Mohamed G. Gouda						     CS 313K
Fall 2012							     Midterm 1

1. Let the domains of x and y be the set of all integers. Compute the truth 
values of the following quantified predicates:
  All x Exist y (x*y = x)
  Exist y All x (x*y = x)
  Exist x All y (x*y != x)
  All y Exist x (x*y != x)

Sol:
  All x Exist y (x*y = x) = T
  Exist y All x (x*y = x) = T
  Exist x All y (x*y != x) = F
  All y Exist x (x*y != x) = F

2. Give a two-sided inference proof that proves (m < n) <=> (m+n)/2 < n. The 
domains of m and n are the set of real numbers.

Sol:
  m < n
<=> {Add n to both sides of "<"}
  m+n < 2n
<=> {Divide both sides of "<" by 2}
  (m+n)/2 < n

3. Give a by-contradiction proof to prove (x is integer and y is not integer)=>
(x+y = z is not integer). The domains of x, y, and z are the set of all real 
numbers.

Sol:
  (x is integer and y is not integer and x+y = z is integer)
=> {(x is integer and z is integer)=>(z-x is integer)}
  (x is integer and y is not integer and x+y = z is integer and z-x = y is 
  integer)
<=> {(y is not integer and y is integer) <=> F}
  F

4. Give an indirect inference proof to prove:
(3n + 2 is odd)=>(n is odd)
where the domain of n is the set of all positive numbers.

Sol:
  n is even
<=> {Definition of even}
  n = 2k, for some integer k
<=> {Compute 3n + 2}
  3n + 2 = 6k + 2, for some integer k
<=> {Arithmetics}
  3n + 2 = 2(3k + 1), for some integer k
=> {3k + 1 is integer}
  3n + 2 = 2l, for some integer l
<=> {Definition of even}
  3n + 2 is even

5. Give a by-contradiction proof to prove (a^2 + b^2) >= 2ab where the domains 
of a and b are the set of all real numbers.

Sol:
  (a^2 + b^2) < 2ab
<=> {Arithmetics}
  (a-b)^2 < 0
<=> {T <=> (a-b)^2 >= 0}
  (a-b)^2 < 0 and (a-b)^2 >= 0
<=> {P and not P <=> F}
  F

6. Give an induction proof to prove that, for every non-negative integer n, 
r^0 + r^1 + ... + r^n = (1 - r^(n+1))/(1-r)
where the domain of r is the set of all real numbers excluding 1.
 
Sol:
Let P(n) be the predicate
r^0 + r^1 + ... + r^n = (1 - r^(n+1))/(1-r)

Base Case: n = 0: 
P(0) <=> r^0 = (1-r)/(1-r) <=> 1 = 1 <=> T

Induction Step: 
for every n >= 0, P(n) => P(n+1)

  P(n)
=> r^0 + r^1 + ... + r^n = (1 - r^(n+1))/(1-r)
=> {Add r^(n+1) to both sides of "="}
  r^0 + r^1 + ... + r^(n+1) = (1 - r^(n+1))/(1-r) + r^(n+1)
=> {Arithmetics}
  r^0 + r^1 + ... + r^(n+1) = ((1 - r^(n+1)) + (1-r)*r^(n+1))/(1-r) 
=> {Arithmetics}
  r^0 + r^1 + ... + r^(n+1) = (1 - r^(n+1) + r^(n+1) - r^(n+2))/(1-r) 
=> {Arithmetics}
  r^0 + r^1 + ... + r^(n+1) = (1 - r^(n+2))/(1-r) 
=> {Definition of P(n+1)}
  P(n+1)