How to convert a decimal number into a fraction ?
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example :
given decimal number: 1.13207...
r = 1.13207...
0
r = 1/(O.13207...) = 7.57142...
1
r = 1/(0.57142...) = 1.75
2
r = 1/(0.75) = 1.33333...
3
r = 1/(0.33333...) = 3
4
p = r * r * r * r * r = 60
0 1 2 3 4
q = r * r * r * r = 53
1 2 3 4
fraction: p/q = 60/53
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Why does this work ?
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The product of all the remainders in a finite
continued-fraction equals the most simple numerator.
p/q = [a ;a ,a ,a ,...a ]
0 1 2 3 n
= [a ;a ,a ,a ,...r ]
0 1 2 3 n
...
= [a ;a ,a ,r ]
0 1 2 3
= [a ;a ,r ]
0 1 2
= [a ;r ]
0 1
= r
0
p = r *r *r *r *...*r
0 1 2 3 n
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Theorems:
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finite continued fraction :
[a ;a ,a ,a ,...,a ,a ,a ,...,a ,a ]
0 1 2 3 (k-1) k (k+1) (n-1) n
A finite continued fraction is rational,
so it can be represented as a normal fraction.
[a ;a ,a ,a ,...,a ,a ,a ,...,a ,a ] = p/q
0 1 2 3 (k-1) k (k+1) (n-1) n
remainder :
r = [a ,a ,...,a ,a ]
k k (k+1) (n-1) n
r = a + 1/r (1)
k k (k+1)
each remainder of a continued fraction can
be seen as another continued fraction.
When the continued fraction is finite, so are
the remainders, and so they can be represented
by normal fractions.
r = p /q (2)
k k k
The last remainder of a finite continued fraction is
a whole number so its denominator equals 1.
r = p /q = p /1
n n n n
q = 1 (3)
n
The relation between numerators and denominators
of succesive remainders :
r = a + 1/r (1)
k k (k+1)
p /q = a + 1/(p /q ) (2)in((1)
k k k (k+1) (k+1)
p /q = a + q /p
k k k (k+1) (k+1)
p /q = (a *p + q ) / p
k k k (k+1) (k+1) (k+1)
and so
p = a *p + q
k k (k+1) (k+1)
and
q = p or p = q (4)
k (k+1) k (k-1)
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Proof with nth order continued fraction:
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What we want to proof:
r *r *r *...*r *r =? P
0 1 2 (n-1) n 0
implement:(2)
(p /q )*(p /q )*(p /q )*...*(p /q )*(p /q ) =? p
0 0 1 1 2 2 (n-1) (n-1) n n 0
implement:(3)
(p /q )*(p /q )*(p /q )*...*(p /q )*(p /1 ) =? p
0 0 1 1 2 2 (n-1) (n-1) n n 0
(p /q )*(p /q )*(p /q )*...*(p /q )*p =? p
0 0 1 1 2 2 (n-1) (n-1) n 0
implement:(4)
(p /q )*(p /q )*(p /q )*...*(p /q )*q =? p
0 0 1 1 2 2 (n-1) (n-1) (n-1) 0
(p /q )*(p /q )*(p /q )*...*p =? p
0 0 1 1 2 2 (n-1) 0
implement:(4)
(p /q )*(p /q )*(p /q )*...*q =? p
0 0 1 1 2 2 (n-2) 0
...
(p /q )*(p /q )*(p /q )*q =? p
0 0 1 1 2 2 2 0
(p /q )*(p /q )*p =? p
0 0 1 1 2 0
implement:(4)
(p /q )*(p /q )*q =? p
0 0 1 1 1 0
(p /q )*p =? p
0 0 1 0
implement:(4)
(p /q )*q =? p
0 0 0 0
p = p
0 0
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Proof with 3th order (n=3) continued fraction:
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r *r *r *r =? P
0 1 2 3 0
(p /q )*(p /q )*(p /q )*(p /q ) =? p
0 0 1 1 2 2 3 3 0
(p /q )*(p /q )*(p /q )*(p /1) =? p
0 0 1 1 2 2 3 0
(p /q )*(p /q )*(p /q )*(q /1) =? p
0 0 1 1 2 2 2 0
(p /q )*(p /q )*p =? p
0 0 1 1 2 0
(p /q )*(p /q )*q =? p
0 0 1 1 1 0
(p /q )*p =? p
0 0 1 0
(p /q )*q =? p
0 0 0 0
p = p
0 0
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Peter Mulkers
Belgium
P.Mulkers@gmx.net
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