Proceeding as in the last example, we obtain 1 for the place of tens of the root; this doubled gives the number 2, by which the two first figures 22 of the remainder are to be divided. Now 22 contains 2 eleven times, but the root can neither be more than 10, nor 10; even 9 is in fact too large, for if we write 9 by the side of 2, and multiply 29 by 9, as the rule requires, the result is 261, which cannot be subtracted from 224. We are, therefore, to consider the division of 22 by 2, only as a means of approximating the units, and it becomes necessary to diminish the quotient obtained, until we arrive at a product, which does not exceed the remainder 224. The number 8 answers to this condition, since 8 X 28 224; therefore, the root sought is 18. By resolving the square of 18 into its three parts, we find and it may be seen, that the 6 tens, contained in the square of the units, being united to 160, double the product of the tens by the units, alters this product in such a manner, that a division of it by double the tens will not give exactly the units. 92. It will not be difficult, after what has been said, to extract the square root of a number, consisting of three or four figures; but some further observations, founded upon the principles above laid down, may be necessary to enable the reader to extract the root of any number whatever. No number less than 100 can have a square consisting of more than four figures, since that of 100 is 10000, or the least number expressed by five figures. In order, therefore, to analyze the square of any number exceeding 100, of 473, for example, we may resolve it into 470+3, or 47 tens plus 3 units. To obtain. its square from the formula, a2+2ab+b2, we make a 47 tens 470 units, b = 3 units, then In this example, it is evident that the square of the tens has no figure inferior to hundreds, and this is a general principle, since tens multiplied by tens, always give hundreds, (Arith. 32). It is therefore in the part 2237, which remains on the left of the proposed number, after we have separated the tens and units, that it is necessary to seek the square of the tens; and as 473 lies between 47 tens, or 470, and 48 tens, or 480, 2237 must fall between the square of 47 and that of 48; hence the greatest square contained in 2237, will be the square of 47, or that of the tens of the root. In order to find these tens, we must evidently proceed, as if we had to extract the square root of 2237 only; but instead of arriving at an exact result, we have a remainder, which contains the hundreds arising from double the product of the 47 tens multiplied by the units. We first separate the two last figures 29, and in order to extract the root of the number 2237, which remains on the left, we further separate the two last figures 37 of this number; the proposed number is then divided into portions of two figures, beginning on the right and advancing to the left. Proceeding with the two first portions as in the preceding article, we find the two first figures 47 of the root; but we have a remainder 28, which, joined to the two figures 29 of the last portion, contains double the product of the 47 tens by the units, and the square of the units. We separate the figure 9, which forms no part of double the product of the tens by the units, and divide 282 by 94, double the 47 tens; writing the quotient 3 by the side of 94, and multiplying 943 by 3, we obtain 2829, a number exactly equal to the last remainder, and the operation is completed. 93. In order to show, by what method we are to proceed with any number of figures, however great, I shall extract the root of 22391824. Whatever this root may be, we may suppose it capable of being resolved into tens and units, as in the preceding examples. As the square of the tens has no figure inferior to hundreds, the two last figures 24 cannot make a part of it; we may therefore separate them, and the question will be reduced to this, to find the greatest square contained in the part 223918, which remains on the left. This part consisting of more than two figures, we may conclude, that the number, which expresses the tens in the root sought, will have more than one figure; it may therefore be resolved, like the others, into tens and units. As the square of the tens does not enter into the two last figures 18 of the number 223918, it must be sought in the figures 2239, which remain on the left; and since 2239 still consists of more than two figures, the square, which is contained in it must have a root, which consists of at least two; the number which expresses the tens sought will therefore have more than one figure; it is then, lastly, in 22 that we must seek the square of that, which represents the units of the highest place in the root required. By this process, which may be extended to any length we please, the proposed number may be divided into portions of two figures from right to left; it must be understood, however, that the last figure on the left may consist of only one figure. 16 63,9 60 9 301,8 87 943 9462 Having divided the proposed number into portions as below, we proceed with the three first portions, as 22,39,18,24 4732 in the preceding article; and when we have found the three first figures 473 of the root, to the remainder 189, we bring down the fourth portion 24, and consider the number 18924, as containing double the product of the 473 tens already found by the units sought, plus the square of these units. We separate the last figure 4; divide those, which remain on the left, by 946, double of 473, and then make trial of the quotient 2, as in the preceding examples. 1892,4 0000 0 Here the operation, in the present case, terminates; but it is very obvious, that if we had one portion more, the four figures already found 4732, would express the tens of a root, the units of which would remain to be sought; we should proceed, there fore to divide the remainder now found, together with the first figure of the following portion, by double of these tens, and so on for each of the portions to be successively brought down. 0 00 0 94. If, after having brought down a portion, the remainder, joined to the first figure of this portion, does not contain double of the figures already found, a cypher must be placed in the root; for the root, in this case, will have no units of this rank; the following portion is then to be brought down, and the operation to be continued as before. The example subjoined will illustrate this case. The quantities to be subtracted are 49.42,09 703 not put down, but the subtractions are suppos- 04,20, 9| 1403 ed to be performed mentally, as in division. 95. Every number, it will be perceived, is not a perfect square. If we look at the table given, page 100, we shall see that between the squares of each of the nine primitive numbers, there are intervals comprehending many numbers, which have no assignable root; 45, for instance, is not a square, since it falls between 36 and 49. It very often happens, therefore, that the number, the root of which is sought, does not admit of one; but if we attempt to find it, we obtain for the result the root of the greatest square, which the number contains. If we seek, for example, the root of 2276, we obtain 47, and have a remainder 67, which shows, that the greatest square contained in 2276, is that of 47 or 2209. As a doubt may sometimes arise, after having obtained the root of a number, which is not a perfect square, whether the root found be that of the greatest square contained in the number, I shall give a rule, by which this may be readily determined. As the square of a + b is if we make b = 1, the a2 +2ab+ b2, square of a + 1 will be a quantity which differs from a2, the square of a, by double of a plus unity. Therefore, if the root found can be augmented by unity, or more than unity, its square, subtracted from the proposed number, will leave a remainder at least equal to twice this root plus unity. Whenever this is not the case, the root obtained will be, in fact, that of the greatest square contained in the number proposed. 96. Since a fraction is multiplied by another fraction, when their numerators are multiplied together, and their denominators together, it is evident that the product of a fraction multiplied by itself, or the square of a fraction is equal to the square of its numerator, divided by the square of its denominator. Hence it follows, that to extract the square root of a fraction, we extract the square root of its numerator and that of its denominator. Thus the root ofis, because 5 is the square root of 25, and 8 that of 64. It is very important to remark, that not only are the squares of fractions, properly so called, always fractions, but every frac tional number, which is irreducible, (Arith. 59) will, when multiplied by itself, give a fractional result, which is also irreducible. 97. This proposition depends upon the following; Every prime number P, which will divide the product AB of two numbers A and B, will necessarily divide one of these numbers. Let us suppose, that it will not divide B, and that B is the greater; if we designate the entire part of the quotient by q, and the remainder by B', we have B = q P+ B', multiplying by A, we obtain AB = qAP + AB', and dividing the two members of this equation by P, we have from which it appears, that if AB be divisible by P, the product AB' will be divisible by the same number. Now B', being the remainder after the division of B by P, must be less than P; therefore B' cannot be divided by P; if we divide P by B' we have a quotient q and a remainder B"; if further we divide P by B", we have a quotient q" and a remainder B"", and so on, since P is a prime number. We have, therefore, the following series of equations; P = q B' + B", P=q"B" + B'", &c. multiplying each of these by A, we obtain AP = q'AB' + AB", AP = q"AB" + AB'", &c. From these results it is evident, that if AB' be divisible by P, the products AB", AB"", &c. will also be divisible by it. But the remainders B', B", B", &c. are becoming less and less, continually, |