Project 1.2 Decimal Expansions of Rational Numbers

Project 1.2 Decimal Expansions of Rational Numbers

By

Steven Hedrick[W1][W2]

Mitchell Jordan

Nathan Walter

Summary:

This project determines under what conditions the fractions of rational numbers will repeat or terminate. It also shows that a repeating decimal can only be displayed as a rational number with a degree of error.

Steven Hedrick: summary, problem and solution for problem 1, and program

Mitchell Jordan: problem and solution for problem 2 and Power Point presentation

Nathan Walter: problem and solution for problem 3

Solutions:

1. Under what conditions will the decimal expansion of p/q terminate? Repeat? Hint: Since p/q = p(1/q), it is sufficient to investigate the decimal expansions of 1/q. Calculate 1/q for enough positive integers q to form a conjecture as to whether the decimal expansion will terminate or repeat. What is your conjecture?

1/q will terminate when q is in the form of 2x5y and x and y are positive integers.

1. If a number is a repeating decimal when expanded, it cannot be easily displayed as a rational number, but it is possible, with a degree of error.

If a decimal repeats at a certain point, the number should be multiplied so that that repeating point becomes a ones-place number.

Ex: 3.135135135… should be multiplied by 1000

To become 3135.135135…

By assigning the original repeating expansion the variable r, we are able to calculate a way to display the number rationally

1)r = 3.135135…

2)1000r = 3135.135135…

3)1000r – r = 3132

4)Eg. 999r = 3132 and r = 3132/999 = 116/37

3. Problem: the purpose of problem 3 was to express various repeating decimals in the the rational number form of p/q. The problem would also go further to ask that the repeating decimal 0.99999… be shown to represent the number 1

1. 13.201201…

First I set r=13.201201…

I then multiply r by 1000 to get 1000r=13201.201…

I then subtract r to get a whole number 1000r-r=13188

999r=13188  r/999=13188/999

So after simplifying  4396/333

1. 0.2727

R=.2727…

100r=27.2727

100r-r=27

So finally we come to the rational number r= 27/100

1. r=0.2323…

100r=23.2323…

100r-r=23

Therefore 99r would be equal to 23/100

1. 4.16333333….

R=4.1633333….

1000r=4163.333333….

1000r-100r=3747

900r=3747

Finally we can conclude that r=3747/900 or  1249/300

The final part of problem number 3 asked that I prove 0.99999…. is another representation of the number 1. To solve this I will utilize the formula provided by the book for rationalizing a repeating decimal:

a)r=0.9999….

b)10r=9.9999

c)10r-r=9

d)9r(1/9)=9(1/9)

e)R=9/9 or 1

f)From this we can derive that r is both equal to 0.9999…. and 1, therefore 0.9999repeating is equal to 1.0. this rational number has more than one decimal representation but I could not find any others that are similar to this circumstance.

[W1]Good job

[W2]