242 



SCIENCE- G OSSIP. 



MATHEMATICS 



CONDUCTED BY PROFESSOR G. H. BRYAN, SC.D., F.R.S. 



Mathematics. — Dr. G. H. Bryan, F.E.S., having 



kindly undertaken to contribute occasional columns 

 devoted to popular mathematics, we have pleasure 

 in presenting the first of them. [Ed. Science- 

 Gossip.] 



Simple Kule for Squaring a Number. — 

 Most people know the ordinary rule for extracting 

 a square root, but few are aware that by a kind 

 of reverse process the square of any number can 

 be simply found. Thus, supposing we have to find 

 the square of 3,456, the work stands as follows : — 



3456 



x 3456 





3000 



x 3000 



= 9000000 



64 



x 4 



= 256 



685 



x 5 



= 3425 



6906 



x 6 



41436 



Answer = 11943936 



It will be noticed in each line of the first column 

 we double the last figure of the preceding line 

 and bring down one more figure. The numbers in 

 the first column are the divisors which would 

 occur in forming the square root of 11,943,936, 

 and they are written down exactly as in ordinary 

 evolution, when the square root comes out to 

 be 3,456. 



Tests of Divisibility. — The ordinary method 

 of " casting out the nines " and " casting out the 

 elevens" is well known. It is, however, less 

 generally recognised that a test of divisibility 

 by 7 and 13 may be found in a somewhat similar 

 form. We have 7 x 13 = 91 = 100 - 10 + 1, and 

 7 x 13 x 11 = 1,001. It follows that numbers greater 

 than 1,000 and less than 1,000,000 will be divisible 

 by 7 or 13 if the difference between the " thou- 

 sands " and the units be so divisible. Thus 

 235,683 is divisible by 7 or 13 if 683-235 = 448 is 

 so divisible. This is divisible by 7, therefore 

 235,683 is divisible by 7. If the number is greater 

 than 1,000,000, we point off the thousands in the 

 ordinary way and subtract the alternate groups 

 from the remaining ones. Thus if the number be 

 23,749,424,563,215 we have 



749 

 563 



1312 difference 650, 



which is divisible by 13 ; hence the original number 

 is divisible by 13. The above test of divisibility is 

 made to depend on the divisibility of a number of 

 three figures ; but a further simplification may be 

 made by adding the number in the hundreds place 

 to the tens place and subtracting it from the units 



place. Taking 351 we add the 3 to the 5 and 

 subtract 3 from the units place, giving 81 — 3 or 78, 

 which is divisible by 13, therefore 351 is divisible 

 by 13. 



Marks for Mistakes in Euclid. — A corre- 

 spondent of the " Mathematical Gazette " laments 

 the absence of general agreement as to the deduc- 

 tion of marks for mistakes in mathematical answers. 

 He suggests 'as an experiment that the readers of 

 the " Gazette " should send the editor postcards 

 stating the deductions they would make in proofs 

 of two propositions in Euclid (I. 37 and IV. 4) for 

 certain specified mistakes. The results, which are 

 promised for a subsequent issue, cannot fail to be 

 interesting. 



Proof of Pascal's Theorem. — This theorem 

 states that if an irregular hexagon is inscribed in 

 a circle, and pairs of opposite sides are produced 

 to meet, their points of intersection will lie in a 

 straight line. A simple " Euclidean " proof, involv- 

 ing nothing beyond the properties of cyclic quadri- 

 laterals, is given by Mr. K. F. Davis, M.A., in 

 the " Educational Times." In Mr. Davis's figure 

 ABCDEF is a cyclic hexagon, of which BA and 

 DE produced meet in G, and AF, CD in K. CB 

 produced meets KG produced in H, and the circum- 

 circle of DFK meets GK in P, the figure being- 

 drawn so that the points on GK occur in the order 

 H, G, P, K. Then, firstly, P D B G are cyclic for 

 DPG = supplement of DPK = supplement of 

 DFK = DFA = supplement of DBA (i.e. of 

 DBG). Secondly, P F B H are cyclic for FPH 

 = supplement of FPK = FDK = supplement of 

 FDC = FBC = supplement of FBH. From the 

 first result BDE or BDG = BPG or BPH, and 

 this by the second result equals BFH. But from 

 the original circle BDE and BFE are supple- 

 mentary. Therefore BFH and BFE are sup- 

 plementary. Therefore A, F, E are in one straight 

 line ; that is, EF passes through H, which proves 

 the theorem. 



The Parallelogram of Velocities. — There 

 are few subjects so imperfectly treated in most 

 text-books as the composition and resolution of 

 velocities. The ordinary statement of the Paral- 

 lelogram -of Velocities is a contradiction of all 

 common sense. " If a body is moving simultaneously 

 with two different velocities represented by two 

 sides of a parallelogram, it will have a single 

 velocity represented by the diagonal," or words to 

 that effect. The idea of a body moving with two- 

 different velocities at the same time is as absurd 

 as that of a person being in two places at the same 

 moment. The usual illustration, which does duty as; 

 a so-called proof of the law, is that of a ball moving 

 along a groove with one velocity while the groove is 

 moving with another velocity. But the ball itself 

 does not actually possess either of these velocities, for 

 the first is merely the ball's velocity relative to the 

 groove, and the second is the velocity of the groove. 

 What the parallelogram of velocities tells us is 

 that, if the velocity of A relative to B is repre- 

 sented by one side of a parallelogram, and the 

 velocity of B relative to a fixed base — say C — is 

 represented by the other side, then the velocity 

 of A relative to C is represented by the dia- 

 gonal. Some teachers think that the notion of 

 relative velocity is too difficult for a beginner 

 to understand, but that is no reason for substi- 

 tuting something which nobody can possibly 

 understand. 



