Test 2 Practice Problems

Either draw a graph with the specified properties or explain why no such graph exists. Graph with four vertices of degrees 1, 1, 1, and 4.

*Impossible* odd number of odd vertices

Either draw a graph with the specified properties or explain why no such graph exists. Graph with four vertices of degrees 1, 2, 3, and 3.

*Impossible* odd number of odd vertices

Determine if the given relation is reflexive, symmetric, transitive, or none of these. C is the circle relation on the set of real numbers: For all x, y ∈ R, xCy ⇔ x^2 + y^2 = 1

*Not Reflexive* C is reflexive ⇔ for all real numbers x, xCx. By definition of C this means that for all real numbers x, x^2 + x^2 = 1. But this is false. As a counterexample, take x = 0. Then x^2 + x^2 = 0^2 + 0^2 ≠ 1 *Symmetric* C is symmetric ⇔ for all real numbers x and y, if xCy then yCx. By definition of C this means that for all real numbers x and y, if x^2 + y^2 = 1 then y^2 + x^2 = 1. But this is true because by commutativity of addition, x^2 + y^2 = y^2 + x^2 for all real numbers x and y. *Not Transitive* C is transitive ⇔ for all real numbers x, y, and z, if xCy and yCz then xCz. By definition of C this means that for all real numbers x, y, and z, if x^2 + y^2 = 1 and y^2 + z^2 = 1 then x^2 + z^2 = 1. But this is false. As a counterexample, take x = 0, y = 1, and z = 0. Then x^2 + y^2 = 1 because 0^2 + 1^2 = 1 and y^2 + z^2 = 1 because 1^2 + 0^2 = 1, but x^2 + z^2 ≠ 1 because 0^2 + 0^2 = 0 ≠ 1.

Determine if the given relation is reflexive, symmetric, transitive, or none of these. O is the relation defined on Z as follows: For all m, n ∈ Z, mOn ⇔ m − n is odd

*Not Reflexive* O is reflexive for all integers m, mOm. By definition of O this means that for all integers m, m — m is odd. But this is false. As a counterexample, take any integer m. Then m — m = 0, which is even, not odd. *Symmetric* Suppose m and n are any integers such that mOn. By definition of O this means that m — n is odd, and so by definition of odd m — n = 2k + 1 for some integer k. Now n — m = —(m — n). Hence by substitution, n — m = —(2k + 1) = 2(—k — 1) + 1. It follows that n — m is odd by definition of odd (since —k — 1 is an integer), and thus nOm by definition of O. *Not Transitive* O is transitive ⇔ for all integers m, n, and p, if mOn and nOp then mOp. By definition of O this means that for all integers m, n, and p, if m — n is odd and n - p is odd then m — p is odd. But this is false. As a counterexample, take m = 1, n = 0, and p = 1. Then m— n = 1 - 0 = 1 is odd and n — p = 0 - 1 = —1 is also odd, but m — p = 1 — 1 = 0 is not odd. Hence mOn and nOp but m does not relate to p.

Is this function onto? f: R → R+ f(x) = x^2

*Onto* f(sqrt(y)) = y for all y ∈ R+

Show that the relation R is an equivalence relation on the set A and find the distinct equivalence classes of R A = {(1,3), (2,4), (−4, −8), (3,9), (1,5), (3,6)}. R is defined on A as follows: For all (a, b), (c, d) ∈ A, (a, b)R(c, d) ⇔ ad = bc

*Reflexive* (a, b)R(a, b) since ab = ab *Symmetric* Since if (a, b)R(c, d) then (c, d)R(a, b) (in both cases ad = bc) *Transitive* If (a, b)R(c, d) and (c, d)R(e, f) then we have ad = bc and cf = de. We can conclude from this that af = be: {ad = bc and cf = de so adcf = bcde so (adcf)/(dc) = (bcde)/(cd) so af = be} So (a,b)R(e,f) Classes are {(1,3), (3,9)},{(2,4), (−4, −8), (3,6)},{(1,5)}

Show that the relation R is an equivalence relation on the set A and find the distinct equivalence classes of R A = {−4, −3, −2, −1,0,1,2,3,4}. R is defined on A as follows: For all (m, n) ∈ A, mRn ⇔ 4|(m^2 - n^2)

*Reflexive* 4|(m^2 - m^2) or 4|0 *Symmetric* If 4|(m^2 - n^2) then 4|(n^2 - m^2) or 4| − (m^2 - n^2) *Transitive* If 4|(m^2 - n^2) and 4|(n^2 - p^2) then 4|(m^2 - n^2)+(n^2 - p^2) or 4|(m^2 - p^2). Classes are [0] = {−4, −2,0,2,4} and [1] = {−3, −1,1,3}

Show that the relation R is an equivalence relation on the set A and find the distinct equivalence classes of R Let A be the set of all statement forms in three variables p, q, and r. R is the relation defined on A as follows: For all P and Q in A, PQR ⇔ P and Q have the same truth table.

*Reflexive* Because any statement form in three variables has the same truth table as itself. *Symmetric* Because for all statement forms S and T in three variables, if S has the same truth table as T, then T has the same truth table as S. *Transitive* Because for all statement forms S, T, and U in three variables, if S has the same truth table as T and T has the same truth table as U, then S has the same truth table as U. R is an equivalence relation because it is reflexive, symmetric, and transitive. There are 28 = 256 distinct equivalence classes, one for each distinct truth table for a statement form in three variables. Each equivalence class consists of all statement forms in three variables with a given truth table.

Is the following relation reflexive, symmetric, transitive? Is it an equivalence relation? If so, what are the classes? Show the digraph for the relation. R defined on A = {0,1,2,3,4,5,6,7,8,9,10} xRy ⇔ 5|(x − y)

*Reflexive* Every element is related to itself. *Symmetric* If there is a forward arrow, then there is a backwards arrow. *Transitive* Since if xRy and yRz, then xRz. *Classes* [0]={0,5,10}, [1]={1,6}, [2]={2,7}, [3]={3,8}, [4]={4,9} See Practice Exam 2 Solutions Q4 on D2L for the digraph.

Determine if the given relation is reflexive, symmetric, transitive, or none of these. D is the relation defined on R as follows: For all x, y ∈ R, xDy ⇔ x*y ≥ 0

*Reflexive* Since x * x = x^2 ≥ 0 *Symmetric* Since if x * y ≥ 0, then y * x ≥ 0 *Not Transitive* x = −1, y = 0, z = 1 then xDy, yDz but x is not related to z

Determine if the given relation is reflexive, symmetric, transitive, or none of these. Let X = {a, b, c} and P(X) be the power set of X (the set of all subsets of X). A relation E is defined on P(X) as follows: For all A,B ∈ P(X), AEB ⇔ the number of elements in A equals the number of elements in B.

*Reflexive* |A| = |B| *Symmetric* If |A| = |B| then |B| = |A| *Transitive* If |A| = |B| and |B| = |C| then |A| = |C|

Show that the relation R is an equivalence relation on the set A and find the distinct equivalence classes of R A is the "absolute value" relation defined on R as follows: For all x, y ∈ R, xAy ⇔ |x| = |y|

*Reflexive* |x| = |x| *Symmetric* Since if |x| = |y| then |y| = |x| *Transitive* If |x| = |y| and |y| = |z| then we have |x| = |z| Classes are {−x, x} for all x ∈ R.

Is this function one-to-one? t: R → R t(x) = x/(1 + x^2)

*not one to one* f(1/2) = f(2) = 2/5

Determine if the following function is one-to-one and/or onto: f: Z → Z+, f(n) = n^2

*not one to one* f(1) = f(-1) *not onto* no element in Z is mapped to 2

Is this function onto? f: Z → Z f(n) = n^3

*not onto* There exists n ∈ Z such that n^3 = 2 (The cube root of 2 is not an integer)

Is this function one-to-one? f: Z → Z f(n) = 3n - 5

*one to one* f(n1) = f(n2) → 3(n1) - 5 = 3(n2) - 5 → 3(n1) = 3(n2) → n1 = n2

Is this function one-to-one? g: R → R g(x) = x^3

*one to one* f(x1) = f(x2) → x(sup(3)sub(1)) = x(sup(3)sub(2)) → x1 = x2

Is this function one-to-one? h: R\{0} → R\{0} h(x) = 1/x\

*one to one* h(x1) = h(x2) → 1/(x1) = 1/(x2) → x1 = x2

Determine if the following function is one-to-one and/or onto f: R\{0} → R, f(x) = 1/x

*one to one* 1/(x1) = 1/(x2) → x1 = x2 *not onto* there is no element in R\{0} mapped to 0

Either draw a graph with the specified properties or explain why no such graph exists. Graph with five vertices of degrees 1, 2, 3, 3, and 5.

See Practice Exam 2 Solutions Q5 a on D2L for the drawn graph.

Either draw a graph with the specified properties or explain why no such graph exists Graph with four vertices of degrees 1, 2, 3, and 4.

See Practice Exam 2 Solutions Q5 d on D2L for the drawn graph.

1. Write A x A 2. Make a graph for the defined relation 3. Write R as a subset of A x A A = {(1,3), (2,4), (−4, −8), (3,9), (1,5), (3,6)}. R is defined on A as follows: For all (a, b), (c, d) ∈ A, (a, b)R(c, d) ⇔ a*d = b*c

See this link for the result to parts 1 and 3 https://pastebin.com/UTSCQTAq For part 2, see Homework 4 Solutions on D2L, Problem Q3 b.

1. Write A x A 2. Make a graph for the defined relation 3. Write R as a subset of A x A A = {−4, −3, −2, −1,0,1,2,3,4}. R is defined on A as follows: For all (m, n) ∈ A, mRn ⇔ 4|(m^2 - n^2)

See this link for the result to parts 1 and 3. https://pastebin.com/84AmGpyq For part 2, see Homework 4 Solutions on D2L, Problem Q3 a.