Discrete exam 2
( A ∩ B ) ∪ ( A ∩ B )=
( A ∩ B )
If f is the absolute value function f: ℝ → ℝ; f(x) = |x|, determine the pre-image f⁻¹([-2,1)).
(-1,1) ⁻¹([-2,1)) is by definition the set of x values that satisfy -2 ≤ |x| < 1. Since any absolute values is at least zero, and therefore also at least -2, that inequality is equivalent to |x| < 1. This, in turn, is equivalent to -1 < x < 1.
A ∪ ∅ ∪ ( B ∩ ∅ ) =
A
A constant function is..
increasing and decreasing
Solve 1 < ⌈ 2x+1⌉ < 6.
(0,2] 1 < ⌈ 2x+1⌉ < 6 is equivalent to 1 < 2x+1 ≤ 5, which is equivalent to 0 < x ≤ 2.
Consider the statements: 1. Every set is a subset of itself. 2. Every set is a proper subset of itself.
1 Is true, 2 is false
Solve the inequality 1 ≤ ⌊2x+1⌋ ≤ 6.
1 ≤ ⌊2x+1⌋ ≤ 6 is equivalent to 1 ≤ 2x+1 < 7, which is equivalent to 0 ≤ x < 3.
True or false 1. ∅ ∈ ∅ 2. ∅ ⊆ ∅.
1. False the empty set cannot be an element of the empty set because it has no elements. 2. True. In general, the empty set is a subset of any set.
True or false 1. ∅ ∈ 𝒫 (∅) 2. ∅ ⊆ 𝒫 (∅).
1. True. ∅ is an element of the power set { ∅,{ ∅}} 2. True. ∅ is a subset of any set
If A and B are sets, simplify A ∩ (A ∪ B).
A (Absportion law)
( A - B ) ∪ ( A ∩ B ) ∪ ( B - A )=
B
B - ( A - B ) =
B
You can always force the function to become surjective.
By redefining the codomain of a function to make it equal to its range
True or false? If a and b are real numbers and ⌊ ⌋ represents the floor function, then ⌊ab⌋ = ⌊a⌋ ⌊b⌋.
False, An example quickly shows that multiplying first and then rounding down is not the same as rounding down and then multiplying: a = 1.5 b = 2.5 ab = 3.75 ⌊a⌋ = 1 ⌊b⌋ = 2 ⌊ab⌋ = 3 but ⌊a⌋⌊b⌋ = 2.
True or False? If a set is empty, so is its power set.
False, The power set of the empty set is {∅}, which is a set containing the empty set.
True or false. The interval [1,2] is equal to the statement 1 ≤ x ≤ 2.
False, a set cannot be equal to a statement
True or false? If a set A has 5 elements, and a set B has 7 elements, then the union A∪B must have 5+7=12 elements.
False, while there could be 12 elements there could also be UP TO 12 elements because A and B could be sharing the same elements.
True or false? {∅} is the empty set.
False. that is the set of the empty set
If A has 5 elements, how many elements does the power set of A have?
If A has n elements, then the power set of A has 2ⁿ elements. 2^ 5 = 32
The function f: [0, ∞) → ℝ; f(x) = x² + 1 is injective, surjective, bijective, neither?
Injective but not surjective The function is injective because for non-negative numbers x and y, x² + 1 = y² + 1 implies x=y. It is not surjective because there are negative numbers not being outputted in the codomain.
The function f: ℝ→ ℝ ; f(x) = x² + 1 is : injective, surjective, bijective or neither?
Neither, it is not injective because f(-1) and (1) produce the same output. It is not surjective because the codomain contains numbers less than 1 that are not outputted because we are always adding 1
Which statements are true: The rational numbers are a subset of the real numbers. The rational numbers are a proper subset of the real numbers. The real numbers are a subset of the rational numbers. The real numbers are a proper subset of the rational numbers.
The first two
How many elements are in the intersection of (2,4) and (3,5) ?
The intersection of (2,4) and (3,5) is the interval (3,4). This interval contains infinitely many real numbers between 3 and 4, such as 3.05, π, the square root of 10 and 3.99.
Which are complements of each other in the universal set of U The set of positive real numbers, the set of negative real numbers (U = the set of real numbers). The set of even integers, the set of odd integers (U = the set of all integers). The set of rational numbers, the set of irrational numbers (U = the set of all real numbers).
The last two
What is the power set of the power set of {1} ?
The power set of {1} is {∅,{1}}. If we let A = ∅ and B = {1}, then the power set of {1} is {A,B}. We find the power set of that to be { ∅, {A}, {B}, {A,B} }. Substituting the definitions of A and B, we find the power set of the power set of {1} to be { ∅, {∅}, {{1}}, {∅, {1}}}.
How can we calculating the sum 3⁴ + 3⁵ + .. + 3⁹ computationally efficiently?
The sum has only 6 terms. It is an efficient solution to evaluate each term individually and add them.
True or false? The sum of the squares of the first n positive integers is n(n+1)(2n+1)/6.
True. This is the summation of square formula
If f: [-2,2] → B; f(x) = x² is surjective, then B =
[0,4]
What sequence is aₙ = 2+5n.
arithemetic with common difference 5
Surjective
describes a mapping in which each element of the range is the target of some element of the domain. Also, onto.
Injective
f(a)=f(b) then x1 = y2
What sequence is this aₙ = 3·11ⁿ.
geomotric sequence of common quotient of 11 The sequence is geometric because it has the general form of a geometric sequence: aₙ = a·qⁿ.
Any function f: {0} → {0} is..
increasing, strictly increasing, decreasing, strictly decreasing, constant, surjective, injective
You can always modify a non-injective function f: A→B to become injective by
replacing A by a suitable proper subset of A.
Is the function f: [0, 1] → ℝ; f(x) = x² injective? Is it surjective? Prove both of your answers based on the definitions of injective and surjective.
the function is injective. Suppose f(a)=f(b) for some a and b in [0,1]. By definition of f, that means a² = b². By applying the square root function to both sides, we get |a|=|b|. Since a and b are not negative, the absolute values have no effect. Thus a = b. The function is not surjective. Let a ∈ [0,1]. Then a² ≥ 0. Hence y values less than 0, which occur in the codomain, do not occur as outputs of f.
If we visualize ℝ² as the plane, then ℤ² is
the grid of points (x,y) with integer coordinates. ℤ² means ℤ×ℤ, which by definition of Cartesian product is all points (x,y) with integers x and y. That set of points forms a grid in the plane.
We say that two sets are disjoint iff..
the intersection is the empty set
If the universal set is [0,2], what is the complement of (0,1)?
{0} ∪ [1,2] (0,1) is the set of real numbers strictly between 0 and 1, i.e. the set of real x that satisfy 0 < x < 1. By de Morgan, the complement of that is all x in the universal set that satisfy x ≥ 1 or x ≤ 0. The universal set is the set of real numbers that satisfy 0 ≤ x ≤ 2. Combining these two conditions, we find that the complement contains the number 0 and the numbers x ≥ 1.
If f is the ceiling function from ℝ to ℝ, what is f((1/2, 3/2))?
{1,2} The ceiling of a real number is always an integer. Thus, the ceiling-image of any set of real numbers is always a set of integers.
{∅} ∪ ∅.
{∅} Taking the union of the empty set has no effect on the empty any set
Use a summation formula we learned in class to compute the sum of the integers from 1000 to 5000. Write the answer in un-simplified form.
½·5000·5001 - ½·999·1000 The sum of the integers from 1 to n is ½·n·(n+1).
Given a universal set U, the complement of U is..
∅ The complement of the universal set is by definition every element in the universal set that is not in the universal set. There are no such elements.
