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Assignment 4
MATH 1200
SOLUTION
Due Monday October 21
Instead of representing real numbers in decimal notation, for any other natural number there is a
corresponding base n representation. For n = 10 this is the usual decimal expansion. So, the statement
that every real number x in the interval [0, 1] has a base 10 representation means that there are integers
ki such that each ki is between 0 and 9 and
x=
k3
k2
k1
+ 3 + ...
+
10 102
10
And we write this as x = 0.k1 k2 k3 ....
Instead of base 10, let’s consider base 7. It is also the case that every real number has a base 7
representation. Let’s use the following notation for the base 7 representation of a number between 0 and
1:
x2
x3
x1
+ 2 + 3 + ... where x1 , x2 , ... ∈ {0, 1, ..., 6}
7
7
7
(1) Find the familiar decimal notation for each of the following numbers given in base 7 form:
(a) (0.111.....)7
(0.x1 x2 x3 ....)7 =
Solution:
(0.111.....)7 =
1
1
1
1
1
1
= = 0.1666666....
+
+ 3 + ... = ·
7 72
7
7 (1 − 17 )
6
(b) (0.12121212.....)7
Solution:
1
2
1
2
1
2
+
+ 3 + 4 + 5 + 6 + ... =
7 72
7
7
7
7
7
2
7
2
7
2
9
9
9
= 2 + 2 + 4 + 4 + 6 + 6 + ... = 2 + 4 + 6 + ... =
7
7
7
7
7
7
7
7
7
9 49
9
1
1
9
1
9
=
= 2 · (1 + 2 + 4 + ...) = 2 ·
·
=
= 0.1875
7
7
7
7 (1 − 712 )
49 48
48
(0.121212.....)7 =
(2) Find the first few digits in the base 7 representation of the following numbers. Try to find the
full base 7 representation, i.e., all the digits.
(a) 17
SOLUTION: 17 is already represented as a sum of reciprocals of powers of 7, so its base 7
representation is (0.1)7 .
(b)
1
2
SOLUTION: We want to write 12 = x71 + x722 + x733 + ... where x1 , x2 , ... ∈ {0, 1, ..., 6} I.e.,
find x1 , x2 , .....
1/2 lies between 3/7 and 4/7, so the first digit must be 3.
1/2 lies between 3/7 + 3/49 = 24/49 and 3/7 + 4/49 = 25/49 so the second digit is also 3.
One can continue in this way but perhaps one can guess that every digit is 3?
Well,
3/7 + 3/72 + 3/73 + ... = 3/7 · (
So all the digits are 3.
1
) = 3/7 · 7/6 = 3/6 = 1/2
1 − 71
2
(c)
3
5
SOLUTION: Proceed as above. Since 4/7 < 3/5 < 5/7. The first digit is 4. Now find x
between 0 and 6 such that 4/7 + x/49 ≤ 3/5 < 4/7 + (x + 1)/49.
Well since 29/49 < 3/5 < 30/49 and 4/7 + 1/49 = 29/49 the second digit is 1.
So, the base 7 expansion of 3/5 is of the form (0.41???????)7 . Hmmmm wonder what the
rest of the digits are.....
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