## A Treatise on Elementary and Higher Algebra |

### From inside the book

Results 6-10 of 55

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**Conse**- quently , by adding the preceding results , we get m2 + ( m - 1 ) + ( m - 2 ) 2 + ( m − 3 ) 2 + + 12 = m + ( m — 1 ) + ( m − 2 ) + ( m − 3 ) + + 1 + m ( m − 1 ) + ( m − 1 ) + 2 x 1 . 1x - - ( m − 2 ) + ( m → 2 ) ( m − 3 ) ... Page 130

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**conse**- quently a'b must be divisible by p . Hence , in ab we may suppose a to be less than p ; and since p is a prime number , it will not be divisible by a without a remainder ; hence , if we put m1 for the quotient of the division of ... Page 211

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**conse**- quently , since B and A are given , C is found ; and the terms A and C of the single ratio Again , if n = 3 , we have reduced . By putting are known . A A A A C (金) C B we get C = Bx A ' A B = = X X ( 1 ) , to be B B B ' B C D ... Page 214

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**conse**- quently , since A " × D " = B " × C " , we have D " = B " × C " n Bn x Cn BX C A " , or DW = a result obtained ac- , A An cording to the common method of proceeding . It may be added that if A , B , C , D are not numbers ( nor ... Page 221

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**conse**- quently , when a is unlimitedly great , it is clear that 1 1 = 1 is the limit of the proposed ratio . And it is clear that this ratio can be immediately obtained from the proposed ratio by omitting a and b , on account of their ...### Contents

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### Other editions - View all

### Common terms and phrases

algebraic arithmetical progression ascending powers becomes binomial called changing the sign clear clearly coefficients common logarithm compound quantity conse consequently corresponding cube root cx² decimal places denote derived function difference divide dividend division equa equal ratios equal roots evident EXAMPLES exponent expressed extract the square factors find the number follows fraction given equation gives greater greatest common divisor Hence imaginary roots inequality integer least common multiple less logarithm monomial multiplicand multiplier negative roots nth root number of terms numbers or quantities odd number polynomial positive integer positive roots proportion proposed equation quadratic quently quotient real roots remainder remaining roots represent result right member rule second term solution square root subtract supposed surds thence third term tion unknown letter whole number

### Popular passages

Page 374 - If A and B together can perform a piece of work in 8 days, A and C together in 9 days, and B and C in 10 days : how many days would it take each person to perform the same work alone ? Ans.

Page 307 - Multiply the complete divisor by the second term of the root, and subtract the product from the remainder.

Page 152 - A man and his wife usually drank out a cask of beer in 12 days ; but when the man was from, home, it lasted the woman 30 days ; how many days would the man be in drinking it alone ? Ans.

Page 141 - II. Divide the greater number by the less, writing the quotient between the verticals, the product under the dividend, and the remainder below. III. Divide the less number by the remainder, the last divisor by the last remainder, and so on, till nothing remains. The last divisor will be the greatest common divisor sought.

Page 347 - Three lines are in harmonical proportion, when the first is to the third, as the difference between the first and second, is to the difference between the second and third ; and the second is called a harmonic mean between the first and third. The expression 'harmonical proportion...

Page 161 - ... multiply each numerator by all the denominators, except its own, for a new numerator, and under it write the common denominator.

Page 215 - Or, four terms are in harmonical proportion, when the first is to the fourth as the difference of the first and second is to the difference of the third and fourth.

Page 375 - A, B, C, D, E play together on this condition, that he who loses shall give to all the rest as much as they already have. First A loses, then B, then C, then D, and at last also E. All lose in turn, and yet at the end of the 5th game they all have the same sum, viz. each $32. How much had each when they began to play ? Ans.

Page 209 - The first term, the last term (or the extremes) and the ratio given, to find the sum of the series. RULE. Multiply the last term by the ratio, and from the product subtract the first term ; then divide the remainder by the ratio, less by 1, and the quotient will be the sum of all the terms.

Page 173 - ... proportion, the sum of the extremes is equal to the sum of the means. Thus...