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Q3E

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Discrete Mathematics and its Applications
Found in: Page 818
Discrete Mathematics and its Applications

Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

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Short Answer

  1. Show that \((1 \cdot 1){\bf{ + }}(\overline {0 \cdot 1} {\bf{ + }}0){\bf{ = }}1\).
  2. Translate the equation in part \((a)\) into a propositional equivalence by changing each \(0\) into an \({\bf{F}}\), each \(1\) into a \(T\), each Boolean sum into a disjunction, each Boolean product into a conjunction, each complementation into a negation, and the equals sign into a propositional equivalence sign.

a) The given \((1 \cdot 1){\bf{ + }}(\overline {0 \cdot 1} {\bf{ + 0}}){\bf{ = 1}}\) is proved.

b) The translate of the given equation \((1 \cdot 1){\bf{ + }}(\overline {0 \cdot 1} {\bf{ + 0}}){\bf{ = 1}}\) is \(({\bf{T}} \wedge {\bf{T}}) \vee (\neg ({\bf{F}} \wedge {\bf{T}}) \vee {\bf{F}}) \equiv {\bf{T}}\)

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Definition

The complement of an element: \({\bf{\bar 0 = 1}}\) and \({\bf{\bar 1 = }}0\)

The Boolean sum + or \(OR\) is 1 if either term is 1.

The Boolean product \( \cdot \) or \(AND\) is 1 if both terms are 1.

Step 2: (a) Using the Boolean product and sum

Using the Boolean product and sum

\(\begin{array}{c}(1 \cdot 1){\bf{ + }}(\overline {0 \cdot 1} {\bf{ + }}0){\bf{ = 1 + (\bar 0 + 0)}}\\{\bf{ = 1 + (1 + 0)}}\\{\bf{ = 1 + 1}}\\{\bf{ = 1}}\end{array}\)

Therefore, you get \((1 \cdot 1){\bf{ + }}(\overline {0 \cdot 1} {\bf{ + 0}}){\bf{ = 1}}\).

Step 3: (b) Using the Boolean product and sum

\((1 \cdot 1){\bf{ + }}(\overline {0 \cdot 1} {\bf{ + 0}}){\bf{ = 1}}\)

Replace 0 by F

Replace 1 by T

Replace + by \( \vee \)

Replace \( \cdot \) by \( \wedge \)

Replace - by \(\neg \) (placed in front of the expression that used to be underneath)

Replace = by \( \equiv \)

Therefore, you get \(({\bf{T}} \wedge {\bf{T}}) \vee (\neg ({\bf{F}} \wedge {\bf{T}}) \vee {\bf{F}}) \equiv {\bf{T}}\).

Most popular questions for Math Textbooks

Construct a \({\bf{K}}\)-map for \({\bf{F(x,y,z) = xz + yz + xy\bar z}}{\bf{.}}\) Use this \({\bf{K - }}\)map to find the implicants, prime implicants, and essential prime implicants of \({\bf{F(x,y,z)}}\).

use the laws in Definition \(1\) to show that the stated properties hold in every Boolean algebra.

Show that in a Boolean algebra, the complement of the element \(0\) is the element \(1\) and vice versa.

Find the cells in a \(K{\bf{ - }}\)map for Boolean functions with five variables that correspond to each of these products.

\(\begin{array}{c}a){x_1}{x_2}{x_3}{x_4}\\b){{\bar x}_1}{x_3}{x_5}\\c){x_2}{x_4}\\d){{\bar x}_3}{{\bar x}_4}\\e){x_3}\\f){{\bar x}_5}\end{array}\)

Which of these functions are self-dual?

\(\begin{array}{l}\left. {\bf{a}} \right)\;{\bf{F}}\left( {{\bf{x,y}}} \right) = x\\\left. {\bf{b}} \right)\;{\bf{F}}\left( {{\bf{x,y}}} \right) = {\bf{xy + \bar x\bar y}}\\\left. {\bf{c}} \right)\;{\bf{F}}\left( {{\bf{x,y}}} \right) = {\bf{x + y}}\\\left. {\bf{d}} \right)\;{\bf{F}}\left( {{\bf{x,y}}} \right) = {\bf{xy + \bar xy}}\end{array}\)

Let \({\bf{x}}\) and \({\bf{y}}\) belong to \(\left\{ {{\bf{0,1}}} \right\}\). Does it necessarily follow that \({\bf{x = y}}\) if there exists a value \({\bf{z}}\) in \(\left\{ {{\bf{0,1}}} \right\}\) such that,

\(\begin{array}{l}{\bf{a) xz = yz?}}\\{\bf{b) x + z = y + z?}}\\{\bf{c) x}} \oplus {\bf{z = y}} \oplus {\bf{z?}}\\{\bf{d) x}} \downarrow {\bf{z = y}} \downarrow {\bf{z?}}\\{\bf{e) x}}|{\bf{z = y}}|z{\bf{?}}\end{array}\)

A Boolean function \({\bf{F}}\) is called self-dual if and only if \({\bf{F}}\left( {{{\bf{x}}_{\bf{1}}}{\bf{, \ldots ,}}{{\bf{x}}_{\bf{n}}}} \right){\bf{ = }}\overline {{\bf{F}}\left( {{{{\bf{\bar x}}}_{\bf{1}}}{\bf{, \ldots ,}}{{{\bf{\bar x}}}_{\bf{n}}}} \right)} \).
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