Calculating Lengthy Repeating DecimalsDate: 10/26/2006 at 21:14:22 From: Cy Subject: 355/113 Hi, Dr. Math. I'm currently enrolled in 8th grade algebra, and I just had a question I was curious about. You know the approximation of pi, 355/113? Well, I was trying to figure out if I could turn it into a decimal using long division, and find out if the decimal repeats or ever terminates. Can you please show me how to find when (or if) the decimal terminates/repeats? I am currently still using long division to find my answer, and have been unsuccessful. This is what i have so far: 3.141592920346902654867256637168141592920353982300884070709645... Thanks, Cy Date: 10/27/2006 at 00:49:51 From: Doctor Greenie Subject: Re: 355/113 Hello, Cy -- If the decimal representation of a fraction terminates, then we can give the decimal fraction an exact name, using base 10 place values: .103 = 103 thousandths = 103/1000 .48932 = 48932 hundred thousandths = 48932/100000 .15 = 15 hundredths = 15/100 = 3/20 If the common fraction can't be written as an equivalent fraction with a denominator which is a power of 10, then the decimal form of the fraction will not terminate. So the only common fractions which have decimal representations which terminate are those whose denominators contain prime factors of only 5 and/or 2 (the prime factors of the base, 10). So your fraction 355/113 will not terminate. Now... how many digits will the repeating decimal go before it repeats? We can't tell before we start calculating; however, we know that the number of places will be less than the denominator. If you think about the long division process, and about the fact that the division never terminates, then the only possible remainders at each step are 1 through 112. Since we can have at most 112 different remainders in the long division process, we must get a repeated remainder after at most 112 steps; and so the repeating decimal part of 355/113 can be at most 112 digits long. And in fact, with a large prime denominator like 113, it is quite likely that the repeating decimal pattern will be 112 digits long. One way to find the repeating decimal part is long division, as you have attempted. This of course can be very tedious, and arithmetic mistakes are easy to make. You have made an error somewhere, because your result is only correct this far: 3.1415929203 The next digit should be "5"; you show "4". Is there a way to find the repeating decimal faster than long division? Yes, there is. To understand the quicker method for finding the decimal representation of 355/113, let's look at the calculation of the first several decimal places using long division: 3.14159292035... -------------------- 113 ) 355.00000000000000 339 ----- 16 0 11 3 ----- 4 70 4 52 ----- 180 113 ---- 670 565 ---- 1050 1017 ----- 330 226 ---- 1040 1017 ----- 230 226 ---- 400 339 ---- 610 565 --- 45... After the whole number part of the answer is obtained, the remainder is 16. So at this point we are dividing "16" followed by an infinite string of 0's by 113. Then after the 8th decimal place, the remainder is 4. So from this point on, we are dividing "4" followed by an infinite string of zeros by 113. Since 4 is 1/4 of 16, the sequence of digits we will get starting here is the string of digits we have obtained to this point, divided by 4. So we can find the complete repeating decimal representation of 355/113 by using long division or some other technique to get the first several decimal places and then dividing that string of digits by 4. Division by 4 is much easier than division by 113, so the overall time required to find the complete repeating decimal is shortened. Here is how the beginning of the process went when I used the calculator on my PC.... 355/113 = 3.1415929203539823008849557522124 /4 = 0.7853982300884955752212389380531 /4 = 0.19634955752212389380530973451327 /4 = 0.049087389380530973451327433628319 /4 = 0.01227184734513274336283185840708 /4 = 0.0030679618362831858407079646017699 Now we can "splice together" the appropriate digits from these strings. For example, the last digit in the first line is rounded, so we look at the several preceding digits, "5752212", and we look for that sequence of digits in the next line. The digits in the second line following this sequence are "389380531", with the last digit "1" having been rounded; we splice these digits onto the other digits we already have. So then in the third line we find the string of digits "38053" and copy the next several digits from that line, omitting the last digit. And so on.... We get the following: 355/113 = 3.14159 29203 53982 30088 49557 52212 38938 05309 -- 73451 32743 36283 18584 07079 64601 769... I have grouped the digits into groups of 5 so we can keep track of how many decimal places we have gone. At this point, we have found 73 digits--more than half the maximum possible length of the repeating pattern--without the pattern repeating. For reasons I don't think I could explain, that means the repeating sequence of digits will in fact be the full 112 digits long. It also means that somewhere in our string of digits we will find a string of digits which is the "9's complement" of the beginning digits. Here is an example (somewhat familiar to you, I hope) of what this means. If we look at the decimal representation of 1/7... 1/7 = .142857142857.... digits 4-6 in the repeating pattern are the "9's complement" of digits 1-3, meaning 142+857=999. So somewhere in our string of the first 73 digits of the decimal representation of 355/113, we should find a string of digits which is the "9's complement" of the first several digits: 9999999... - 1415929... ---------- 8584070... And indeed we do find this string, starting with the 57th digit. (Note: (112/2)+1 = 57--the 57th digit is the first digit in the second half of the 112-digit repeating pattern; the two 56-digit halves are 9's complements of each other.) And so now we can complete the process of finding the 112-digit repeating pattern for 355/113 by using the 9's complement of the first 56 digits: 355/113 = 3.14159 29203 53982 30088 49557 52212 38938 05309 -- 73451 32743 36283 1 -- 85840 70796 46017 69911 50442 47787 61061 94690 -- 26548 67256 63716 8... If you are really good with mental math (specifically, dividing by 4), you can find this complete repeating pattern by taking the digits you already have and dividing by 4 to get subsequent digits.... Finally, when I first started working on your problem and wanted to check my calculations, I looked into finding repeating decimal patterns by using a spreadsheet, since spreadsheets are good tools for performing repeated calculations. It turned out to be relatively easy to set up a spreadsheet to mimic the long division process. (If you don't know anything about spreadsheets, try to find someone who does so they can show you this process....) Here's what we do on the spreadsheet to find the repeating decimal for 355/113: (1) In cell B1, enter "16". We don't really care about the whole number part of the answer, "3"-- we are only interested in the remainder after the first step, which is 16. (2) In cell A2, enter the following formula: =FLOOR(10*B1/113,1) This formula multiplies the remainder from cell B1 (16) by 10, divides it by 113, and rounds down to the nearest whole number. This mimics the process of finding the next digit of the answer. (16 times 10 = 160; 160/113 = 1 remainder 47; next digit is "1", and the remainder is 47.) (3) In cell B2, enter the following formula: =10*B1-113*A2 This formula finds the new remainder by multiplying the previous remainder by 10 and subtracting 113 times the digit you just found. (16 times 10 = 160; 160-113(1) = 47.) (4) Highlight cells A2 and B2, drag down until you have reached to or past cells A113 and B113, and use the "edit/fill/down" feature (control-D in Microsoft Excel) to copy the formulas in cells A2 and B2 to the remaining rows of columns A and B. The sequence of digits in cells A2 through A113 are the digits of the repeating decimal representation of 355/113. You will see that the digits begin repeating with cell A114. Thanks for submitting your question. I got a lot of good mental exercise out of it; and I discovered an easy way to use a spreadsheet to mimic the long division process and thus provide a quick path to finding the decimal representation of any repeating fraction. I hope all this helps. Please write back if you have any further questions about any of this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
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