A FEW HINTS TO TEACHERS. In few words, I wish to call the attention of the readers of the "Journal" to the importance of an improvement in the mode of teaching the useful and beautiful art of writing, as generally practiced in our Common Schools, especially in country districts. I find in many schools that it is a custom, which prevails to a great extent, to allow the pupils to write when they please, where they please, and as much as they please; the Teachers finding, as they say, no time to set apart to devote exclusively to the instruction of a writing class, and in so doing, virtually practicing upon that threadbare error, that "writing is of secondary importance." Many seem to think, and in fact say, that pupils will learn to write without any particular instruction from the Teacher, merely by having copies to imitate. Now, let me ask, is it not necessary that learners be instructed to sit in certain positions, in order that they may write with ease? And how very important it is that children be shown how to hold the pen, even in their earliest efforts to learn to write. Is it not evident, that the muscles of the hand and arm should be so trained, that the writer gain complete command over the motions of the same? If so, in what other way can the Teacher of fifty or sixty scholars so effectually accomplish these ends, as to improve a certain portion of time each day in giving general and individual instruction to a class in penmanship, excluding all other exercises during the time? While giving instruction daily to writing classes, the Teacher should never allow his pupils to grasp the pen tightly; for if they do so they soon exhaust the muscular power of the hand and arm, or at least greatly weaken it. The consequence is unsteadiness of the hand, and an entire unfitness to execute with neatness or ease the daily task in penmanship. Again, a decided improvement might be effected in teaching the art of writing, by Boards of Education recommending and adopting some one system of writing, to the exclusion of all others. Writing books containing printed copies should be used in every school; and every scholar, after being thoroughly drilled in the use of the pen, whether he has taken lessons in writing or not, in the old "hap-hazard" way, should be instructed in the first lessons of that system; and let him not leave the first book of the series until he has gained a practical knowledge of all it contains, and can write every mark and letter with facility and correctness. Then let each book be taken up in order and The Teacher should always mastered before the next is called into use. insist on a thorough knowledge of the whole course, if at all practicable. A series of writing books should contain no less than twelve books or grades of lessons. The importance of this mode of instructing in penmanship is evident. Almost every one who has ever been connected with schools, in any way, knows how great the evils are arising from frequently changing Teachers, each of whom has a new system of writing, or, more likely, writes without system. Hence the necessity of school boards taking this matter into consideration, and as speedily and certainly as possible, correcting the evil by adopting a good system of writing, and adhering to it, and not changing upon the suggestion of every third class pedagogue. Cincinnati, Sept. 10, 1857. JOHN R. STARKEY. Mathematical Department. PROF. W. H. YOUNG, ATHENS, EDITOR. [All communications for this Department should be addressed to the Editor, Ohio University, Athens, O.; and to be in time, must be mailed by the first of the month preceding that in which they are expected to appear.] SOLUTIONS OF QUESTIONS PUBLISHED IN AUGUST. No. 13. [The solutions furnished to this problem are so various and the results so different, while no one is satisfactory, that we would advise correspondents to try again. The nature of the problem seems to require a somewhat tedious approximation; yet we think a shorter method will meet the case, even without a resort to the Calculus. The "Salineville" correspondents assume a value for the arc in terms of the sine, versine, and chord. Will they furnish the authority, or the reasoning?] No. 14. If a solid globe of glass be blown into a hollow sphere sphere one-eighth of an inch in thickness, what will be the diameter of the sphere? SOLUTION BY A. SCHUYLER. Let x= the diameter of the hollow sphere. Then x 1 the diameter of the hollow space within the shell. D3. Therefore x3 ¦ π (x − 1)3 = 1728. Developing and π 10 reducing, and we have x2-x inches. Ans. 6 ACKNOWLEDGMENTS.-Both questions were solved by A. Schuyler, James McClung, Jos. Turnbull; No. 13, by J. L. C., and S. Harvey; No. 14, by Omega, and Jas. Rutherford. J. C. Anderson solved the questions published in July. Prof. McFarland suggests that the answer given, in the July number, to No. 6, is the horizontal “skip;" while the question calls for the distance skipped, measured on BC. Mr. Schuyler calls our attention to No. 8. We only give the rule, and result; not claiming a demonstration, which, to be complete, would be lengthy, and, we think, uninteresting to the majority. EXPLANATION.-Despite our best endeavors to secure entire accuracy in representing the correspondents of the Mathematical Department, the September issue is full of mistakes. As we had promised, on a former occasion, the like should not occur again, it is due to ourself to explain. The proof, on being mailed to us for correction, was not prepaid, and, of course, never left the post-office at Columbus. When it did not come to hand at the proper time, we concluded, relying on our former very particular instructions to that effect, that the "Mathematics" would be omitted. The Journal, however, soon arrived, mathematics, mistakes and all. We plead a loud Not guilty. solution of No. 10, by Jas. McClung, especially suffered. We regret this very much, for we regard the solution a most beautiful application of the Mathematical Zero. An acquaintance with the Binomial Theorem will be sufficient to detect the errors, which are many. In the 2d line of solution No. 12, read "middle point of the line AB.” J. B. Dunn's article on 66 Contractions," should have been set up as a communication. It contains several slight and readily detected typographical errors. QUESTIONS FOR SOLUTION. The No. 18. Arithmetical Question.-Divide $400 between A., B. and C., so that A. shall have one-fourth of the whole more than B., and B. one-third of the remainder more than C. J. L. C. No. 19. Having given the perpendicular (a) between two parallel OHIO JOURNAL OF EDUCATION. chords, which include one-third the area of the circle, it is required to find the radius of that circle. JAS. MCCLUNG. No. 20. Undecimation, or Proof of Multiplication and Divis ion.-Take any two numbers and multiply as usual. Then prove as follows: Subtract the left hand figure of the multiplicand from the one next to it on the right, borrowing 11 if necessary; and take this remainder from the next on the right, this remainder from the next, and so on, borrowing 11 whenever the figure to be subtracted is too large. Set down or retain in the mind the final remainder. Proceed in the same manner with the multiplier, and multiply the final remainder by that already obtained from the multiplicand. If this product contain two figures, undecimate as before, and take the remainder, thus obtained, from the left hand figure of the product, this remainder from the next, going from right to left. If the final remainder be zero, the product may be regarded correct. Examples: (Ex. 1.) 926 180 24 2 4 (Ex. 2.) 327 8 46434 X 166680 Explanation.-1st. Begin with multiplicand: 9 from 13 (borrowing 11)=4, 4 from 6 = 2, which is set down on the right. Again, 1 from 8 = 7, 7 from 11 (borrowing) = 4; set down as before. 4× 2=8. Now begin with the product on the right: 8 from 11 (borrowing) 1, 1 from 65, 5 from 6 = 1, and 1 from 1=0. Where the same figure is repeated, both may be omitted, as may be seen above with the 6s. 4 = 3, 3 from 8 2d. 3 from 13 (borrowing) = 10, 10 from 18 = 8. Again, 1 from 3, 3 from 13 = 10, 8×10 = = 80. Here we have, undecimating, 8 from 11=3; and then beginning on the right of the product, we have = 1, 1 from 3: 2, 2 from 4 = 3 from 4 4 = 0. 2, 2 from 6=4, and 4 from Required a demonstration of the rule. This method is equally applicable to division, where we consider the Divisor and Quotient as the factors, and the Dividend as the product. A remainder, if there be one, must be subtracted from the dividend before undecimating. In tracing out this property of the number 11, we have found much that is valuable, curious, and puzzling-some of which we shall notice hereafter. IS THERE ANY SUCH POWER OF A QUANTITY AS THE ZERO POWER? THEOREM FIFTH-Any quantity whose exponent is zero, is equal to unity. Vide Ray's Algebra, Part 2d, page 44th. The principle enunciated in the theorem is one universally admitted by the authors of Algebras at the present time, or at least it is incorporated into their Algebras. Though it may seem like presumption for me to differ with such mighty authority, I wish to present, through your valuable paper, a few objections that have arisen in my mind to this principle, and which I wish some one to remove, for they must be removed before I can assent to it. 1st. It conflicts with the definition of an exponent, viz: An exponent is a number showing how many times a quantity has been used as a factor, or multiplied into itself, to produce a given quantity. Hence, in the zero power, we have the number used no times as a factor, or we have no factor, and consequently no number. 2d. It conflicts with the principle, that there can be no abstract negative quantity unconnected with any other quantity. For the same process of reasoning that gives us a2, gives us also a -1, a-2, etc., as powers of a, in which the exponents -1, -2, etc., are abstract negative quantities, standing unconnected with any other quantity. We would like also to know the difference between a2 and a -2, if the number is used as a factor as many times as there are units in the exponent. True equations sometimes give negative results; but the result shows an absurdity in the statement of the problem, or a mistake in the solution, and I cannot see why it does not in this case. 3d. For every other power of quantities, we have a corresponding root. Thus, for the second power or square, we have the square root; for the cube, the cube root, etc.; but what would the zero root of a quantity be? 4th. By this theorem, I can prove that any quantity, no matter how small, is equal to any other quantity, no matter how large, and that all quantities are equal to each other. Let us take two identical equations: (1000) (1000)2, and 12 - 1o. = = Dividing both sides of each equation by the numbers squared in each case, and we have first, (1000)1=1000, and 11= 1; then (1000)° = 1, and 1o 1, according to the theorem; hence (1000)o 1o. Now if the axiom, that like powers of equal quantities are equal, be true, then its converse, that if like powers of two quantities are equal, |