TEST 2 MAT 243

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

(-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.

Solve the inequality 1 < ⌈ 2x+1⌉ < 6.

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

True or false? {∅} is the empty set.

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

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

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

True

Solve the inequality 1 ≤ ⌊2x+1⌋ ≤ 6

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

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

[0,4] f being surjective means that B must be the range of f. When you square all the numbers in the interval [-2,2], you get exactly the numbers in the interval [0,4]. That means [0,4] is the range of f.

Given a positive integer n, evaluate

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

We say that two sets are disjoint iff..

their intersection is the empty set. Two sets are disjoint means that they have no elements in common. Visually, their Venn diagrams don't overlap.

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

32 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 The general relationship is |A × B| = |A| |B| .

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. A is a subset of B means that everything in A is also in B. Therefore, every set is a subset of itself. A is a proper subset of B means that A is a subset of B, but there is also something in B that is not in A. A set is never a proper subset of itself.

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

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

If A and B are sets, simplify A ∩ (A ∪ B).

A This is the absorption law for sets.

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}

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 four choices result in a bijection.

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

Pick all that apply. A constant function is always..

increasing decreasing

The function f: [0, ∞) → ℝ; f(x) = x² + 1 is

injective but not surjective The function is injective because for non-negative numbers x and y, x² + 1 = y² + 1 implies x=y. The function is not surjective because the range is [1, ∞), but the codomain is ℝ.

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

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

Simplify [-1, 3] ∪ (-2, 0) ∪ [1, 4).

(-2, 4) On problems such as this one, visualize the intervals by sketching them.

Consider the statements 1. ∅ ∈ ∅ 2. ∅ ⊆ ∅.

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.

Given A = {1,2} and B = {1,3}, find A × B.

A × B = { (1,1), (1,3), (2,1), (2,3) } A × B is the set of all ordered pairs of the form (a,b), where a is in A and b is in B. You must not confuse (a,b) with {a,b}. (a,b) is an ordered pair, like a point in the plane. {a,b} is a set, i.e. an unordered collection, of up to two objects.

Suppose A and B are sets. Match sets that must be equal, based on the given information. A ∪ ∅ ∪ ( B ∩ ∅ ) B - ( A - B ) ( A ∩ B ) ∪ ( A ∩ B ) ( A - B ) ∪ ( A ∩ B ) ∪ ( B - A )

A ∪ ∅ ∪ ( B ∩ ∅ ) = A B - ( A - B ) = B ( A ∩ B ) ∪ ( A ∩ B ) = A ∩ B ( A - B ) ∪ ( A ∩ B ) ∪ ( B - A ) = A ∪ B A ∪ ∅ ∪ ( B ∩ ∅ ) = A ∪ ( B ∩ ∅ ) because a union with the empty set has no effect. A ∪ ( B ∩ ∅ ) = A ∪ ( ∅ ) because an intersection with the empty set is always empty. A ∪ ( ∅ ) = A because a union with the empty set has no effect. To simplify B - ( A - B ), we remember that the set difference A - B can be expressed as A ∩ Bᶜ (complement of B). Therefore, B - ( A - B ) = B ∩ (A -B)ᶜ = B ∩ ( A ∩ Bᶜ )ᶜ. By De Morgan, ( A ∩ Bᶜ )ᶜ = Aᶜ ∪ B. Therefore, B - ( A - B ) = B ∩ (Aᶜ ∪ B). By the absorption law, B ∩ (Aᶜ ∪ B) = B. ( A ∩ B ) ∪ ( A ∩ B ) = A ∩ B by the idempotent law. To simplify ( A - B ) ∪ ( A ∩ B ) ∪ ( B - A ), we again express the set difference using the complement: A - B = A ∩ BᶜB - A = Aᶜ ∩ B. Now we reverse distribute and use the identity law: ( A - B ) ∪ ( A ∩ B ) = (A ∩ Bᶜ) ∪ ( A ∩ B ) = A ∩ (Bᶜ ∪ B) = A ∩ U = A. Therefore, using the associative law, ( A - B ) ∪ ( A ∩ B ) ∪ ( B - A ) = A ∪ ( B - A ) = A ∪ ( Aᶜ ∩ B ). Finally, we use the distributive law, followed by the identity law: A ∪ ( Aᶜ ∩ B ) = (A ∪ Aᶜ) ∩ (A ∪ B ) = U ∩ (A ∪ B ) = A ∪ B.

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: Let a = 1.5 b = 2.5Then ab = 3.75 and ⌊a⌋ = 1 ⌊b⌋ = 2 ⌊ab⌋ = 3 but ⌊a⌋⌊b⌋ = 2. Thus, ⌊a⌋⌊b⌋≠ ⌊ab⌋ when a = 1.5 b = 2.5.

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 with one element, and therefore not empty.

Is the following a correct proof that for every integer x, x + y = 2 has an integer solution y? Suppose x = 1. Then x + y = 2 has y = 1 as a solution. Select True for yes, False for no.

False We cannot prove a statement for all x by treating one example case of x.

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

False 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)}.

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

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 }.

Is the following a correct proof that there is an integer solution of 3n + 5 = 8? Suppose 3n+5=8 for some integer n. Then 3n=3 and n=1. Select True for yes, False for no.

False The alleged proof commits the cardinal mistake of assuming the conclusion. You can't prove existence of a solution if you start with the assumption of existence.

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.

Check the properties that the function f: [-2, 0) → [0,4], f(x) = x² has.

Injective Decreasing Strictly decreasing The function is not surjective because the range is (0,4], but the codomain is [0,4]. The 0 is included in the codomain but not in the range. f is strictly decreasing - the easiest way to verify this is by evaluating the derivative, f'(x) = 2x, which is negative on the domain of f. Since f is strictly decreasing, it is also decreasing, and injective.

Consider the sequence aₙ = 2+5n.

It is arithmetic with common difference 5. If you know that the general form of an arithmetic sequence is aₙ = a + nd, where d is the common difference, then you see that aₙ = 2 + 5n is an arithmetic sequence with common difference d = 5. If you didn't know this, then you could confirm it algebraically. The sequence is arithmetic because aₙ₊₁ - aₙ is always the same number: aₙ₊₁ - aₙ = (2 + 5(n + 1)) - (2 + 5n) = 5. This shows not only that the sequence is arithmetic, but arithmetic with common difference 5. You can see that aₙ = 2 + 5n is not geometric by looking at the first three terms: a₀ = 2 a₁ = 7 a₂ = 12. The ratio a₁/a₀ is 7/2, but the ratio a₂/a₁ is 12/7, a different number.

Consider the sequence aₙ = 3·11ⁿ

It 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.

For all positive integers n, there is an even integer k such that n - 1/n < k < n + 2 + 1/n.

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.

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 which implies a² - b² =0. Factoring the right hand side of the equation, we obtain (a-b)(a+b)=0. Thus, either a=b or a=-b. Since a and b are in [0, 1], a=-b is only possible when a=b=0. Thus, we have proved that" if f(a)=f(b), then a = b" which proves that f is injective. 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.

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. 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 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). A and B are complements of each other in U iff A ∪ B = U. The set of positive real numbers and the set of negative real numbers are not complements of each other in the set of real numbers because their union is missing the number 0. The set of even integers and the set of odd integers are complements of each other in the set of all integers because their union is the set of integers. Every integers is even or odd. The set of rational numbers and the set of irrational numbers are complements of each other in the set of all real numbers because their union is the set of real numbers. Every real number is rational or irrational

Consider the statements 1. ∅ ∈ 𝒫 (∅) 2. ∅ ⊆ 𝒫 (∅). The notation 𝒫 (S) means the power set of S.

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

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

You can always modify a non-injective function f: A→B to become injective, by replacing A by a suitable proper subset of A. By redefining the codomain of a function to make it equal to its range, you can always force the function to become surjective.

Let f:R→R;f(x)=x². Evaluate the two sets f((-1,1)) and f⁻¹((0,1]).

f((-1,1))=[0,1); f⁻¹((0,1])=[-1,0) U (0,1] Let f : A → B. Then 1.The image of a subset X of A is the set of outputs of the inputs in X. Formally, f(X)={f(x)|x in X}. 2. The preimage of a subset Y of B is the set of inputs in A whose outputs lie in Y. Formally, f⁻¹(Y)={x| f(x) in Y}

Pick all that apply. Any function f: {0} → {0} is..

increasing strictly increasing decreasing strictly decreasing constant surjective injective Functions defined on singletons (one-element sets) are always increasing, decreasing, strictly increasing and strictly decreasing. This may seem intuitively strange (how can you say a function has these properties when its graph consists of only one point?) but it makes sense based on the formal definition. f increasing means that for all a and b in the domain of f, a < b implies f(a) ≤ f(b). Written symbolically, f increasing is the case if and only if ∀a,b ( a < b → f(a) ≤ f(b) ). The for all quantification produces only a single pair of values: a = b = 0. In that case, the premise of the conditional, a < b, is false. With a false premise, the conditional a < b → f(a) ≤ f(b) is true. Thus the entire quantified statement is true. f strictly increasing means that for all a and b in the domain of f with a < b, you have f(a) < f(b). Symbolically written, f is strictly increasing if and only if ∀a,b ( a < b → f(a) < f(b) ). This is true for the same reason that was just given for increasing. You can similarly justify why f is decreasing and strictly decreasing. f is constant because its range only has one value. f is injective because each output only comes from one input. The sole output 0 comes from only one input, 0. f is surjective because its range is {0}, which is equal to the codomain.

The function f: ℝ→ ℝ ; f(x) = x² + 1 is

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.

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.

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. 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.

What is the power set of the power set of {1} ?

{∅, {∅}, {{1}}, {∅, {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}}}.

Simplify {∅} ∪ ∅.

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

Suppose A is a set. Simplify A × ∅.

∅ The cartesian product A × ∅ is by definition the set of all ordered pairs (a,b) with a ∈ A and b ∈ ∅. Since there are no b ∈ ∅, there are no such ordered pairs. Thus A × ∅ is empty.

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.