Description
Problem 1: Sets
Which of the following sets are equal? Show your work step by step.
- {t : t is a root of x2 – 6x + 8 = 0}
- {y : y is a real number in the closed interval [2, 3]}
- {4, 2, 5, 4}
- {4, 5, 7, 2} – {5, 7}
- {q: q is either the number of sides of a rectangle or the number of digits in any integer between 11 and 99}
(Solution)
1
– Homework #2 2
Problem 2: Cartesian Product of Sets (15 points)
Explain why (A × B) × (C × D) and A × (B × C) × D are not the same.
(Solution)
| Problem 3: Cartesian Product of Sets in Algorithms |
(25 points)
Let A, B and C be sets which have different cardinalities. Let (p, q, r) be each triple of A × B × C where p ∈ A, q ∈ B and r ∈ C. Design an algorithm which finds all the triples that are satisfying the criteria: p ≤ q and q ≥ r. Write the pseudo code of the algorithm in your solution.
For example: Let the set A, B and C be as A = { 3, 5, 7 }, B = { 3, 6 } and C = { 4, 6, 9 }. Then the output should be : { (3, 6, 4), (3, 6, 6), (5, 6, 4), (5, 6, 6) }.
(Note: Assume that you have sets of A, B, C as an input argument.)
(Solution)
for write a condition do
end
When you want to write a while loop, you can use:
while write a condition do
If you need to return, use return end
For any additional things you have to do while writing your pseudo code, Google ”How to use algorithm2e in Latex?”.
– Homework #2 3
Problem 4: Relations (3+3+3+3+3+3+3=21 points)
Determine whether the relation R on the set of all integers is reflexive, symmetric, antisymmetric, and/or transitive, where (x, y) ∈ R if and only if (a) x 6= y.
(Solution)
- xy ≥ 1.
(Solution)
- x = y + 1 or x = y – 1.
(Solution)
- x is a multiple of y.
(Solution)
- x and y are both negative or both nonnegative.
(Solution)
- x ≥ y2. (Solution)
- x = y2.
(Solution)
| Problem 5: Functions
If f and f ◦ g are one-to-one, does it follow that g is one-to-one? Justify your answer. (Solution) |
(15 points) |
| Problem 6: Inverse of Functions | (7+7=14 points) |
Let f be the function from R to R defined by f(x) = x2. Find
(a) f−1 ({ x | 0 < x < 1 })
(Solution)
(b)f−1 ({ x | x > 4 })







