Problem 1.
Let f(x)=ax + b and g(x)=cx+d, where a,b,c and d are constants. Determine for which
constants a,b,c and d it is true that fg = gf
fg(x) = f(cx+d)=a(cx+d) + b = acx +ad + b
gf(x) = g(ax+b) = c(ax+b) + d = acx + cb + d
So, fg = gf if ad+b = cb+d.
Problem 2.
Show that function f(x)=ax + b from R to R is invertible, where a,b are constants, with a
 0, and find the inverse of f.
Solution: y = ax + b so x = [y-b]/a which means g(x)= [x-b]/a is the inverse.
Problem 3.
Let f be the function from R to R defined be f(x) = x2. Find
a) f-1({1}) = {1, -1}
b) f-1({x : 0  x  1}) = {x : -1  x  0}  {x: 0  x 1}
c) f-1({x : x  4}) = {x : x  2}
Problem 4.
Show that if x is a real number and n is an integer, then x + n = x + n
Let x = m + p where m is integer and p is greater than 0 and less than 1.
Then m + p = m + 1. So, x + n = m + n +1.
Now, x + n = (m + n) + p = m + n +1
Problem 5.
Show that if x is a real number and n is an integer, then
a) x  n if and only if x  n
b) n  x if and only if n  x
Problem 6.
Can you conclude that A = B if A and B are two sets with the same power set?
2A = 2B  A = B
equivalent to A  B  2A  2B
Example: A = {a,b,c}, 2A = {0,{a},{b},{c},{a,b},{a,c},{b,c},{a,b,c}}
Problem 7.
Explain why (A  B)  (C  D), A  (B  C)  D are not the same.
Problem 8.
Assume that:
1) Jack owns the dog.
2) Every dog owner is an animal lover.
3) No animal lover kills an animal.
4) Either Jack or John killed the cat who is named Tuna.
Did John kill the cat?
Rule of inference – resolution: [A  B]  [B  C]  [A  C]
Solution
Introducing predicates and constants:
J1 – Jack , J2 – John, T – Tuna, D - dog
K(x,y) - x killed y
A(x) – x is an animal
L(x,y) - x loves y
O(x,y) - x owns y
Facts:
O(J1,D), A(T), A(D),
O(x,D)  A(y)  L(x,y)
A(y)  L(x,y)  K(x,y)
K(J1,T)  K(J2,T)
Proof by contradiction Assume that K(J2,T)
All facts convert to disjuncts:
O(J1,D), A(T), A(D),
O(x,D)  A(y)  L(x,y)
A(y)  L(x,y)  K(x,y)
K(J1,T)  K(J2,T)
K(J2,T)
L(x,T)  K(x,T) (3)
A(y)  L(J1,y) with A(T)
K(J1,T) (3)
L(J1,T)
L(J1,T)
FALLS