Decimal Expansion of Fractions Brent Murphy

PQPQ Problem: Under What conditions will the decimal expansion of p/q terminate? Under what conditions will it repeat? p/q can be investigated as p*(1/q).

Terminating When placing 1 over q as a fraction the decimal expansion will terminate if it can be factored down to a prime number < 5, unless q is also a multiple of 3 and a prime number > 5 is not a factor.

Examples of Terminating decimal expansion ½ and 1/5 terminate. 1/10 will terminate because 10 can be factored down to 2 and 5 and it is not a multiple of 3. 1/16 terminates. 16 can be factored down to four 2’s, and is not a multiple of 3. 1/25 terminates. 25 can be factored down to 5 and 5 and is not a multiple of 3.

Repeating Decimal Expansion If q is a multiple of 3 it will repeat. If q can be factored down to a prime number > 5 then it will repeat.

More examples of repeating Decimals 1/29 repeats. 29 is a prime number > 5. 1/35 repeats. 35 can be factored to 5 and 7. 7 is a prime number greater than 5, so it must repeat the decimal when placed under a number as a fraction. 1/14 repeats. It can be factored to a 2 and 7. 7 is prime and > 5 so 1/14 repeats.

Examples of repeating Decimals 1/3 repeats. 1/9 repeats. 9 is a multiple of 3. 1/15 repeats. 15 can be factored to 3 and 5. The fact that it is a factor of 5 (one of the first 3 prime numbers) would cause it to terminate if it were not a multiple of is a multiple of 3 so it repeats.

Problem 2 Suppose you are given a decimal expansion of a fraction. How can you represent that as p/q? For Terminating decimals For Repeating decimals

Terminating Decimals Terminating decimals are a little easier than repeating decimals. If you make the decimal a whole number by multiply it by 10^x, where x is the number of decimal places needed to move to make the decimal a whole number. Put that number over 10^x and reduce.

Examples of terminating decimals to fractions Example 1: N= 3.74 N*10^2 = /10^2= 374/100 = 187/50 Example 2: N = N*10^3 = 4,169 N/10^3= 4,169/1000

Repeating Decimals Repeating decimals are a little harder and require a little more logic. N= a repeating decimal number. Declare a variable (M) that M= N*10^x, where x is the number of decimal places needed to move the decimal to where it begins to repeat. Note: This step may not be needed if the decimal begins to repeat after immediately. In this case, assume M=N

Repeating decimals Step 2 Next, multiple M by 10^y, where y is the number terms in the geometric sequence before it begins to repeat once more. Then, setup an equation where 10^yM = the number M multiplied by 10^y.

Repeating Decimals Step 3 Subtract M from both sides leaves the right side of the equation as a whole number. Divide that whole number by 10^yM – M and it will create a fraction. Remember, that M = N*10^y. You are looking for N, not M so you must setup and equation for N*10^y and then divide the fraction found for M by N*10^y and it will give you the fraction for N. Reduce and Enjoy!!!

Examples of repeating decimals N = N = M M*10^3 = 3, M – M = 3, M = 3,132 M = 3,132/999 M = 116/37 M= N so N = 116/37

Repeating Decimal Example N = M = = 100N 10M = 5, M – M = 4,647 9M = 4,647 M = 4647/9 M = 1549/3 100N = 1549/3 N = 1549/300

Problem 3 Express as rationals: A) … B).27… C).23… D) … Show that.99…. Represents 1.0

A N = … 1000N = 13, N – N = 13, – N 999N = 13,188 N = 13,188/999 N = 4,396/333

B N =.27… 100N = N – N = 27 99N = 27 N = 27/99 N = 3/11

C N = N = N – N = 23 99N = 23 23/99 = N

D N = … M= 100N M = M = 4, M – M = 3,747 9M = 3,747 M = 3,747/9 = 1249/3 100N = 1249/3 N = 1249/300

Show that.9999 represents 1.0 N =.999… 10N = N – N = 9 9N = 9 N = 1.0