Boolean Algebra - Discrete Mathematics - Solved Exam, Exams of Discrete Mathematics

This exam paper is very helpful for the student of discrete mathematics. The major points in these exam paper are: Boolean Algebra, Boolean Equality, N-Bit Binary Numbers, Complement Representations, Boolean Expression, N-Digit System, Sum of Products, Boolean Transform, Proof of Consensus Theorem, Recursive Function

Typology: Exams

2012/2013

Uploaded on 04/23/2013

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1 Please Notice
Only final answers available.
Solutions and proofs not included.
Proofs for recursive sequences please refer to questions 6.34 and 6.35
in Shaum’s series.
For each sequence, the RANK for the first number is 0.
2 Binary Number Systems
1.
10 001010 110101
5000101 111010
(10) + (5) = 110101 + 111010 = 101111 + 1 = 110000
2.
Definition x10nx
Range [5×10n1,5×10n11]
Arithmetic 21610 + 6510 = 106216 + 65 = 999849
3.
Definition x8nx
Range [4×8n1,4×8n11]
Arithmetic 1208278=
861208+ 86278=
7776608+ 7777518=
17776318=
7776318
1
3 Boolean Algebra
2. ab0+bc +a0c0
1As we have only 6-digit, the highest is then filtered out.
1
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1 Please Notice

ˆ Only final answers available. ˆ Solutions and proofs not included. ˆ Proofs for recursive sequences please refer to questions 6.34 and 6.35in Shaum’s series. ˆ For each sequence, the RANK for the first number is 0.

2 Binary Number Systems

Definition Range [− 5 − ×x 10 ⇒n (^) − (^101) , n 5 −× x 10 n− (^1) − 1] Arithmetic − 21610 + 65 10 = 10^6 − 216 + 65 = 999849 3.Definition −x ⇒ 8 n (^) − x Range [− 4 × 8 n−^1 , 4 × 8 n−^1 − 1] Arithmetic 86 − 120 − 1208 − 278 = 777660 8 + 8^6 −^278 = 17776318 8 + 777751 =^8 = 77763181

3 Boolean Algebra

  1. ab′^ + bc + a′c′ (^1) As we have only 6-digit, the highest is then filtered out. 1
  1. x′y′z′
  2. x + y
  3. ab + b′d′^ + c′^ + ad
  4. x′^ + y + z′

4 Recursive Function

4.1 Recursive function: formulation

1.(a) 591 (b) 324561 2.(a) H(6, S, E, G) ⇒ H H(5(4, E, S, G, S, E, G)) ⇒⇒ H H(3(2, S, G, E, G, S, E)) ⇒⇒

G −→^2 E (b) G −→^2 E S(4, 1), E(3), G(6, 5 , 2) (c) GS(4) −→^1 , E S(3), G(6, 5 , 2 , 1)

(d) S −→^1 E