Mohamed G. Gouda						     CS 313K
Fall 2012							     Homework 4

1. Let A be any set. Which of the following six statements is true? can be true?
or is false?
   (a) {} in A
   (b) {} sub A
   (c) {{}} sub A
   (d) {} in PS(A)
   (e) {} sub PS(A)
   (f) {{}} sub PS(A)
Note that {} denotes the empty set and {{}} denotes a set that contains the 
empty set as an element:

Sol:
(a) can be true because {} can be an element in A
(b) is true
(c) can be true because {} can be an element in A
(d) is true
(e) is true
(f) is true

2. Let A be the set of all real numbers other than (-1), and B the set of all 
real numbers. Define f:A->B be the function f(x) = 2x/(x+1). Show, by direct 
inference, that if f is injective. Show, by guessing, that f is not surjective.

Sol:
Proving that f is injective

 f(x) = f(y)
=> {Definition of f}
 2x/(x+1) = 2y/(y+1)
=> {Arithmetics}
 (xy + x) = (yx + y)
=> {xy = yx}
 x = y

Proving that f is not surjective

Guess that there is no x such that f(x) != 2. Then, prove this fact by 
contradiction.
 f(x) = 2
=> {Definition of f}
 2x/(x+1) = 2
=> {Arithmetics}
 2x = 2(x+1)
=> {Arithmetics}
 0 = 2
=> F

3. Let f:A->B and g:B->C be two functions. Show, by direct inference, that 
       (f is surjective) and (g is surjective) 
       => (g.f is surjective)

Sol:
 (z in C)
=> {g is surjective}
 (Exist y in B, g(y) = z)
=> {f is surjective and y in B}
 (Exist x in A, f(x) = y)
=> {Definition of "."}
 (Exist x in A, g.f(x) = g(f(x)) = g(y) = z)

4. Let R be the set of all real numbers and let f:R->R be the function 
f(x) = x^3. Show, by direct inference, that f is bijective (i.e. f is 
both injective and surjective).

Sol:
Proving that f is injective:

 f(x) = f(y)
=> {Definition of f}
 x^3 = y^3
=> {Arithmetics}
 x = y

Proving that f is surjective:

 y in R
=> {Let x be the real number y^(1/3)}
 Exist x in R, x^3 = y
=> {Definition of f}
 Exist x in R, f(x) = y

5. Let R be the set of all real numbers and let f:R->R and g:R->R be two
functions. 
(a) Use guessing to define functions f and g such that f.g = g.f
(b) Use guessing to define functions f and g such that f.g != g.f

Sol:
(a) Let f(x) = x and g(x) = x+1. In this case, 
f.g(x) = f(g(x)) = f(x+1) = x+1
g.f(x) = g(f(x)) = g(x) = x+1
Thus, f.g(x) = g.f(x)

(b) Let f(x) = x^2 and g(x) = x+1. In this case,
f.g(x) = f(g(x)) = f(x+1) = (x+1)^2
g.f(x) = g(f(x)) = g(x^2) = x^2 + 1
Thus, f.g(x) != g.f(x)

6. Let f:A->B be a function that has an inverse f' and let B1 and B2 be 
two subsets of set B. Show, by a two-sided inference proof, that 
    	    f'(B1 ^ B2) = f'(B1) ^ f'(B2)

Sol:
 x in f'(B1 ^ B2)
<=> {Definition of f'}
 f(x) in B1 ^ B2
<=> {Definition of ^}
 (f(x) in B1) and (f(x) in B2)
<=> {Definition of f'}
 (x in f'(B1)) and (x in f'(B2))
<=> {Definition of ^}
 x in (f'(B1) ^ f'(B2))