Luoman Yu
Solutions courtesy of Luoman Yu
Assignment 3 - Numbers
Exercise 9 Given the infinite decimal 0.12121212 · · · find the equivalent fraction.
Solution: 0.12121212 · · ·=
12 4
=
99 33
Exercise 10 An even integer is always of the form 2n where n is an integer. Show that if n2 is
even then n is even
Solution: If n is not even
Suppose n=2m+1
Then n2= (2m+1)2=4m2+4m+1=2(2m2+2m) +1 and n is odd
So if n2 is even then n is even
Please observe that the above establishes the contra-positive of the given statement – we have
shown the two are equivalent As you can see this is a very easy proof - Luoman has proved:
if n odd, then n2 is odd. This is a very important technique. Some such proof is necessary
here.
Exercise 11 Show that if x and y are rational, then xy is rational. What about the case when x and
y are irrational? Is the product irrational?
Solution: 1. if x and y are rational
a
,
b
ac
Then xy 
,
bd
Suppose x 
y
c
,
d
Since xy can express as a fraction, so it is rational
2. If x and y are irrational
y 2
Suppose, x  2 ,
Then xy=2, and it is a rational.
So when x and y are irrational, the product may not be irrational.
Exercise 12 Given two intervals I and J of any sort, their intersection consists of the points
common to both and their union consists of the points that lie in one or the other or in both of the
intervals. The intersection is written I  J and the union is written I  J. With this in mind write
the following as an interval
1. [3,5)  (5,0)
2. ( ,2)  (1, )
3. (2,  )  [0, 2]
4. (−1, 2)  (−2, 3]
Solution:
1. [-3, 0)
2. (, )
3. None
4. (−2, 3]
Exercise 13 Describe with interval notation the set of x described as
1. the set of x such that |x − 2| < 1 and |x − 1|  1.5
2. the set of x such that |x − 2| > 1
3. the set of x such that |x − 2| > 1 or |x − 2| < 0.5
4. the set of x such that |x| > 1
5. the set of x such that |x| > 1 and |x|  2
Solution:
1. (1, 3)  [0.5, 2.5] = (1, 2.5]
2. (,1)  (3, )
3. (,1)  (3, )  (1.5,2.5)
4. (,1)  (1, )
5. (( ,1)  (1, ))  [2,2]  [2,1)  (1,2]
Exercise 14 Compute each of the following
1. (2 + 3i) + (3 + 4i)
2. (2 + 3i) − (2 + 3i)
3. (2 + 3i)(3 + 4i)
2  3i
3  4i
5. (1  2i )(i ))
4.
Solution:
1. (2 + 3i) + (3 + 4i) = 5+7i
2. (2 + 3i) − (2 + 3i) = 0
3. (2 + 3i)(3 + 4i) = 6+8i+9i+12i2 = 6+17i-12 = -6+17i
4.
2  3i 2  3i 3  4i
6  8i  9i  12i 2
6  i  12 18  i

*



2
3  4i 3  4i 3  4i 9  12i  12i  16i
9  16
25
5. (1  2i)(i))  2i 2  i  2  i