THE SQUARE ROOT ALGORITHM Divide the digits into groups of two, starting at the decimal point: ____________ V 10'75'80.00 The first group may contain either one or two digits. Find the largest integer (call it a) whose square is less than or equal to the first group, and write it above that group: 3 ____________ V 10'75'80.00 Subtract a^2 from that group and bring down the next group, as if you were doing long division: 3 ____________ V 10'75'80.00 9 _____ 1 75 Now the fun begins. Multiply the answer so far by 20, and divide it into the number at the bottom. Call the quotient (without the remainder) b, and write it above the second group: 3 2 ____________ V 10'75'80.00 9 _____ 60)1 75 (To understand what is going on in this step and the next, stare awhile at Figure 5.1 of Katz. The extra factor of 10 in the divisor arises because the digit a contributes 10a units on the scale of the digit b being considered now.) Multiply b by the divisor (60) and add b^2. Subtract this number from the bottom line, and bring down the next group: 3 2 ____________ V 10'75'80.00 9 _____ 60)1 75 1 24 -------- 51 80 Multiply the entire answer so far by 20 and divide to get c. Multiply c by the divisor and add c^2. 3 2 8 ____________ V 10'75'80.00 9 _____ 60)1 75 1 24 -------- 640)51 80 51 84 (= 640 x 8 + 64) Oops! We overshot, so we need to reduce c by 1. (This can't happen to a competent student in long division, but it can happen here because of the need to accommodate the c^2 term.) 3 2 7 ____________ V 10'75'80.00 9 _____ 60)1 75 1 24 -------- 640)51 80 45 29 (= 640 x 7 + 49) -------- 6 51 00 We can keep going to calculate places after the decimal point: 3 2 7. 9 ____________ V 10'75'80.00 9 _____ 60)1 75 1 24 -------- 640)51 80 45 29 -------- 6540)6 51 00 5 89 41 ---------- 61 59 00, etc. However, you must not do this unless you know that the digits after the decimal point in the original number really are zero; the number of genuinely significant figures in the answer can't exceed the number of groups (half the number of significant digits) in the original number. That's because an error in the root gets multiplied by the big, accurate part in computing the square. My calculator says: _______ V 107580 =327.9939 and (327.9)^2 = 107518.41, (327.99)^2 = 107577.44 We can see how the error in the square is diminishing.