• :00Days
  • :00Hours
  • :00Mins
  • 00Seconds
A new era for learning is coming soonSign up for free
Log In Start studying!

Select your language

Suggested languages for you:
Deutsch (DE)
Deutsch (UK)
Deutsch (US)

Americas

Europe

Answers without the blur. Sign up and see all textbooks for free! → Illustration

Q14SE

Expert-verified
Discrete Mathematics and its Applications
Found in: Page 844
Discrete Mathematics and its Applications

Discrete Mathematics and its Applications

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

Answers without the blur.

Just sign up for free and you're in.

Register for Free I’ll do it later
Illustration

Short Answer

Show that each of these identities holds.

\({\bf{a)}}\) \({\bf{x}} \odot {\bf{x = 1}}\)

\({\bf{b)}}\) \({\bf{x}} \odot {\bf{\bar x = 0}}\)

\({\bf{c)}}\) \({\bf{x}} \odot {\bf{y = y}} \odot {\bf{x}}\)

\({\bf{a)}}\) The given equation \({\bf{x}} \odot {\bf{x = 1}}\) holds.

\({\bf{b)}}\) The given equation \({\bf{x}} \odot {\bf{\bar x = 0}}\) holds.

\({\bf{c)}}\) The given equation \({\bf{x}} \odot {\bf{y = y}} \odot {\bf{x}}\) holds.

See the step by step solution

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 \({\bf{ + }}\) or \({\bf{OR}}\) is \({\bf{1}}\) if either term is \({\bf{1}}\).

The Boolean product \( \bullet \) or \({\bf{AND}}\) is \({\bf{1}}\) if both terms are \({\bf{1}}\).

The \({\bf{NOR}}\) operator \( \downarrow \) is \({\bf{1}}\) if both terms are \({\bf{0}}\).

The \({\bf{XOR}}\) operator \( \oplus \) is \({\bf{1}}\) if one of the terms is \({\bf{1}}\) (but not both).

The \({\bf{NAND}}\) operator \(\mid \) is \({\bf{1}}\) if either term is \({\bf{0}}\).

The Boolean operator \( \odot \) is \({\bf{1}}\) if both terms have the same value.

Step 2: Using the Boolean operator

It is given that \({\bf{x}} \odot {\bf{x = 1}}\).

Now create a table for all possible values of \({\bf{x, y}}\) and \({\bf{z}}\).

The Boolean operator \( \odot \) is \({\bf{1}}\) if both terms have the same value.

\(\begin{array}{*{20}{r}}{\bf{x}}&{{\bf{ x}}}&{{\bf{ x}} \odot {\bf{x}}}&{\bf{1}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}\\{\bf{1}}&{\bf{1}}&{\bf{1}}&{\bf{1}}\end{array}\)

Then note that the last two columns of the table are identical.

Hence, it gives \({\bf{x}} \odot {\bf{x = 1}}\).

Step 3: Using the Boolean operator

(b)

It is given that \({\bf{x}} \odot {\bf{\bar x = 0}}\).

Now create a table for all possible values of \({\bf{x, y}}\) and \({\bf{z}}\).

The Boolean operator \( \odot \) is \({\bf{1}}\) if both terms have the same value.

\(\begin{array}{*{20}{r}}{\bf{x}}&{{\bf{ \bar x}}}&{{\bf{ x}} \odot {\bf{\bar x}}}&{\bf{0}}\\{\bf{0}}&{\bf{1}}&{\bf{0}}&{\bf{0}}\\{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\end{array}\)

Then note that the last two columns of the table are identical.

Hence, it gives \({\bf{x}} \odot {\bf{\bar x = 0}}\).

Step 4: Using the Boolean operator

It is given that \({\bf{x}} \odot {\bf{y = y}} \odot {\bf{x}}\).

Now create a table for all possible values of \({\bf{x, y}}\) and \({\bf{z}}\).

The Boolean operator \( \odot \) is \({\bf{1}}\) if both terms have the same value.

\(\begin{array}{*{20}{r}}{\bf{x}}&{{\bf{ y}}}&{{\bf{ x}} \odot {\bf{y}}}&{{\bf{ y}} \odot {\bf{x}}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}\\{\bf{0}}&{\bf{1}}&{\bf{0}}&{\bf{0}}\\{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{1}}&{\bf{1}}\end{array}\)

Then note that the last two columns of the table are identical.

Hence, it gives \({\bf{x}} \odot {\bf{y = y}} \odot {\bf{x}}\).

Most popular questions for Math Textbooks

Exercises 14-23 deal with the Boolean algebra \(\left\{ {{\bf{0,1}}} \right\}\) with addition,multiplication, and complement defined at the beginning of this section. In each case, use a table as in Example \(8\).

21. Verify De Morgan's laws.

Show that if \(F\) and \(G\) are Boolean functions of degree \(n\), then

\(\begin{array}{l}a)F \le F{\bf{ + }}G\\b)FG \le F\end{array}\)

Show that \({\bf{x\bar y + y\bar z + \bar xz = \bar xy + \bar yz + x\bar z}}\).

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 \({\bf{F}}\left( {{{\bf{x}}_{\bf{1}}}{\bf{, \ldots ,}}{{\bf{x}}_{\bf{n}}}} \right){\bf{ = }}\overline {{\bf{F}}\left( {{{{\bf{\bar x}}}_{\bf{1}}}{\bf{, \ldots ,}}{{{\bf{\bar x}}}_{\bf{n}}}} \right)} \).

In Exercises 35–42, Use the laws in Definition \(1\) to show that the stated properties hold in every Boolean algebra.

Show that in a Boolean algebra, if \(x \vee y{\bf{ = }}0\), then \(x{\bf{ = }}0\) and \(y{\bf{ = }}0\), and that if \(x \wedge y{\bf{ = }}1\), then \(x{\bf{ = }}1\) and \(y{\bf{ = }}1\).
Icon

Want to see more solutions like these?

Sign up for free to discover our expert answers
Get Started - It’s free

Recommended explanations on Math Textbooks

Decision Maths

Read explanation

Calculus

Read explanation

Probability and Statistics

Read explanation

Statistics

Read explanation

Mechanics Maths

Read explanation

Geometry

Read explanation

Pure Maths

Read explanation

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.