MAT 243 Practice for Test 2
Simplify [-1, 3] ∪ (-2, 0) ∪ [1, 4).
(-2,4) On problems such as this one, visualize the intervals by sketching them.
Given the formula f(x) = x^2, pick all domain/codomain pairs A,B that would make f: A → B bijective. a. A = {1}, B = {1} b. A = [-1,0], B = [0,1] c. A = [-1,0), B = (0,1] d. A = {-1, -1/2, 1/4, 1/8}, B = {1/64, 1/16, 1/4, 1}
ALL
Pick all that apply. Any function f: {0} → {0} is.. a. increasing b. strictly increasing c. decreasing d. strictly decreasing e. constant f. surjective g. incjective
ALL Functions deNned 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.
True or false, 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 }.
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 Nrst and then rounding down is not the same as rounding down and then multiplying: Let a = 1.5 b = 2.5 Then 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.
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? The following is a valid proof of the theorem that for every integer n, there is an integer k such that n < k < n+2. Assume n is an integer. Let k be an integer such that n < k < n+2. This shows that for every n, an integer k such that n < k < n+2 exists.
False The proof is invalid because the argument given is circular reasoning. The existence of k is not shown, merely postulated.
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.
Identify all mistakes made in the following proof that for each integer n, there is an integer k such that n < k < n+2. 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. a. 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. b. 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". c. 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. d. There is nothing wrong with this proof.
a, b, c
What's wrong with this proof that an even number plus an even number is even? Suppose n and m are even numbers. Then n is 2k and m is 2k as well. It follows that their sum is 2k plus 2k. By using the laws of algebra, we can show that 2k plus 2k is 4k, or 2 times 2k. Hence, n plus m is even by definition of even number. a. The quantiacation "for some integer k" is missing from "n = 2k". Without the quantification, the statement "n = 2k" is meaningless because k is an undeaned variable. b. Variable name literalism: while an even number can always be written as 2k for some integer k, two potentially different even numbers will not be equal to 2k for the same k. A correct proof must use two different variable names like this: Then n = 2p and m = 2q for some integers p, q. c. There is a meaningless "can" statement we can show. In proofs, you don't declare that you can show. You have to do the showing. d. nothing e. The proof is much too verbose. A good proof is concise. We don't usually spell out standard arithmetic operators in English: we usually write + instead of "plus". We also don't spell out obvious things like the fact that we are using the laws of algebra when we're doing algebra.
a, b, c, e
Which methods may we use for correctly evaluating the sum 3⁴ + 3⁵ + .. + 3⁹ ? a. We rewrite the sum as (3⁰ + 3^1 + 3^2 + 3^3 + 3⁴ + 3⁵ + ... + 3⁹) - (3⁰ + 3^1 + 3^2 + 3^3) and apply the geometric summation formula to each term. This produces the solution. b. The sum has only 6 terms with relatively low exponents. It is still practical to evaluate each term individually and add them. c. We can factor out the common power 3⁴ and thereby write the sum as 3⁴ (3⁰ + 3. + 3. + 3. + 3⁴ + 3⁵). Then we apply the geometric summation formula to simplify this into d. Using laws of exponentiation, we can just add up the exponents. The sum is equal to 3^(4+5+6+....+8+9) = 3^(39).
a,b,c You can't add the exponents in a geometric summation. Consider the example 2. + 2⁴ . That's 8 + 16, which is 24. If you tried to evaluate this using the incorrect law of exponentiation that you can just get the exponents, you get 2⁷, which is 128.
What is wrong with this argument? Check all that apply. Suppose x = 2 2x = 4 2x + 1 = 5 a. The logical "glue" is missing. Lines 2 is a consequence of line 1, and line 3 is a consequence of line 2. Words and phrases like "Thus", "Therefore", "It follows", etc. need to be used to make it clear that we are drawing conclusions. b. nothing c. Punctuation is missing. Each sentence needs to be terminated with a period (.). d. The assumption that x is an integer was not stated. Line 1 should have been "Suppose x is an integer and x = 2".
a,c
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 Failure to be injective means that a function has "too many" inputs. You can always artificially create injectivity by restricting the function to a subset of its domain where outputs are unique. Failure to be surjective means that a function has "too many" potential outputs. You can always artificially create surjectivity by changing the codomain to the range.
Suppose A and B are sets. Match sets that must be equal, based on the given information. a. A ∪ ∅ ∪ ( B ∩ ∅ ) b. B - ( A - B ) c. ( A ∩ B ) ∪ ( A ∩ B ) d. ( A - B ) ∪ ( A ∩ B ) ∪ ( B - A )
a. A 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. b. B 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. c. A ∩ B ( A ∩ B ) ∪ ( A ∩ B ) = A ∩ B by the idempotent law. d. A ∪ B To simplify ( A - B ) ∪ ( A ∩ B ) ∪ ( B - A ), we again express the set difference using the complement: A - B = A ∩ Bᶜ B - 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.
Given A = {1,2} and B = {1,3}, and A x B. a. A x B = { (1,1), (1,3), (2,1), (2,3) } b. A x B = { {1,1}, {1,3}, {2,1}, {2,3} } c. A x B = { (1,1), (2,3) } d. A x B = { {1,1}, {2,3} }
a. A x B = { (1,1), (1,3), (2,1), (2,3) } A x 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.
Consider the statements 1. ∅ ∈ 𝒫 (∅) 2. ∅ ⊆ 𝒫 (∅). The notation 𝒫 (S) means the power set of S. a. They are both true. b. 1. is true, 2. is false. c. 1. is false, 2. is true. d. They are both false
a. 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. a. [0, 3) b. [0,3] c. [0, 5/2) d. [0, 5/2]
a. [0, 3) 1 ≤ ⌊2x+1⌋ ≤ 6 is equivalent to 1 ≤ 2x+1 < 7, which is equivalent to 0 ≤ x < 3.
We say that two sets are disjoint iff.. a. their union is the universal set. b. their intersection is the empty set. c. they are not equal.
a. 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.
Given a universal set U, the complement of U is.. 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.
Pick all that apply. A constant function is always... a. strictly increasing b. increasing c. strictly decreasing d. deceasing e. none of the other options
b,d Recall that f increasing means that for all a and b in the domain of f, a < b implies f(a) ≤ f(b). Written symbolically, f is increasing if and only if ∀a,b ( a < b → f(a) ≤ f(b) ). For a constant function f, the conclusion of the conditional, f(a) ≤ f(b), is always true because f(a) = f(b) for all a, b in the domain. Thus the entire quantified statement is true. f strictly increasing means that for all a and b in the domain of f, a < b implies f(a) < f(b). For a constant function, f(a) < f(b) is always false, and if there are at least two different numbers in the domain, then there is at least one case of a,b in the domain where a < b. Thus, there is at least one case of a,b in the domain where the conditional is false (true premise but false conclusion). Thus, the for all.. quantification of the conditional is also false. We conclude that a constant function with a domain that has at least two elements is not strictly increasing. Therefore, it is not true that a constant function is always strictly increasing. We can similarly justify the answers for decreasing and strictly decreasing.
Consider the sequence aₙ = 2+5n. a. It is arithmetic with common difference 2. b. It is arithmetic with common difference 5. c. It is geometric with common quotient 2. d. It is geometric with common quotient 5. e. It is neither arithmetic nor geometric. f. It is both arithmetic and geometric. The common quotient is 2, the common difference is 5. g. It is both arithmetic and geometric. The common quotient is 5, the common difference is 2. h. 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.
b. 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 conNrm 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.
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 a. 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. b. 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. c. 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. d. 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. e. 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. 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.
b. SUN Star's apparent counter-example to injectivity ignores the fact that the domain is [0,1]. There is no f(-1). The explanation of why f is not surjective is correct, but it does not constitute proof. Star would have had to show that elements of the codomain are missing from the range, not just plausibly claim it. Similarly, Sky's argument is not proof founded on the definitions. It "delegates" the properties to be proved to other, similarly unproved properties. Blaze's attempted proof of injectivity is question-begging (assuming the conclusion). Applying the square root function implicitly assumes that g : [0, 1] → [0, 1]; g(x) = x² is bijective , because otherwise, it would have no inverse. So the attempted proof implicitly argues that f is injective because g is bijective, and thus also injective. How do we know g is injective? Because of the exact same algebra that we would need to carry out (which is shown in Sun's proof) to show that f is injective. Assuming that g is injective is therefore equivalent to assuming that f is injective, i.e. assuming the conclusion. Shadow Moon's argument for injectivity is based on misunderstanding of that property. That the same input results in the same output is a property of any function. Blaze, Sun and Shadow Moon all give the same correct proof of surjectivity. Flow's argument is not incorrect, but it appeals to powerful (and actually fairly diwcult to prove) calculus facts. It is not the elementary proof that was asked for (elementary meaning based on the deNnitions of injective and surjective.)
If f: [-2,2] → B; f(x) = x^2. is surjective, then B = a. [-4,4] b. [0,4] c. [4,4] d. The answer cannot be determined base on the given information.
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.
The function f: [0, ∞) → ℝ; f(x) = x² + 1 is a. surjective but not injective b. injective but not surjective c. bijective d. neither surjective nor injective
b. 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.
Suppose A is a set. Simplify A x ∅ a. A b. ∅ c. {(A , ∅)} d. {(A , {∅})}
b. ∅ The cartesian product A x ∅ 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 x ∅ is empty.
If f is the absolute value function f: ℝ → ℝ; f(x) = |x|, determine the pre-image f⁻¹([-2,1)). a. (1,2] b. [0,1) c. (-1,1) d. [0,2]
c. (-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.
Consider the statements 1. ∅ ∈ ∅ 2. ∅ ⊆ ∅. a. They are both true. b. 1. is true, 2. is false. c. 1. is false, 2. is true. d. They are both false.
c. 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.
Consider the sequence aₙ = 3.11ⁿ. a. It is arithmetic with common difference 3. b. It is arithmetic with common difference 11. c. It is geometric with common quotient 3. d. It is geometric with common quotient 11. e. It is neither arithmetic nor geometric. f. It is both arithmetic and geometric. The common quotient is 3, the common difference is 11.
d. 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.
The function f: ℝ→ ℝ ; f(x) = x^2 + 1 is a. surjective but not injective b. injective but not surjective c. bijective d. neither surjective nor injective
d. 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.
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. {∅, {∅}, {{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}}}.
If A and B are sets, |A| = 10 and |B| = 5, then |A x B| = ?
50 The general relationship is |A x B| = |A| |B| .
Given a positive integer n, evaluate 1/(n*(n+1))
n/(n+1) This telescoping sum is evaluated in the lecture.
What do you get when you index shift the sigma sum so that k starts at 0? 9 Σ (𝑘 + 7)^3 𝑘=5
4 Σ (𝑘 + 12)^3 𝑘 = 0 Index shifting really means making a substitution for k and then renaming the new variable back to k. To get the index started at 0, we need to introduce the new variable j = k - 5. k = 5 means j = 0. k = 9 means j = 4. k + 7 = (j + 5) + 7 = j + 12. Thus, in terms of j, the sum is 4 Σ (j + 12)^3 j = 0 We get the correct answer from there by renaming the j back to k.
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.
True or false? If A and B are sets, then A x B = B x A.
False You can see that A x B = B x A cannot be a set identity by counter-example. Suppose A = {1} and B = {2}. Then A x B = {(1,2)} while A x B = {(2,1)}.
True or false? {∅} is the empty set.
False {∅} is a set that contains one element and is therefore not empty.
Check all correct ways of evaluating 1000 Σ k^2 + 6𝑘 + 9. k=5
Both are correct
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.
Write a short proof up to three sentences for the statement ∃x∀y(x+y = y). The domain of discourse is the real numbers. First write your proof on a piece of scratch paper, then assemble the proof below.
First sentence: Let x=0. Second sentence: Assume y is an arbitrary real number. Third sentence: Then x+y=y.
Write a short proof up to three sentences for the statement ∀x∃y(y < 2x+1 and y is odd). In your proof you can use without proof simple theorems such as "even+even is even", "even + odd is odd" etc. The domain of discourse is the integers. First write your proof on a piece of scratch paper, then assemble the proof below.
First sentence: Suppose x is an arbitrary integer. Second sentence: Select y=2x-1. Third sentence: Then y is odd and y=2x-1 < 2x+1.
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.
On a piece of scratch paper write a short proof for the following statement. 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 Discussions. 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.
True 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.
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 formula for squares we covered in class.
What is wrong with this proof that integer solutions of n + 1 - k = 0 exist? Suppose n = 7 and k = 8. Since k = n+1. Proving that n + 1 - k = 0. a. "Since" introduces a subordinate clause, not a main clause. The subordinate clauses describes the reason for what follows in the main clause. "Since k =n+1, we have n + 1 - k = 0." would have been an acceptable phrase. b. nothing c. We cannot use the -ing form for the main verb in a main clause. d. The word 'proving' suggests that what comes after that is supposed to be a concluding statement reiterating the theorem that has been proved. However, n + 1 - k = 0 is not the theorem; it's not even a proposition, because k and n are undefined variables in that statement. What are n and k? Do they have values? Is this supposed to be true for all n and k? Or is merely the existence of integers n,k being asserted that make the equation true? Without that information, the equation is not a meaningful statement. A correct anal conclusion would have been: "This proves that n + 1 - k = 0 has integer solutions n,k."
a, c, d
Check the properties that the function f: [-2, 0) → [0,4], f(x) = x^2 has. a. Injective b. Surjective c. Bijective d. Increasing e. Decreasing f. Strictly increasing g. Strictly decreasing
a, e, and g 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.
Solve the inequality 1 < ⌈ 2x+1⌉ < 6. a. (0,2] b. (0,5/2] c. (-1/2,5/2] d. (1/2, 2]
a. (0,2] 1 < ⌈ 2x+1⌉ < 6 is equivalent to 1 < 2x+1 ≤ 5, which is equivalent to 0 < x ≤ 2.
If f is the ceiling function from ℝ to ℝ, what is f((1/2, 3/2))? a. {1,2} b. (1,2) c. [1,2] d. the empty set
a. {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.
Use a summation formula we learned in class to compute the sum of the integers from 1000 to 5000. Write the answer in unsimplified 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. 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).
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 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.
If the universal set is [0,2], what is the complement of (0,1)? a. (1,2] b. [1,2] c. {0} ∪ [1,2] d. (0,1,2]
c. {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.
Let f:R→R;f(x)=x.. Evaluate the two sets f((-1,1)) and f⁻.((0,1]). a. f((-1,1))=[0,1); f⁻.((0,1])=(-1,1) b. f((-1,1))=[0,1]; f⁻.((0,1])=(-1,0) U (0,1) c. f((-1,1))=[0,1); f⁻.((0,1])=(0,1) d. f((-1,1))=[0,1); f⁻.((0,1])=(-1,0) U (0,1) e. f((-1,1))=[0,1); f⁻.((0,1])=[-1,0) U (0,1] f. f((-1,1))=(0,1); f⁻.((0,1])=(0,1) g. f((-1,1))=[0,1]; f⁻.((0,1])=(-1,1)
e. f((-1,1))=[0,1); f⁻.((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⁻.(Y)={x| f(x) in Y}.
If we visualize ℝ^2. as the plane, then ℤ^2. 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).
ℤ^2. means ℤ x 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.
True or false. 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.
Consider the statements: 1. Every set is a subset of itself. 2. Every set is a proper subset of itself. a. 1. is true, 2. is false. b. 1. is false, 2. is true. c. They are both true. d. They are both false.
a. 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.
Simplify {∅} ∪ ∅. a. ∅ b. {∅} c. { ∅, {∅}} d. {∅, ∅}
b. {∅} Taking the union with an empty set has no effect on any set.
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 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, simplify A ∩ (A ∪ B). A B The Universal Set The Empty Set
A This is the Absorption Law for sets
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.
