Cmsc 175 Discrete Mathematics

CmSc 175 Discrete Mathematics Study Guide Unit Exam 3

Unit exam 3 will cover proof problems on equivalence relations and partial order relations, proof problems on Math induction, and counting.

Here are sample problems:

A. Equivalence and partial order relations

1. Let R  A x A be a binary relation as defined below. In which cases R is a partial order? A total order? a. A = the positive integers; (a,b)  R iff b is divisible by a

b. A = N x N; ((a,b),(c,d) )  R iff a  c or b  d.

c. A = N; (a,b)  R iff b = a or b = a+1

d. A is the set of all English words; (a,b)  R iff a is no longer than b

2. Define a relation R on N x N : R = { ((a, b), (c, d)) | a + d = b + c} Show that R is an equivalence relation on N x N

3. Define a relation R on N x N : R = { ((a, b), (c, d)) | a + b = c + d} Show that R is an equivalence relation on N x N

4. Define a relation R on N x N : R = { ((a, b), (c, d)) | ad = bc} Show that R is an equivalence relation on N x N

5. Define a relation R on N x N : R = { ((a, b), (c, d)) | ab = cd} Show that R is an equivalence relation on N x N

6. Define a relation R on N : R = { (a, b) |  integer q, b = qa, i.e. b is a multiple of a} Show that R is partial order relation on N

1 7. Let S be a collection of sets. Define a relation R on S: R = {(s1, s2) | s1  s2} Show that R is partial order relation on S

8. Let R1 and R2 be any two partial orders on the same set A. Prove that R1  R2 is a partial order

9. Let R1 and R2 be any two equivalence relations on the same set A. Prove that R1  R2 is an equivalence relation.

10. Let R1 be an equivalence relation on some set A, and R2 be any other relation on the same set.

10.1. Will R1  R2 be necessarily a relation of equivalence? If the answer is “yes” prove that is reflexive, symmetric and transitive. If the answer is “no” show an example where at least one of the properties reflexivity, symmetry and transitivity is missing.

10.2. Will R1  R2 be necessarily a relation of equivalence? If the answer is “yes” prove that is reflexive, symmetric and transitive. If the answer is “no” show an example where at least one of the properties reflexivity, symmetry and transitivity is missing.

10.3. Solve the same problem when R1 is a partial order relation

B. Math induction

1. Let S(n) = 1 + 2 + 3 + … + n

Prove P(n): S(n) = n(n + 1)/2 for all n  1

2. Let S(n) = 1 + 3 + 5 … + (2n – 1)

Prove P(n): S(n) = n2 for all n  1

3. Let S(n) = 2 + 4 + 6 … + 2n

Prove P(n): S(n) = n(n + 1) for all n  1

2 4. Let S(n) = 12 + 32 + 52 + …. + (2n-1) 2 Prove P(n): S(n) = n(2n – 1)(2n + 1)/3 for all n  1

5. Let S(n) = 1 + a + a2 + a3 + …. + a(n-1)

Prove P(n): S(n) = (an – 1)/(a-1), for all n ≥1, a ≠ 1

6. Let S(n) = 1 + 2 + 22 + 23 + …. + 2n

Prove P(n): S(n) = (2n+1 – 1) for all n ≥0

Note that in this problem you have to prove P(0) in the base step

7. Let S(n) = 1 + 5 + 9 + … + (4n – 3)

Prove P(n): S(n) = n(2n – 1) for all n  1

8. Let S(n) = 2 + 6 + 18 + …. + 2*3(n-1)

Prove P(n): S(n) = 3n – 1, for all n ≥1

9. Let S(n) = 3 + 6 + 9 + 12 + 15 + …. + 3n

Prove P(n): S(n) = 3n(n+1)/2 for all n  1

10. Let S(n) = 1*2 + 2*3 + 3*4 + …. + n(n+1)

Prove P(n): S(n) = n(n+1)(n+2)/3 for all n ≥1

11. Let S(n) = 1 / (1*2) + 1 / (2*3) + … + 1 / n(n + 1)

Prove P(n): S(n) = n / (n+1) for all n ≥1

12. Let S(n) = 1 / (1*3) + 1 / (3*5) + … + 1 / (2n – 1)(2n + 1)

Prove P(n): S(n) = n / (2n + 1) for all n  1

13. Let S(n) = 1* 1! + 2*2! + 3*3! + … + n*n!

Prove P(n) : S(n) = (n+1)! – 1 for all n  1

3 14. Let Q(n) = (1 – ½)(1 – 1/3)…..(1 – 1/(n+1))

Prove the statement P(n): Q(n) =1/(n+1), for all n ≥1

15. Let Q(n) = ( 1 – 1/22)*(1 – 1 /32)* ….*(1 – 1/n2)

Prove P(n) : Q(n) = (n+1)/2n for all n  2

C. Counting

All problems in the handout on counting

Sample solutions:

Problem A.5: Define a relation R on N x N : R = { ((a, b), (c, d)) | ab = cd} Show that R is an equivalence relation on N x N

Reflexivity. To show reflexivity, we need to show that ((a,b),(a,b) is in R for any (a,b) in N x N. Since ab = ab (by basic algebra), and by the definition of R, we have ((a,b),(a,b) in R Therefore R is reflexive

Symmetry. In order to show symmetry, we have to show that If ((a,b),(c,d)) is in R, then ((c,d),(a,b)) is also in R.

In order to show that ((c,d),(a,b)) is in R we need to show that cd = ab.

Let ((a,b),(c,d)) be in R. By the definition of R, we have ab = cd , therefore cd = ab. By the definition of R ((c,d),(a,b)) is in R

Therefore R is symmetric

Transitivity. In order to show transitivity, we need to show that If ((a,b),(c,d)) is in R and ((c,d),(e,f)) is in R, then ((a,b),(e,f)) is also in R. In order to show that ((a,b),(e,f)) is in R we need to show that ab = ef

Let . ((a,b),(c,d)) be in R By the definition of R, ab = cd (1)

4 Let ((c,d),(e,f)) be in R. By the definition of R, cd = ef (2)

From (1) and (2) we have ab = cd = ef, therefore ab = ef. Therefore ((a,b),(e,f)) is in R. Therefore R is transitive.

We have shown that R is reflexive, symmetric and transitive, therefore R is a relation of equivalence.

Problem A.6 Define a relation R on N : R = { (a, b) |  integer q, b = qa, i.e. b is a multiple of a} Show that R is partial order relation on N

Anti-symmetry. In order to show that R is anti-symmetric we need to show that If (a,b) is in R and a≠ b, then (b,a) is not in R.

Let (a,b) be in R. By the definition of R, there is an integer q such that b = qa. Solving for a, we get a = b * (1/q) q is a n integer, therefore 1/q is not an integer. Note that if q = 1, then a = b and we want a≠ b

Therefore (b,a) is not in R, therefore R is antisymmetric.

Transitivity.

In order to show that R is anti-symmetric we need to show that If (a,b) is in R and (b,c) is in R, then (a,c) is in R.

Let (a,b) be in R and (b,c) be in R. By the definition of R this means that there are integers q and r such that b = qa, c = rb

Therefore c = rqa. Let p = rq, p is integer. Therefore c = pa, i.e. c is a multiple of a

By the definition of R (a,c) is in R Therefore R is transitive.

Being anti-symmetric and transitive, R is a partial order relation.

5 Problem B.12 Let S(n) = 1 / (1*3) + 1 / (3*5) + … + 1 / (2n – 1)(2n + 1)

Prove P(n): S(n) = n / (2n + 1) for all n  1

Base step: Show that P(1): S(1) = 1/(2 + 1) is true.

1/(2 + 1) = 1/3 By the definition of the sum, S(1) = 1/(1*3) = 1/3

Therefore P(1) is true.

Inductive step

We will show that the conditional P(k)  P(k+1) is true. P(k): S(k) = k / (2k +1)

P(k+1): S(k+1) = (k+1)/(2(k+1) + 1) = (k+1)/(2k + 3)

Assume P(k) is true, i.e. S(k) = k / (2k +1)

S(k+1) = S(k) + 1 / (2(k+1) – 1)(2(k+1) + 1) =

= k / (2k +1) + 1 / (2k + 1)(2k + 3) =

= k(2k + 3)/ (2k + 1)(2k + 3) + 1 / (2k + 1)(2k + 3) =

= [ k(2k + 3) + 1 ]/ (2k + 1)(2k + 3) =

(2k2 +3 k +1) / (2k + 1)(2k + 3)

(2k2 +3 k +1) = (k+1)(2k + 1) (see the note at the end)

(2k2 +3 k +1) / (2k + 1)(2k + 3) = (k+1)(2k + 1) /(2k + 1)(2k + 3) = (k+1) / (2k + 3)

Therefore P(k+1) is true Therefore P(k)  P(k+1) is true By the principle of math induction P(n) is true for all n 1

Note: How did I know that (2k2 +3 k +1) = (k+1)(2k + 1) ? This is not very obvious. I looked at my goal: S(k+1) = (k+1)/(2k + 3) But I have (2k2 +3 k +1) / (2k + 1)(2k + 3) If (2k2 +3 k +1) = (k+1)(2k +1), then I will have

6 S(k+1) = (k+1)(2k +1) / (2k + 1)(2k + 3) = (k+1)/(2k + 3), i.e. I will prove my goal.

What remains is simply to multiply (k+1)(2k +1) and see if it is equal to (2k2 +3k +1)