions, we multiply the two given denominators together; and, secondly, in order to reduce the two fractions to this common denominator, we multiply both terms of each fraction by the denominator of the other fraction. Frequent interpellations and illustrations must satisfy the master that his pupil is always aware why he is operating in such a manner, whilst a great number of examples arranged according to their gradual difficulty give to the pupil the necessary skill and practice in solving these problems.]

M. Bring to the same denominator ļ and }, and }, 15 and is Add together } +, +á šti2}+3%, 88. 83d.+158. 95 d. Subtract 1- 5-11, 275-, 481–199, 2341:-1726, 921–4875, &c.

M. I want you to add together +*+. We have until now only operated with two fractions. What must we do first to add these fractions together?-A. We must give them a common denominator.

M. Bring the first two fractions under the same denominator.-A.

M. Now we have to add 12ths and 5ths, namely, 1 +8+ Bring them to one denominator.-A. 12ths and 5ths may be reduced to 60ths.

M. How did you find the denominator 121-A. By multiplying the two denominators 3 and 4 together.

M. How did you afterwards find the denominator 60 k-A. By multiplying 12 by 5.

M. Instead of multiplying 3 first by 4, and the product by 5, you might at once have multiplied 3 by 4 x 5 or 20, which gives 60; instead of multiplying 4 first by 3, and the product by 5, you might at once have multiplied 4 by 3 x 5 or 15, which equally gives 60. You see at once that the common denominator is found by multiplying one of the denominators by the product of the two remaining ones, or, what is the same, by multiplying the three denominators together. To give the altered fraction the same value as the original one had, we must, of course, multiply each numerator by the same number by which we multiply its denominator, that is to say-by the product of the remaining denominators. Thus, in the above example each term of the first fraction is so multiplied by 4x5 or 20, each term of the second by 3 x 5 or 15, each term of the third by 3 x 4 or 12, which gives the result, f="0, 1=16, =s, and the sum +1+4=16 =17* Now let us add in the same way +5+, then 21+34 +25. The same rule will be good for the addition of more than 3 fractions; the demonstration is the same as above. If there are many fractions, or their terms too great to be easily remembered, we write the fractions under one another, and to the right repeat each numerator with the number by which it is to be multiplied, and the new numerators thus obtained upon which we operate as in the former instances.

* For shortness' sake we give the latter part of this demonstration in continuous form, leaving it, as in many other cases, to the master by suitable questions to elicit the same from his pupil.

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

VIII. To find the least common Denominator. M. You have seen now that an addition of several fractions which have not the same denominator is a rather long operation, and its complication chiefly arises from the large numbers we have to deal with. It is, therefore, important to see whether we may not in many cases abbreviate our operations, or, in other words, find a smaller common denominator than by multiplying all original ones together. Such cases really occur pretty often ; for instance, if we had to add į+ + + + {+11, instead of using the product of all these denominators, 2 x3x 4x6x8 x 12=13,824, as the new common one, the comparatively small number 24 would do for the same purpose, as all the given fractions may be reduced to 24ths. The question is, therefore, how to find the least common denominator. Taking once more the fractions just mentioned, why do you say that 24 may be used as a common denominator for all 1-A. Because I am able to reduce all to 24ths.

M. How do you do that ļ—A. By multiplying the terms of each fraction ; those of į by 12, those of į by 8, those of by 6, &c.

M. How do you find each of these numbers ?--A. By dividing 24 by the respective denominator.

M. Or you may say by seeing how many times each respective denominator is contained in 24. If I introduced the fraction into the above series of fractions, would the number 24 then still do for the common denominator -A. No, because I cannot change 5ths into 24ths.

M. Or because 5 are not exactly contained in 24. What is, therefore, requisite for a number to be a common denominator to several fractions ?-A. The common denominator must exactly contain each special denominator.

M. Or we may say: The common denominator must be a multiple of each original denominator. If we, therefore, have several fractions to bring to the same denominator, we only consider their denominators, and try to find the smallest possible number which contains each of them or is a multiple of each. Before we go any further, let me tell you, that in multiplying several numbers together, the order in which we multiply does not alter the product; thus, 3 x 7 is the same as

7*3 : which you can easily prove by this figure :


If we count the points in each horizontal line we find 5, and as there are 3 lines, it makes 3 x 5 points; but if we count the points in each vertical line we find 3, and as there are 5 vertical lines, it makes 5 x 3. If we had to multiply 3 x 5 x4, we might either first take the 5 points in each horizontal line 4 times, and repeat the product 3 times, thus

..... or first multiply the 3 points in each vertical line by 4, and repeat the product 4 times, thus

The result would be always the same, the number of points being 60. The 4 numbers, 3, 4, 7, 10, we may multiply in 24 different orders, and always obtain the product 840.

M. Let us now bring the following fractions under the least common denominator : ,, Jo, to, 1h, and 1. Repeat what we said before about the requisite of a common denominator.-A. It must contain or be a multiple of each original denominator.

M. What are the original denominators ? Write them in a line. A. 4, 6, 8, 10, 12, 18, 20.

M. Reduce each of them to its lowest factors or numbers by the multiplication of which you obtain it.-A. 4=2 x2, 6=2x3, 8=2x2x2, 10=2 x 5, 12=2 x 2 x 3, 18=2 x3 x3, 20=2 * 2 x 5.

M. The numbers or factors thus obtained are 2, 2, 2, 3, 2, 2, 2, 2, 5, 2, 2, 3, 2, 3, 3, 2, 2, 5.

M. Which of the original denominators contained the factor 2 most ? -A. 8 contained it 3 times, for 8=2 x 2 x 2.

M. Therefore let 2 stand 3 times in the above series, but strike out all remaining 2's. Which original denominator contained the factor 3 the most ?-A. 18, for 18=2x3 x 3.

M. Let the number 3 stand twice in the series, and strike out the remaining ones. The factor 5 is only once required in 10 as well as in 20; we obliterate the last number 5 in our line, which will now look like this :

2, 2, 2, 3, 2, 2, 2, 2, 5, 2, 2, 3, 8, 8, 8, 8, 2, , or 2, 2, 2, 3, 3, 5. M. Multiply the remaining numbers. I say the product must contain all the original denominators or be a multiple of each. As for the first, or 4, this is evident; for we began multiplying 2x2=4, and then multiplied these 4 by several following numbers; of course the product must be a multiple of 4. But as we have seen that the order in which we multiply does not affect the product, we may as well begin multiplying 2 x 3, which gives 6, the second denominator, or 2x2x2=8,

« AnteriorContinuar »