Calculus Project 1.2
By Rob Miller, Brian Debarr,
Karli Jaco and Jeff Nestor
Problem Overview
Problem 1 requires us to explore under
what conditions would p/q terminate and
under what conditions it would repeat.
Solving p/q
We started first changing p/q to p (1/q).
We then performed calculations on this
equation letting all positive integers 2
through 20 represent q.
We also performed a random sampling of
prime and non-prime number higher than
20 in order to test our theory.
We know that any number q raised to the
power of 0 equals 1, which is a factor in all
numbers, so those factors are excluded
purposely.
Evaluating the Expression p/q
So p/q = p (1/q).
Expression
Factors
1/1 = 1
1
1/2 = .5
2
1/3 = .33
3
1/4 = .25
2*2
1/5 = .20
5
1/6 = .166
2*3
1/7 = .142857142857
7
Evaluating the Expression p/q
Expression
Factors
1/8 = .125
2*2*2
1/9 = .11
3*3
1/10 = .1
2*5
1/11 = .0909
11
1/12= .0833
2*2*3
1/13 = .076923076923
13
1/14 = .07142857142857
2*7
1/15 = .066
3*5
Evaluating the Expression p/q
Expression
Factors
1/16 = .0625
2*2*2*2
1/17 = .0588235294117647
17
1/18 = .055
2*2*2*3
1/19 = .052631578947368421 19
1/20 = .05
2*2*5
1/37 = .027027
37
1/55 = .01818
11 * 5
1/64 = .015625
2 * 2 * 2 *2 *2* 2
Results of Expression p/q
We can see that every q that has factors of 2
and 5 or 2 or 5 only will terminate. All other
combinations will repeat, including those that
have in their combination of factors 2 and 5
or 2 or 5.
Expressing Decimals Rationally
Karli’s solution for 3.a
Let
r = 13.201201…
Then
1000r = 13201.201201…
and
1000r – r = 13188
so
999r = 13188 and r =
=
Brian’s solution for 3.b
Let
r = .2727…
Then
100r = 27.2727…
and
100r – r = 27
so
99r = 27 and r =
=
Expressing Decimals Rationally
Rob’s solution for 3.c
Let r = .2323…
Then
100r = 23.2323…
and 100r – r = 23
so 99r = 23 and r =
Jeff’s solution for 3.d
Let r = 4.1633…
Then
1000r = 4163.33…
and 1000r – r = 4159
so 999r = 4159 and r =