MidSem11 Alt Sol

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THE AUSTRALIAN NATIONAL UNIVERSITY

Mid-Semester Examination – Second Semester 2011 – Make-up version 1:solutions

DISCRETE MATHEMATICAL MODELS

(MATH1005)

Writing period: 2 Hours duration

Study period: 15 Minutes duration

No permitted material.Calculators are not allowed.

Notes to candidates :

Begin each question on a new page and label it clearly.

All questions are to be completed in the script book provided, and not on the examination

paper.

The maximum grade obtainable for each question is 2.5.

The maximum grade obtainable for the exam is 30.

Page 1 of 5 - Discrete Mathematical Models - (MATH1005)



1. Negate the following statement:

∃n ∈ N\{1, 2} ∃x,y,z ∈ N xn + yn = z n.

(∃n ∈ N\{1, 2} ∃x,y,z ∈ N xn+yn = z n) ≡ (∀n ∈ N\{1, 2} ∀x,y,z ∈ N xn+yn = z n).

2. Let p,q,r be statements. Prove the following logical equivalence, then draw the corre-sponding electronic circuit.

( p ∧ q ) ∨ ( p ∧ ¬q ∧ r) ≡ p ∧ (q ∨ r).

p q r p ∧ q ¬q p ∧ ¬q p ∧ ¬q ∧ r q ∨ r ( p ∧ q ) ∨ ( p ∧ ¬q ∧ r) p ∧ (q ∨ r)0 0 0 0 1 0 0 0 0 00 0 1 0 1 0 0 1 0 00 1 0 0 0 0 0 1 0 00 1 1 0 0 0 0 1 0 0

1 0 0 0 1 1 0 0 0 01 0 1 0 1 1 1 1 1 11 1 0 1 0 0 0 1 1 11 1 1 1 0 0 0 1 1 1

(b)

p

q

r

3. Let A be a finite set. Prove that

P (A × {0}) = {∅} ∪ {T × {0} ; T ⊆ A , T = ∅}.

Let S ∈ P (A × {0}) be non-empty. Define T = {a ∈ A ; (a, 0) ∈ S }. Then, for x = (a, 0) ∈A × {0}

x ∈ T × {0} ⇐⇒ a ∈ T ⇐⇒ x ∈ S.

So S ∈ {T × {0} ; T ⊆ A , T = ∅}. This proves that

P (A × {0}) ⊆ {∅} ∪ {T × {0} ; T ⊆ A , T = ∅}.

Now let T ⊆ A. Then T × {0} = {(a, 0) ∈ A × {0} ; a ∈ T } ⊆ A × {0}.

4. Let A, B ,C be subsets of a universe of discourse U . Find a counterexample to

A ∩ Bc ∩ C = A ∩ B ∩ C c.




Take A = {1}, B = ∅, C = {1, 2}, U = {1, 2, 3}. Then A ∩Bc∩C = {1} and A ∩B ∩C c = ∅.

5. Let U be a non-empty finite set of finite sets. Define R ⊆ U × U by

ARB ⇐⇒ P (A) ⊆ P (B).

Prove that R is a partial order relation.Let A ∈ U . P (A) = P (A), so ARA. Therefore R is reflexive.Let A, B ∈ U and assume that ARB and B RA, i.e. P (A) ⊆ P (B) and P (B) ⊆ P (A). ThenP (A) = P (B). Since A ∈ P (A) = P (B), we have A ⊆ B. Similarly B ⊆ A. ThereforeB = A. This proves that R is antisymmetric.Let A,B ,C ∈ U , and assume that ARB and BRC . Then P (A) ⊆ P (B) ⊆ P (C ). SoP (A) ⊆ P (C ), and thus ARC . This proves that R is transitive.The relation R is thus a partial order relation.

6. Let F ⊆ Z× Z3 be defined by

xF y ⇐⇒ x ≡ y (mod 3).

(a) Prove that F is a function.

(b) Determine whether or not F is onto.

(c) Determine whether or not F is one-to-one.

(a) Let x ∈ Z. There exists q ∈ Z and y ∈ Z3 such that x = 3q +y, by the division algorithm.Therefore x ≡ y (mod 3), i.e. xF y. Now, let x ∈ Z and y, z ∈ Z3 be such that xF y andxF z . Then x ≡ y (mod 3) = z (mod 3), so x = z . The relation F is thus a function.(b) We have F (1) = 1, F (2) = 2, and F (3) = 3, so

∀y ∈ Z3 ∃x ∈ Z F (x) = y,

i.e F is onto.(c) We have 4 = 1 + 3 ≡ 1 (mod 3), so F (4) = 1 = F (1). Therefore F is not one-to-one.

7.

(a) Compute A116 × 1316.

(b) Write AE 316 in base 2.

(c) Write −7 as four binary digits, using the usual encoding of Z.




(a) A116 × 1316 = 1E 316 + A1016 = B F 316.(b) AE 316 = 1010111000112.(c) −7 = −1112 = −(01102 + 00012). Therefore −7 is encoded as 11001.

8.

(a) Compute 643 × 327 (mod 5).

(b) Find a number x ∈ Z such that x ≥ 26 and

xd ≡ x (mod 11)

for all d ∈ N.

(a) 643 × 327 (mod 5) = 3 × 6 (mod 5) = 1 (mod 5).

(b) We want xd−1

≡ 1 (mod 11). 34 ≡ 1 (mod 11) so 34d−1

≡ 1d−1

(mod 11) = 1 (mod 11).

9. Let (xn)n∈N ⊆ Q be such thatxn+1 = x2

n ∀n ∈ N ∪ {0},

x0 = 2.

Prove that xn = 22n ∀n ∈ N.Let P (n) : xn = 22n. We have x1 = x2

0 = 4 = 221, so P (1) holds.

Let k

∈N

and assume P

(1) ∧P

(2) ∧...

∧P

(k

).Then xk+1 = x2k

= (22k

)2, since P (k) holds.Therefore xk = 22×2k = 22k+1, i.e. P (k + 1) holds.By induction, this gives xn = 22n ∀n ∈ N ∪ {0}.

10. Let (xn)n∈N ⊆ R be such that

xn+2 − 4xn+1 + xn = 0 ∀n ∈ N ∪ {0},

x0 = 2,

x1 = 2√

3.

Prove that xn = (2 +√

3)n − (2 − √ 3)n ∀n ∈ N.

Let P (n) : xn = (2 +√

3)n − (2 − √ 3)n. We have x1 = 2

√ 3 = (2 +

√ 3)1 − (2 − √

3)1.

Let k ∈ N and assume P (1) ∧ P (2) ∧ ... ∧ P (k).Then xk+1 = 4xk − xk−1 = 4(2 +

√ 3)k − (2 − √

3)k − (2 +√

3)k−1 + (2 − √ 3)k−1, since P (k)

and P (k − 1) hold. Therefore

xk+1 = (2+√

3)k−1(4(2+√

3)−1)−(2−√

3)k−1(4(2−√

3)−1) = (2+√

3)k−1(7+4√

3)−(2−√

3)k−1(7−




Now remark that (2 +√

3)2 = 7 + 4√

3 and (2 − √ 3)2 = 7 − 4

√ 3. This implies that

xk+1 = (2 +√

3)k+1 − (2 − √ 3)k+1, i.e. P (k + 1) holds.

By induction, this gives xn = (2 +√

3)n − (2 − √ 3)n ∀n ∈ N.

11. Let (x, y) ∈ Q2

. Compute 1 22 1

3 x

y

.

1 22 1

3 x

y

=

5 44 5

1 22 1

x

y

=

13 1414 13

x

y

=

13x + 14y

14x + 13y

.

12.

(a) Prove that A = 2 2

2 3 is invertible.

(b) Compute A−1.

(c) Find an inverse to

2 2 0

2 3 00 0 1

of the form

a b 0

c d 00 0 1

.

(a) det(A) = 6 − 4 = 2 = 0. Therefore A is invertible.(b)

A−1

= 1

2 3

−2

−2 2

.

(c)

1

2

3 −2 0

−2 2 00 0 2

2 2 0

2 3 00 0 1

=

1 0 0

0 1 00 0 1

,

so

2 2 0

2 3 00 0 1

−1

= 1

2

3 −2 0

−2 2 00 0 2

.


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MidSem11 Alt Sol