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:
- 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?
Values for which 1/q will terminate or repeat
1/1 / Term. / 1/11 / Repeat / 1/21 / Repeat / 1/31 / Repeat / 1/41 / Repeat
1/2 / Term. / 1/12 / Repeat / 1/22 / Repeat / 1/32 / Term. / 1/42 / Repeat
1/3 / Repeat / 1/13 / Repeat / 1/23 / Repeat / 1/33 / Repeat / 1/43 / Repeat
1/4 / Term. / 1/14 / Repeat / 1/24 / Repeat / 1/34 / Repeat / 1/44 / Repeat
1/5 / Term. / 1/15 / Repeat / 1/25 / Term. / 1/35 / Repeat / 1/45 / Repeat
1/6 / Repeat / 1/16 / Term. / 1/26 / Repeat / 1/36 / Repeat / 1/46 / Repeat
1/7 / Repeat / 1/17 / Repeat / 1/27 / Repeat / 1/37 / Repeat / 1/47 / Repeat
1/8 / Term. / 1/18 / Repeat / 1/28 / Repeat / 1/38 / Repeat / 1/48 / Repeat
1/9 / Repeat / 1/19 / Repeat / 1/29 / Repeat / 1/39 / Repeat / 1/49 / Repeat
1/10 / Term. / 1/20 / Term. / 1/30 / Repeat / 1/40 / Term. / 1/50 / Term.
1/q will terminate when q is in the form of 2x5y and x and y are positive integers.
- 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
- 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
- 0.2727
R=.2727…
100r=27.2727
100r-r=27
So finally we come to the rational number r= 27/100
- r=0.2323…
100r=23.2323…
100r-r=23
Therefore 99r would be equal to 23/100
- 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]