Download Math 1710 – Assignment #1

Math 1710 – Assignment #2 (Numbers)- Solutions by: Xi Cheng Liu
Exercise 9- Solution:
x = 0.12121212…
100x – x = 12.12121212…. – 0.12121212… = 12
x = 12/99 = 4/33
Exercise 10- Solution:
2 a1
2
p1

p2
2a 2

n

the smallest prime number

n

n
=
=
p1  p2


pr
2 ar
and
n
2
=2
k (n
2
is even)
p1 must be 2
pr
=2 
p2

pr
is even.
This is a good proof and uses the Fundamental Theorem of arithmetic - that being said there is a very simple
first principles proof. Here’s how it goes.
The problem is to show that: n2 even implies n is even. Let p be the proposition: “n2 is even” and
let q be the proposition “n is even”. So we need to prove that: p implies q. We however know that this is
the same as proving ~p implies ~q ( the contrapositive). So lets do this. That is assume n is not even
and prove that this means that n2 is not even. This is now easy.
If n is not even then it is odd which means that n = 2m + 1 for some integer m. This in turn means that
n2 = (2m + 1)2 = 4m2 +4m +1 = 4(m2 +m) +1 - which is an odd number because 4(m2 +m) is even and
an even number plus 1 is an odd number. Thus, we have shown that n not even implies n2 is not even.
Exercise 11- Solution:
1) x, y  Q => x = a/b, y=c/d => xy = ac/bd => xy  Q
2) x,y are irrational numbers, let x= 2 , y =
xy =
2
8 =
8,
16 = 4, 4  Q, so the product of 2 irrational numbers may not be irrational.
Exercise 12- Solution:
1. [-3,5)  (-5,0) = [-3,0)
2. (-  ,2)  (1,  ) = (-  ,  )
3. (2,  )  [0,2] = φ
4. (-1,2)  (-2,3] = (-2,3]
Exercise 13- Solution:
1. |x-2| < 1 => -1 < x-2 < 1 => x  (1, 3)
|x-1|  1.5 => -1.5  x-1  1.5 => x  [-0.5, 2.5]
(1, 3)  [-0.5, 2.5] = (1, 2.5]
So x  (1, 2.5]
since ac/bd is a quotient of integers
2. |x-2| >1 => (x-2 < -1)  (x-2 >1) => (x<1)  (x>3)
So x  (-  ,1)  (3,  )
3. |x-2| >1 => (x-2 < -1)  (x-2 >1) => (x<1)  (x>3) => x  (-  ,1)  (3,  )
4.
5.
|x-2| <0.5 => -0.5 < x-2 < 0.5 => x  (1.5, 2.5)
((-  ,1)  (3,  ))  (1.5, 2.5)  (-  ,1)  (1.5, 2.5)  (3,  )
So x  (-  ,1)  (1.5, 2.5)  (3,  )
|x| >1 => (x<-1)  (x>1)
So x  (-  ,-1)  (1,  )
|x| >1 => (x<-1)  (x>1) => x  (-  ,-1)  (1,  )
|x|  2 => -2  x  2 => x  [-2, 2]
((-  ,-1)  (1,  ))  [-2, 2]  [-2, -1)  (1, 2]
So x  [-2, -1)  (1, 2]
Exercise 14- Solution:
1. (2+3i)+(3+4i) = (2+3)+(3+4)i=5+7i
2. (2+3i)-(2+3i) = (2-2)+(3-3)i=0
3. (2+3i) (3+4i) = (2*3)-(3*4)+(2*4+3*3)i = -6+17i
4.
2 * 3  3 * 4  (3 * 3  2 * 4)i
18  i
2  3i
2  3i 3  4i
=
=
=

9  16
25
3  4i
3  4i 3  4i
5.
(1  2i )(i ) =  2  i = -2+i
Exercise 15- Solution:
If x  0 and y  0, ||x| -|y|| = |x-y|
If x<0 and y<0, ||x| -|y|| = |y-x| = |x-y|
If x  0 and y<0, ||x| -|y|| = |x+y| < |x-y|
If x<0 and y  0, ||x| -|y|| = |-x-y| = |y+x| < |y-x| = |x-y|
Therefore,  x,y  R, ||x| -|y||  |x-y|
|