Test 2 Practice

If f is the absolute value function f: ℝ → ℝ; f(x) = |x|, determine the pre-image f⁻¹( [-2,1) ). (1,2] [0,1) (-1,1) [0,2]

(-1,1) f⁻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.

Simplify [-1, 3] ∪ (-2, 0) ∪ [1, 4).

(-2, 4)

Pick all that apply. A constant function is.. strictly increasing increasing strictly decreasing decreasing none of the other options.

increasing, decreasing

Solve the inequality 1 < ⌈ 2x+1⌉ < 6. (0,2] (0,5/2] (-1/2,5/2] (1/2, 2]

(0,2] 1 < ⌈ 2x+1⌉ < 6 is equivalent to 1 < 2x+1 ≤ 5.

True or false? If A and B are sets, then A × B = B × A.

You can see that A × B = B × A cannot be a set identity by counter-example. Suppose A = {1} and B = {2}. Then A × B = { (1,2) } while A × B = { (2,1) }.

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. 1. is false, 2. is true. They are both true. They are both false.

1. is false, 2. is true.

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. 1. is false, 2. is true. They are both true. They are both false.

1. is true, 2 is false

How many elements does the set { 0, { 1, {2,3}} } contain? 1 element 2 elements 3 elements 4 elements

2 elements The two elements { 0, { 1, {2,3}} } contains are 0 and { 1, {2,3}}.

If A has 5 elements, how many elements does the power set of A have?

32.0 (with margin: 0.0) If A has n elements, then the power set of A has 2ⁿ elements.

If A and B are sets, |A| = 10 and |B| = 5, then |A × B| = ?

50 Correct Answers 50.0 (with margin: 0.0)

The interval [1,2] is equal to the statement 1 ≤ x ≤ 2. True False

A set cannot be equal to a statement. Sets and statements are different categories of objects. It is true that the statement defines the membership condition for the set, i.e. it defines whether a number x is in the set. Thus, [1,2] = { x | 1 ≤ x ≤ 2 }.

True or False? If a set is empty, so is its power set. True False

False The power set of the empty set is {∅}, which is a set with one element, and therefore not empty.

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 the union could well have 12 elements, all we can guarantee is that it has up to 12 elements, because there could be elements that A and B have in common.

How many elements are in the intersection of (2,4) and (3,5) ?

Infinitely Many The intersection of (2,4) and (3,5) is the interval (3,4). This interval contains infinitely many real numbers, such as 3.05, π, the square root of 10 and 3.99.

Consider the sequence aₙ = 3·11ⁿ. It is arithmetic with common difference 3. It is arithmetic with common difference 11. Is is geometric with common quotient 3. Is is geometric with common quotient 11. It is neither arithmetic nor geometric. It is both arithmetic and geometric. The common quotient is 3, the common difference is 11.

Is is geometric with common quotient 11. You can see that aₙ = 3·11ⁿ is not arithmetic by computing the first three terms: a₀ = 3 a₁ = 33 a₂ = 363. The difference between the first two terms is 30, but the difference between the 2nd and 3rd term is 330. The sequence is geometric because it has the general form of a geometric sequence: aₙ = a·qⁿ.

What do you get when you index shift the sigma sum LaTeX: \sum_{k=5}^9 (k+7)^3 ∑ k = 5 9 ( k + 7 ) 3 so that k starts at 0?

LaTeX: \sum_{k=0}^4 (k+12)^3 ∑ k = 0 4 ( k + 12 ) 3

The two sets {1, 2} and {2, 1} are equal. True False

The set is an unordered data structure. Both notations represent the set that contains the two numbers 1 and 2.

Consider the statements 1. ∅ ∈ 𝒫 (∅) 2. ∅ ⊆ 𝒫 (∅). The notation 𝒫 (S) means the power set of S. They are both true. 1. is true, 2. is false. 1. is false, 2. is true. They are both false.

They are both true. The power set of ∅ is { ∅, { ∅ } }. ∅ is an element of that. ∅ is a subset of any set.

Solve the inequality 1 ≤ ⌊2x+1⌋ ≤ 6. [0, 3) [0,3] [0, 5/2) [0, 5/2]

[0, 3) 1 ≤ ⌊2x+1⌋ ≤ 6 is equivalent to 1 ≤ 2x+1 < 7.

If f : [-2,2] → B; f(x) = x² is surjective, then B = [-4,4] [0,4] [4,4]

[0,4]

What is the power set of the power set of {1} ? a. {1} b. {∅,1} c. {∅,{1}} d. {∅, {∅}, { {1} }, {∅, {1} } } e. {∅, {∅}, {1}, {∅, {1} } } f. {∅, {∅}, {1}, {∅, 1} }

d. The power set of of the power set of {1} is {∅, {∅}, { {1} }, {∅, {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}} }.

True or false? If a and b are real numbers and ⌊ ⌋ represents the floor function, then ⌊ab⌋ = ⌊a⌋ ⌊b⌋. You Answered Correct Answer

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? {∅} is the empty set. True False

false {∅} is a set that contains one element and is therefore not empty.

The function f: [0, ∞) → ℝ; f(x) = x² + 1 is surjective but not injective injective but not surjective bijective neither surjective nor injective

injective but not surjective

1/1*2 + 1/2*3 + 1/3*4...+1/n(n+1) Given a positive integer n, evaluate

n/n+1 This telescoping sum is evaluated in the lecture

Simplify {∅} ∪ ∅.

{ ∅ } Taking the union with an empty set has no effect on any set.

We say that two sets are disjoint iff..

their intersection is the empty set.

True or false? The sum of the squares of the first n positive integers is n(n+1)(2n+1)/6. True False

true

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} (1,2) [1,2] the empty set

{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. The real numbers x that satisfy 1/2 < x ≤ 1 have a ceiling of 1. The real numbers x that satisfy 1 < x < 3/2 have a ceiling of 2.

Consider the statements 1. ∅ ∈ ∅ 2. ∅ ⊆ ∅. They are both true. 1. is true, 2. is false. 1. is false, 2. is true. They are both false.

1. is false, 2. is true. The empty set cannot be an element of the empty set because the empty set has no elements. The empty set is a subset of the empty set because all elements of the empty set are in the empty set. If you think that the last statement is false, then you must think that the negation is true. The negation of all elements of the empty set are in the empty set is there is an element in the empty set that is not in the empty set. Thus, the burden of proof is on you to produce such an element. However, you can't do that, because you can't find any elements in the empty set. Thus, you are forced to concede that all elements of the empty set must be in the empty set. In general, the empty set is a subset of any set.

Evaluate the sumation from k = 5 to 1000 for K^2 + 6k + 9

336845374.0 (with margin: 0.0)

Given the formula f(x) = x², pick all domain/codomain pairs A,B that would make f: A → B bijective. A = {1}, B = {1} A = [-1,0], B = [0,1] A = [-1,0), B = (0,1] A = {-1, -1/2, 1/4, 1/8}, B = {1/64, 1/16, 1/4, 1}

All are correct

Consider the sequence aₙ = 2+5n. It is arithmetic with common difference 2. It is arithmetic with common difference 5. Is is geometric with common quotient 2. Is is geometric with common quotient 5. It is neither arithmetic nor geometric. It is both arithmetic and geometric. The common quotient is 2, the common difference is 5. It is both arithmetic and geometric. The common quotient is 5, the common difference is 2. We can't determine whether it is arithmetic or geometric since the definition did not include the information whether it is 1-based or 0-based.

It is arithmetic with common difference 5.

The following is a possible proof writing assignment for test 2. Develop a solution on paper. You cannot enter your solution on this practice test, but on test 2, you will be asked to enter your solution into a text box. After you are done with this practice test, you will find a reference solution to this proof in the detailed response feedback. If you are not certain whether the proof you developed was correct, post it on piazza. For all positive integers n, there is an even integer k such that n - 1/n < k < n + 2 + 1/n. Answer True if you developed a solution on paper for this problem as instructed.

Possible solution: Assume n is an arbitrary positive integer. If n is even, then choose k=n+2, which is also even. Since n - 1/n < n < n + 1/n, it follows that n - 1/n < n < k = n+2 < n+2 +1/n. If n is odd, then choose k = n+1, which is even. Since n - 1/n < n < n + 1/n, n - 1/n < n < k= n+1 < n+2 + 1/n. Thus, we have shown that either way, an even k exists that satisfies n - 1/n < k < n+2 + 1/n.

Determine which one of the following students answered the following problem correctly: 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. Star: the function is neither injective nor surjective. It is not injective because f(-1)=f(1). It is not surjective because the negative numbers are missing from its range. Sun: 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. Sky: the function is injective because it is strictly increasing. It is not surjective because the range is [0,1], which is a proper subset of the codomain. Blaze: the function is injective. Suppose f(a)=f(b). By definition of f, that means a² = b². By applying the square root function to both sides, we get a = b. The function is not surjective. Let a ∈ [0,1]. Then a² ≥ 0. Hence y values less than 0 do not occur as outputs of f. Shadow Moon: the function is injective. Suppose a=b. Then f(a)=f(b). Thus, the function satisfies the definition of injectivity. 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. Flow: the function is injective because its derivative f(x)=2x is positive for all x > 0. The mean value theorem we know from calculus 1 then implies that the function is injective. The function is not surjective. Using calculus again, the absolute maximum of the function is 1 and the absolute minimum is 0. We can now apply the intermediate value theorem from calculus since f is continuous. Therefore, the range of f is [0,1], which is not the stated codomain.

Sun

Which methods can we use for calculating the sum 3⁴ + 3⁵ + .. + 3⁹ computationally efficiently? We rewrite the sum as (3⁰ + 3¹ + 3² + 3³ + 3⁴ + 3⁵ + ... + 3⁹) - (3⁰ + 3¹ + 3² + 3³) and apply the geometric summation formula to each term. This produces the solution LaTeX: \frac{3^{10}-1}{2}-\frac{3^4-1}{2} 3 10 − 1 2 − 3 4 − 1 2 . The sum has only 6 terms. It is an efficient solution to evaluate each term individually and add them. We can factor out the common power 3⁴ and thereby write the sum as 3⁴ (3⁰ + 3¹ + 3² + 3³ + 3⁴ + 3⁵). Then we apply the geometric summation formula to simplify this into LaTeX: 3^4\cdot\frac{3^6-1}{2} 3 4 ⋅ 3 6 − 1 2 . You Answered Using laws of exponentiation, we can just add up the exponents. The sum is equal to LaTeX: 3^{4+5+6+....+8+9}=3^{39}.

The sum has only 6 terms. It is an efficient solution to evaluate each term individually and add them.

he function f: ℝ→ ℝ ; f(x) = x² + 1 is surjective but not injective injective but not surjective bijective neither surjective nor injective

neither surjective nor injective The function is not injective because there are distinct inputs that have the same output, such as -1 and 1. The function is not surjective because the codomain contains numbers less than 1, which are not produced as outputs.

Check all that apply. a. The rational numbers are a subset of the real numbers. b. The rational numbers are a proper subset of the real numbers. c. The real numbers are a subset of the rational numbers. d. The real numbers are a proper subset of the rational numbers.

a, b The real numbers comprise the rational numbers (the quotients of integers) as well as those numbers that cannot be written as quotients of integers, aka the irrational numbers. That makes the rational numbers a proper subset of the real numbers. A proper subset of a set S is always also a subset of S.

Check all true statements. a. You can always modify a non-injective function f: A→B to become injective, by replacing A by a suitable proper subset of A. b. You can always modify a non-injective function f: A→B to become injective, by replacing B by the range of f. c. By redefining the codomain of a function to make it equal to its range, you can always force the function to become surjective. d. By modifying the codomain of a function to be equal to the domain, you make the function bijective.

a, c

If A and B are sets, simplify A ∩ (A ∪ B). A B The Universal Set The Empty Set

a. This is the absorption law for sets.

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. a. ½·5000·5001 - ½·999·1000 b. ½·5000·5001 - ½·1000·1001 c. 5000·5001 - 999·1000 d. 5000·5001 - 1000·1001

a. You find this sum as the difference of the sum of the integers from 1 to 5000 and the sum of the integers from 1 to 999 (NOT from 1 to 1000). The sum of the integers from 1 to n is ½·n·(n+1).

If we visualize ℝ² as the plane, then ℤ² is a. the grid of points (x,y) with integer coordinates. b. all points (x,y) with x and y being perfect squares (squares of integers).

a. 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.

Given a universal set U, the complement of U is.. Correct Answer a. ∅ b. U c. The answer cannot be determined because we don't know what U is.

a. 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.

Identify all mistakes made in the following proof that for each integer n, there is an integer k such that n < k < n+2. a. Suppose n is an arbitrary integer. Therefore, k = n + 1 for all integers n. This means that n < n + 1 < n + 2. This proves that an integer k exists. b. Suppose n is an arbitrary integer. Therefore, k = n + 1 The choice of k = n + 1 does not follow from the fact that n is an integer. The inappropriate word "Therefore" should be replaced by "Pick", "Select", "Choose" or other words to that effect. Therefore, k = n + 1 for all integers n. After we decided that n is some integer, n is just that - an integer. It's no longer a free variable. It makes no sense to apply universal quantification to it, again just like it makes no sense to say " 2 + 3 = 5 for all integers 5". Therefore, k = n + 1 for all integers n. This means that n < n + 1 < n + 2. It makes no sense to draw a conclusion from a variable definition that does not reference the variable. n < n+1 < n+2 is true for all n regardless of what k is. The proper conclusion to be drawn from the definition of k is that n < k < n+2. This proves that an integer k exists. We know that integers exist. If we're going to make a concluding statement that summarizes the theorem we have proved, we must quote the theorem correctly, not a caricature of our theorem. A correct concluding statement would be: this proves that for every integer n, an integer k exists such that n < k < n+2.

all are correct

Pick all that apply. Any function f: {0} → {0} is.. increasing strictly increasing decreasing strictly decreasing constant surjective injective

all are correct.

Select all sets that are complement pairs (i.e. the two sets are complements of each other). The universal set U is given in each situation. a. The set of positive real numbers, the set of negative real numbers (U = the set of real numbers). b. The set of even integers, the set of odd integers (U = the set of all integers). c. The set of rational numbers, the set of irrational numbers (U = the set of all real numbers).

b, c