Solutions of the examples in Higher algebra by Hall, H. S., , Macmillan and Co. edition, in English. SOLUTIONS OF THE EXAMPLES I2T HIGHEE ALGEBEA H. S. HALL, M.A., S. R KNIGHT, B.A., FOBMEELT SCHOLAR OP TRINITY COLLEGE, CAMBRIDGE, This work forms a Key or Companion to the Higher Algebra, and contains full. So here I am to share these useful solutions with you guys. Higher Algebra Hall and Knight. To make it easier i have included the eBook .pdf.
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Higher algebra by hall and night isn't the right book you are using. First of all its outdated, secondly the topics are to short and too shortly explained and proof are . Solutions for Hall & Knight - Ebook download as PDF File .pdf), Text File .txt) This work forms a Key or Companion full solutions Higher Algebra. June. HALL. an d fo r man y valuab le su ggestio n in every part o f it. HALL.,. May,. H S. S. R. KN IGH T. Co mpariso n between algebraical an d geo metrical defin.
Condition: New. Language: English. Brand new Book. LaTeX Edition This is an economical edition typeset in 8. For This work forms a Key or Companion to the Higher Algebra, and contains full solutions of nearly all the Examples.
From value of Since and have no common factor. With the notation of the preceding example. Thus y of the number in the denary scale. By ascribing to a the we get the corresponding values of 6. Hence the distance between these divisions is. As in the preceding example. Thus the greatest value of a is Now must be since the bells cease in less than 20 minutes. Hence 29? Since zero solutions are inadmissible. Let o. The greatest value of a is 8. Since there are to be 6 solutions.
Siz times. This Example includes Examples V T. Jgl' '. The positive root of a. The positive root of 7a. JaV -an. From From From 1 3. Multiplying up. From Ex. Denote the continued fractions by! In Art. Wehave Subtracting and rearranging. Denoting the value of the second continued fraction by changing a and 6. Solving for x. Page 1. J- 4 here the penultimate convergent solution. Expressing y in terms of x. The factors of are Hence 6. The factors of are 1. Thus the men bought Denote the lengths of the two sides respectively.
Hendriek bought 23 more than Catriin. Let X denote the number of hogs bought by any one of the men. Hence Cornelius bought 8. In this case the values of are the numerators of the even convergents.
Ir-l n-l n — We have Pages The successive orders of differences are i. We have 8. Thus the scale of relation is 1. We By the method of Art. We have 2. And as in Art. We have 4. The series is Consider the successive orders of differences of 3. The term of the series 2.
Divide by 3"-'.. The m"' term of the series 1. Pages — By writing down the series for e''.. The general term of 2. Equating coefficients of x'.
Assume l [CHAP.. Hence Here -jj- is 1. As in Example 3. This last expression consists of two parts. I'rom Fermat's Theorem.. Let r. It N is not prime to 3.
Every number x is one of the forms 3g. Again if n is odd. If If is a is odd. Let m As in the first part of the question the expression is divisible by 5. Thus the form 3n.
N is prime to K -W is not prime to Since a and 6 are both less than m. Since x is odd. By Fermat's Theorem n'' - 1 is divisible by 3. Here both a and b are prime to m. By Wilson's Theorem. Thus 8. Let then. Similarly it is divisible by a a".. Thus the given is divisible expression 7. Let the coefficients be denoted by c Cj.. With the exception of the first term all the terms of the series are prime to p.
Again n'. Thus the given expression is divisible by 8 x 16 or Hence the general solution of the equation a. The highest power required ji'-l equal to the sum of the integral parts of that is. I is an integer. If ' Hence is. Now by b '"' g-l. X stands for any one of these integers.. Now by Art. N" is zero. N-l by a multiple of 2ir.. N-l denote the Then one of these numbers..
N-l x must differ from necessarily in this order N- prime to N. Denote the continued fractions hy a. Proceeding as in Example 2.
Sn 2 Here -g' -g. X denotes the value of the given expression.. This may be proved db initio. Un H and so on as in Art. In the equation of Ex. A cards may queen and a knave can be drawn together in 16 ways. XXXIl] X and x be the respective probabilities of the o 2 and second event. Two dice may be thrown in 36 ways.
With two the first dice 8 can be thrown as follows 6. The table is Call the A and B. Jlo which are as 36 Thus the odds against G are 43 to Thus the odds are 5 to 1 against the event. The total.
The chance of A 2. Thus these chances The 8 volumes can be placed on the 1 the same works will be altogether in 3 x 13 x 4 shelf in [8 ways. With two dice. That is out of the 36 ways in which he can throw the two throw more than 9 in 6 ways. He can throw 12 in way.
There are 7 letters which can be placed altogether in 7 ways. As the two vowels are not to be separated we may consider them as a single letter. B's chance of failure 5 —. The chance that both will not fail is 1. The chance that the first fails is -. A's chance of failure is 4 -. And there are only 5 different ways of placing the coins so tnat the extreme places are occupied by half crowns. The number of favourable ways is the same as the number of wavs in which the 9 other cards forming the hand can be chosen.
The chance of an even number the is first time is Q —. J In order that a majority may be favourable. Therefore the chance of a sovereign 1 from the first 12 compartment s x? Therefore the required chance 8x9 17x17 7. Therefore none of the four numbers must end in 0.
And the is 2ii' If the last digit in the product is not 1. The total there are 8 ways in number of ways in which the coins may which one head can appear.. The sovereign can only be in the second purse events have happened if both the following of the first purse 1 the sovereign was among the 9 coins taken out and put into the second 2 first.
The chance that they black are alternately of different colours beginning with is. The chance [CHAP. Then besides. Four at 4 ' safe. A and B fail. B's chance in any round If a.
The chance that B rides 2 A—o. Call the specified persons A and B. Therefore the odds against A and B sitting together are m. A and B 3 fail and then 2 1 B 22 succeeds. The respective chances of these 5 events are!
Let X be the value in shillings of the unknown coins 5 is. Or more simply thus: Thus 5. Two shillings can be drawn in 3 ways.. One tion on this sovereign and one shilling can be drawn in 6 ways.
The three dice can be thrown in ways. In order to win the set. Therefore by Art. B's chance. B may throw anything from 3 to 8 inclusive. The number ways in which the 3 dice may fall is 6'. If the 3. In order to lose.
And in this last combination of numbers the 5 may occur in any one of the four throws. Thus the whole expectation Or more simply thus: The chance of throwing 7 with two dice is.
With the notation of Art. Now we require B's expectation in the long run. The two dice of may be thrown in 4 x 6 or 24 ways. Let the purse contain n coins in all. Now the chance of a second draw n-1 is n.. The numbers the coefficients ways in which 2.
If 2 is. Then suppose 1 is drawn. Hence the number of ways of making up 6n is the sum of 2n terms. A could lose in two ways. London came from London. There are two cases: There are n equally likely hypotheses.
Thus 12 p. The antecedent chance that the lost card is a spade it is is -. There are three equally likely hypotheses tain 2.
At first. B'b chance of winning his race is -. B'a chance certainty: Now in order that the odds in favour of the event five to one. Let p be the a priori probability of the event. Then the o priori chance is placed. IS the total. The number of ways in which nine things can be arranged. If the first two the third person draws from m -2 tickets.
The chance that the first bag is chosen is. See XXXn. By writing down the different combinations it is easy to see that 1. Again the chance that the is second bag is chosen is. The person's chance is -. Example 4. J Thus their respective chanees are as 7.
Therefore on the whole the average value is 6. When A tries with B. The two dice which 2. Thus the chances of the of throwing 7 are equal.
O when A with D. E are identical. If he fails and all the other 3 persons fail he getp a second throw. Thus by There are two hypotheses. D's E's F'a Now A. In order to win. In order that the i"" [CHAP. Consider six players. B being at the end of 2 sr Similarly the chance that is A and B are adjacent. Now the number of ways in which 7 digits may be arranged so as to make 59 is equal to the coefficient of a. The total number of pairs of positions they can occupy without restriction.
A being at the end of the carriage. But the suin of three digits cannot be equal to In order that a number may be divisible by This coefficient is Wherever A may be seated there is only 1. Each way of arranging the odd digits may be associated with each way of is and arranging the even digits. Now 5" is the number of ways in which an odd digit can be chosen to end the number. There are now two favourable cases.. The the sum. Now 8" is the number of and 5 and of these we have further to ways in which we can exclude exclude the 4" cases in which the last digit can be selected solely from 1.
Putting this in the form x3 l-3a. The number 1 [CHAP. The Example And of these three chances is 1. So that is if x be A's chance of winning. Or we may proceed as follows. This is o. All these cases are equally likely. If it is drawn second. B'a chance the sum event happens.
The chance If it is 6d. Example A throws and the chance that is JS has a throw is o. Then is necessary that one of the random points falls in each division the. Similarly and in each case the remaining thing may be given in 2 ways. Ire chance of this II. And the total number of ways. Then if the first triangle is acute angled. Let A. Hence the required result follows as in [This problem and solution are due to the Eev.
C he the three points. J3j 3. Bj has 2 sovereigns. Draw tangents to the circle at the three random points. Bj has 3 sovereigns. AB triangle DEF. Let AB P. OD be each equal to a. Therefore if we take three rectangular axes OA. OG each equal to a. OD in E. Q be at distances x. Tate a pair of rectangular axes If the 3 parts of the line are x. P2 be the a priori probabilities of drawing i sovereigns from the 1" and 2'"' purses respectively.
F respectively. Let Pi. Therefore x. OD each equal to a. Now the favourable cases are restricted to points within the triangles CEF. This consists of a right-angled isosceles triangle each - a-2b 3b.
Take a. If RS is to lie within PQ satisfied simultaneously. Draw figure. Also let. Similarly for companionship in any the same third class compartment. Draw Then LF'T in the figure. And the favourable eases are re- stricted to points in the triangles GFM.
Now that A and C's ehanee ie the same for every compartment. Pages to Subtracting 1. Add together the second and third row and subtract the sum from the first row. Adding together all Erom Art. Hence k must be equal minant. Assume then that the given determinant and is equal to the product of To find the unknown s quantities we have as in Ex.
But by Ex. Alternative Solution. The preceding equations may be easily solYed by the ordinary thus taking the equations in x. The solution of these and similar equations may however be sometimes more easily obtained by trial from a consideration of the fact that p. The values of x and y are easily found by substitution.
Changing the columns into rows. Multiply the second row by. Thus the second determinant -2x y'. Multiply the columns by 1. Here however we shall give another method. Hence the determinant vanishes in all these cases. The solution is similar to that of the next example. We may express the value of any one of the unlcnown quantities as the quotient of one determinant by another.
Hence the determinant is the product of these six factors and constant. Multiplying the columns by 1. It ia easy to see that k G have the values given in the question. As iu Art. This expression is equal to If andif whence By putting See Solution to Ex. Moreover it is symmetrical and of four dimensions in a. On multiplication. From a formula given in Art. It is easy to prove the converse result by resolving 7 -. Let a: The common denominator is 6.
Sa c2 x. The converse of this result is easily proved by Partial From the given condition. From 16—2. The numerator 6 is. S6cd 6. Further the given expression is of seven dimensions. By writing -i for 6. Thus the expression is divisible by. This theorem is 6. Wehave [CHAP. Put shew that 6 also be deduced from Art. Proceeding as in Ex.
The theorem may Hence 82 s. By proceeding SflS as in Art. With the notation of the preceding Example. But and similarly To find k. Since all the signs are positive. Paces Squaring and adding the a? Thus From and second equations. By multiplication.. Multiplying the Similarly first of these equations by y.
But from the given equations. From hence the third equation. Multiply a: From these last two equations. It remains to eliminate x between this equation and ax'. Corresponding to the first pair of roots there will be a quadratic 4. Let the roots be a. Let a.. From the 1 first two. Let 3a. Let 2a. Let -. Let a-d. Let a-3d. Multiply the equation through by a. Examples 1 12 do not require full Corresponding to 5. Hence it has at least four imaginary roots since it is of the positive roots.
Hence there must be at least two imaginary roots. Corresponding to these we have the quadratic factors a. Therefore it must have at least six imaginary roots. Expand and by y. In the second case assume for the roots. Let a-dd. The H. Therefore the roots are a. Here and f't. Then [CHAP.
Therefore roots. Hence and 4i? On substituting for x we get the required relation. By inspection a. Thus the mean root Example 8 may be 9. Divide all through by a? Assume a-Sd.. Therefore using a. Here we have to increase each root divisor in Horner's process. Here as in Ex. TP If a. Denote the roots by a. The sum of the roots The equation may now be written a.
Thus the values of x are. Also other roots are. Thus the real root is or Pages As usual. Thus the 6. Thus the roota are 2. Multiply the given equation by 8a. Thus xx. Examples 9 17 may be solved by the methods given in Arts.
By inspection factor x The required condition may be obtained by ehminating x between the two equations. From thus these equations.
The first derived equation is 4a.
By the conditions of the question. Denote the product of the two roots by y.. Denote the roots by a and a. We easily obtain?
Solving these as equations in xy. Hence 3a.
If r be the radix. Denoting If 2. Thus A takes A and B run iu a To eliminate i. B seconds. Without altering the value of the whole expression.
Let X and y denote the nnmber of yards that Hence the term required last binomial. Suppose that the waterman can row x miles per hour in still water. From the given equations. When the expression a perfect square. Let r denote the radix of the scale then. From tliis the required result at once follows. Denote his weekly wages by x pence. By transposing and squaring. IS cla-b.. Kepeating the process. Denote the numbers hj a-3d. Also the whole number of permutations without restriction is We should thus obtain 2 5 permutations j the Jive papers.
Let I. C together do the work in x hours. B alone in x -. Hence working together they can do. From o the first two equations by cross multiplication. Suppose that A. Suppose the two mathematical papers A and B were fastened together and considered as one. Eliminating y. Hence numerator and denominator must be divisible by x X.
Subtracting numerator from denominator. By hypothesis. Denote the roots by o and a? Therefore the Qy 9a. Suppose that at first x persons voted for the motion. Let X denote the number of men in the side of the hollow square.
Suppose that y persons changed their minds. Hence 12a. The first place can be filled in 17 ways. Square both sides. After the first operation the first vessel contains wine.
From the data. By then addition. We have -! By [PAGE We have thatis.. By simplifying each side separately. X is if the is common the difference of the A. Let d te the [PAGE. From the second equation. Suppose that n is [PAGe' the number of hours.
The sum of the series 1. It will be observed that the last terms of the first. Putting a. If n-i 2. In tte scale of 7 let the digits beginning from the be x. But a. Sz is an integer o. Since the logarithms of all is negative. Thus a. Then when J is at i or at any previous instant B is xy miles behind A.. B Now therefore the rate of approach of B and the geese is x-- 3 miles per hour.
For the G. B 50 -2a. A are [PAGE waggon 31 meets the waggon. For the A. On is. Wehave 15a. S 1-px. Assume Let then 1-px. Denote the numbers by 2m From hence the given equation. Denote the values of the continued fractions [PAGE by x and y. From these two equations. Let n be the number of persons. This readily follows from the inequalities.
B the number of boarders. The first result follows at once by putting of the given relations. V'' - Buta. Prom the Beoond and third equations. X is real. Suppose that Oj. If tte scale of relation is 1 -px. From the first [PAGE two equations. By If subtraction. Then remembering that a. Multiply the first equation by - the second by 6. The least common multiple is ipq. Let X and y denote the two numbers.
I Let g denote their greatest multiple. The numbers may therefore be denoted by 4p and 4j where p and q have no common factor.
The expression is of four dimensions. Hence the coefficient of a. After multiplying up and transposing. Suppose the number of pounds received for the first lot is expressed by the digits x. The number of pounds digits y. Substituting for x from the first [PAGE we have equation. The first derived function of 12a. Thus Q the roots are. Hence b cannot be equal to is d. The title should be at least 4 characters long. Your display name should be at least 2 characters long.
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