8. The square or cube root of an even number, is even; and the square or cube root of an odd number, is odd. 9. Every square number necessarily ends with one of these figures, 1, 4, 5, 6, 9; or with an even number of ciphers preceded by one of these figures. 10. No number is a square that ends in 2, 3, 7, or 8. 11. A cubic number may end in any of the natural numbers, 1, 2, 3, 4, 5, 6, 7, 8, 9, or 0. 12. All the powers of any number, ending in 5, will also end in 5; and if a number ends in 6, all its powers will end in 6. 13. Every square number is divisible by 3, and also by 4, or becomes so when diminished by unity. Thus, 4, 9, 16, 25, &c., are all divisible by 3, and by 4, or become so when diminished by 1. 14. Every square number is divisible also by 5, or becomes so when increased or diminished by unity. Thus, 36—1, and 49+1, are divisible by 5. 15. Any even square number is divisible by 4. 16. An odd square number, divided by 4, leaves a remainder of 1. 17. Every odd square number, decreased by unity, is divisible by 8. 18. Every number is either a square, or is divisible into two, or three, or four squares. Thus 30 is equal to 25+4+1; 33=16+16+1; 63=49+9+4+1. 19. The product of the sum and difference of two numbers, is equal to the difference of their squares. Thus, (5+3)×(5—3)==16; also 52—32—16. 20. If two numbers are such, that their squares, when added together, form a square, the product of these two numbers is divisible by 6. Thus, 3 and 4, the sum of whose squares, 9+16=25, is a square number, and their product 12, is divisible by 6. Hence, 21. To find two numbers, the sum of whose squares shall be a square number. Take any two numbers and multiply them together; the double of their product will be one of the numbers sought, and the difference of their squares will be the other. Thus, take any two numbers, as 2 and 3; the double of their product is 12, and the difference of their squares is 5; now 12252-169, the square of 13. 22. When two numbers are such, that the difference of their squares is a square number; the sum and difference of these numbers are themselves square numbers, or the double of square numbers. Thus, 8 and 10 give for the difference of their squares 36; and 18, the sum of these numbers, is the double of 9, which is a square number, and 2, their difference, is the double of 1, which is also a square number. 23. If two numbers, the difference of which is 2, be multiplied together, their product increased by unity, will be the square of the intermediate number. 24. The sum or difference of two numbers, will measure the difference of their squares. 25. The sum of two numbers, differing by unity, is equal to the difference of their squares. 26. The sum of two numbers will measure the sum of their cubes; and the difference of two numbers will measure the difference of their cubes. 27. If a square measures a square, or a cube a cube, the root will also measure the root. 28. If one number is prime to another, its square, cube, &c., will also be prime to it. 29. The difference between an integral cube and its root, is always divisible by 6. 30. If any series of numbers beginning from 1, be in continued geometrical proportion, the 3d, 5th, 7th, &c., will be squares; the 4th, 7th, 10th, &c., cubes; and the 7th will be both a square and a cube. Thus, in the series, 1, 2, 4, 8, 16, 32, 64, &c., the 3d, 5th, and 7th terms are squares; the 4th and 7th are cubes; and the 7th is both a square and a cube. EXTRACTION OF THE SQUARE ROOT. 571. To extract the SQUARE ROOT, is to resolve a given number into two equal factors; or, to find a number which being multiplied into itself, will produce the given number. (Art. 564. Obs.) Ex. 1. What is the square root of 36? Solution.-Resolving the given number into two equal factors, we have 36=6×6. Ans. The square root of 36 is 6. 2. What is the length of one side of a square field which contains 529 square rods? Operation. 529(23 4 43)129 129 Since we may not see what the root of 529 is at once, we will separate it into two periods by placing a point over the 9 and another over the 5. Now the greatest square of 5, the left hand period, is 4, the root of which is 2. Placing the 2 on the right of the number, we subtract its square from the period 5, and to the right of the remainder bring down the next period. We then double the 2, the part of the root already found, and, placing it on the left of the dividend for a partial divisor, we perceive it is contained in the dividend, omitting its right hand figure, 3 times. Placing the 3 on the right of the root, also on the right of the partial divisor, we multiply the divisor thus completed by 3, and subtract the product from the dividend. The answer is 23 rods. QUEST.-571. What is it to extract the square root of a number? Note.--Since the root is to contain 2 figures, the 2 stands in tens place, hence the first part of the root found is properly 20; which being doubled, gives 40 for the divisor. For convenience we omit the cipher on the right; and to con.pensate for this, we omit the right hand figure of the dividend. This is the same as dividing both the divisor and dividend by 10, and therefore does not alter the quotient. (Art. 146.) 572. Hence, we derive the following general RULE FOR EXTRACTING THE SQUARE ROOT. I. Separate the given number into periods of two figures cach, by placing a point over the units figures, then over every second figure towards the left in whole numbers, and over every second figure towards the right in decimals. 11. Find the greatest square number in the first or left hand period, and place its root on the right of the number for the first figure in the root. Subtract the square of this figure of the root from the period under consideration; and to the right of the remainder bring down the next period for a dividend. III. Double the root just found and place it on the left of the dividend for a partial divisor; find how many times it is contained in the dividend, omitting its right hand figure; place the quotient on the right of the root, also on the right of the partial divisor; multiply the divisor thus completed by the figure lust placed in the root; subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. IV. Double the root already found for a new partial divisor, divide, &c., as before, and thus continue the operation till the root of all the periods is extracted. If there is a remainder after all the periods are brought down, the operation may be continued by annexing periods of ciphers. PROOF.-Multiply the root into itself; and if the product is equal to the given number, the work is right. (Art. 564.) 573. Demonstration.—Take any number as that in the last example; then separating it into parts, 529 = 500+29. Now the greatest square in 500 is 400, the root of which is 20, with a remainder of 100; consequently, the first part of QUEST.-572. What is the first step in extracting the square root? The second? Third? Fourth? When there is a remainder, how proceed? How is the square root proved? the root must be 20, and the true remainder is 100+29, or 129. And since there are three figures in the given number, there must be two figures in the root; (Art. 562. Obs. 2;) but the square of the sum of two numbers, is equal to the square of the first part added to twice the product of the two parts and the square of the last part; it follows therefore that the remainder 129, must be twice the product of 20 into the part of the root still to be found, together with the square of that part. (Art. 562.) Now dividing 129 by 40 the double of 20, the quotient is 3, which being added to 40 makes 43; finally, multiplying 43 by 3, the product is 129, which is manifestly twice the product of 20 into 3, together with the square of 3. In the same manner the operation may be proved in every case. (For illustration of this rule by geometrical figures, see Practical Arithmetic, p. 318.) 1. The reason for separating the given numbers into periods of two figures each, is that a square number can not have more figures than double the number of figures in the root, nor but one less. It also shows how many figures the root will contain, and thus enables us to find part of it at a time. (Art. 562. Obs. 2.) 2. The reason for doubling that part of the root already found for a divisor, is because the remainder is double the product of the first part of the root into the second part, together with the square of the second part. 3. In dividing, the right hand figure of the dividend is omitted, because the cipher on the right of the divisor being omitted, the quotient would be 10 times too large for the next figure in the root. (Arts. 130, 146.) 4. The last figure of the root is placed on the right of the divisor simply fos convenience in multiplying it into itself. ORS. 1. The product of the divisor completed into the figure last placed in the root, cannot exceed the dividend. Hence, in finding the figure to be placed in the root, some allowance must be made for carrying, when the product of this figure into itself exceeds 9. 2. If the right hand period of decimals is deficient, it must be completed by annexing a cipher to it. 3. There will always be as many decimal figures in the root, as there are periods of decimals in the given number. 574. The square root of a common fraction is found by extracting the root of the numerator and denominator. A mixed number should be reduced to an improper fraction. When either the numerator or denominator of a common fraction is not a perfect square, the fraction may be reduced to a decimal, and the approximate root be found as above. QUEST.-573. Dem. Why do we separate the given number into periods of two figures each? Why double the root thus found for a divisor? Why omit the right hand figure of the dividend? Why place the last figure of the root on the right of the divisor? Obs. How many decimal figures will there be in the root? 574. How is the square root of a common fraction found? Of a mixed number? Required the square root of the following numbers: 3. 2601. 10. 27889 17. 566.44. 24. 25. 32. What is the square root of 119550669121? 33. What is the square root of 964.5192360241 ? 575. When the root is to be extracted to many figures, the operation may be contracted in the following manner. First find half, or one more than half the number of figures required in the root; then having found the next true divisor, cut off its right hand figure, and divide the remainder by it; place the quotient in the root, and continue the operation as in contraction of division of decimals. (Art. 333.) 34. Required the square root of 365 to eleven figures in the Ans. 19.104973174. root. 35. Required the square root of 2 to twelve figures. APPLICATIONS OF THE SQUARE ROOT. 576. A triangle is a figure which has three sides and three angles. When one of the sides of a triangle is perpendicular to another side, the angle between them is called a right-angle. 577. A right-angled triangle is a triangle which has a right-angle. The side opposite the right-angle is called the hypothenuse, and the other two sides, the base and perpendicular. The triangle ABC is right-angled at B, and the side AC is the hypothenuse. Hypothenuse. A Base. Perpendicular. |