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Q25E

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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

Show that these identities hold.

\(\begin{array}{c}a)\;x \oplus y{\bf{ = (x + }}y)\overline {(xy)} \\b)\;x \oplus y{\bf{ = (x\bar y) + }}(\bar xy)\end{array}\)

a) The given identity \(x \oplus y = \left( {x + y} \right)\overline {\left( {xy} \right)} \) is hold

b) The given identity \(x \oplus y = \left( {x\bar y} \right) + \left( {\bar xy} \right)\) is hold

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Definition

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

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

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

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

Step 2:(a) Using the operator

\(x\) and \(y\) can both take on the value of \(0\) or \(1\).

The XOR operator is \(1\) if one of the two elements (but not both) are \(1\).

\(\begin{array}{*{20}{r}}x&y&{x + y}&{xy}&{\overline {\left( {xy} \right)} }&{\left( {x + y} \right)\overline {\left( {xy} \right)} }&{x \oplus y}\\0&0&0&0&1&0&0\\0&1&1&0&1&1&1\\1&0&1&0&1&1&1\\1&1&1&1&0&0&0\end{array}\)

Notes that the last two columns of the table are identical.

Therefore, you get\({\bf{(x + }}y)\overline {(xy)} {\bf{ = }}x \oplus y\).

 Step 3: Using the operator

\(x\) and \(y\) can both take on the value of \(0\) or \(1\).

The \(XOR\) operator is \(1\) if one of the two elements (but not both) are \(1\).

\(\begin{array}{*{20}{r}}x&y&{\bar x}&{\bar y}&{\left( {x\bar y} \right)}&{\left( {\bar xy} \right)}&{\left( {x\bar y} \right) + \left( {\bar xy} \right)}&{x \oplus y}\\0&0&1&1&0&0&0&0\\0&1&1&0&0&1&1&1\\1&0&0&1&1&0&1&1\\1&1&0&0&0&0&0&0\end{array}\)

Notes that the last two columns of the table are identical.

Therefore, you get\((x\bar y) + (\bar xy) = x \oplus y\).

Most popular questions for Math Textbooks

Use a \({\bf{K}}\)-map to find a minimal expansion as a Boolean sum of Boolean products of each of these functions in the variables \({\bf{w, x, y}}\) and \({\bf{z}}\).

\(\begin{array}{l}{\bf{a) wxyz + wx\bar yz + wx\bar y\bar z + w\bar xy\bar z + w\bar x\bar yz}}\\{\bf{b) wxy\bar z + wx\bar yz + w\bar xyz + \bar wx\bar yz + \bar w\bar xy\bar z + \bar w\bar x\bar yz}}\\{\bf{c) wxyz + wxy\bar z + wx\bar yz + w\bar x\bar yz + w\bar x\bar y\bar z + \bar wx\bar yz + \bar w\bar xy\bar z + \bar w\bar x\bar yz}}\\{\bf{d) wxyz + wxy\bar z + wx\bar yz + w\bar xyz + w\bar xy\bar z + \bar wxyz + \bar w\bar xyz + \bar w\bar xy\bar z + \bar w\bar x\bar yz}}\end{array}\)

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

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

Show that in a Boolean algebra, the modular properties hold. That is, show that \({\bf{x}} \wedge {\bf{(y}} \vee {\bf{(x}} \wedge {\bf{z)) = (x}} \wedge {\bf{y)}} \vee {\bf{(x}} \wedge {\bf{z)}}\) and \({\bf{x}} \vee {\bf{(y}} \wedge {\bf{(x}} \vee {\bf{z)) = (x}} \vee {\bf{y)}} \wedge {\bf{(x}} \vee {\bf{z)}}\).

In Exercises 1–5 find the output of the given circuit.

Construct a circuit that computes the product of the two-bitintegers \({{\bf{(}}{{\bf{x}}_{\bf{1}}}{{\bf{x}}_{\bf{o}}}{\bf{)}}_{\bf{2}}}\)and\({{\bf{(}}{{\bf{y}}_{\bf{1}}}{{\bf{y}}_{\bf{o}}}{\bf{)}}_{\bf{2}}}\).The circuit should have four output bits for the bits in the product. Two gates that are often used in circuits are NAND and NOR gates. When NAND or NOR gates are used to represent circuits, no other types of gates are needed. The notation for these gates is as follows:
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