For each number, perform the following bitwise operations.You don't have to show work for each conversion.Format your answers in the following ways, points will be deducted if answers are not formatted correctly.Answers are numbered.Write all hexadecimal numbers using the 0x notation to differentiate between other number systems.Write all binary numbers in groups of 4 bits XXXX XXXX or groups of 8 bits XXXXXXXXGiven the following signed 8-bit binary number: 1110 0101What are the two possible binary numbers after shifting 3 bits to the right?What is the result of (~142) in unsigned 8-bit binary?Hint: Convert 142 to binary then perform one's complement.Please answer 1 and 2.

Answered: For each number, perform the following…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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For each number, perform the following bitwise operations.
- You don't have to show work for each conversion.
Format your answers in the following ways, points will be deducted if answers are not formatted correctly.
- Answers are numbered.
- Write all hexadecimal numbers using the 0x notation to differentiate between other number systems.
- Write all binary numbers in groups of 4 bits XXXX XXXX or groups of 8 bits XXXXXXXX

Given the following signed 8-bit binary number: 1110 0101
What are the two possible binary numbers after shifting 3 bits to the right?
What is the result of (~142) in unsigned 8-bit binary?

Hint: Convert 142 to binary then perform one's complement.

Please answer 1 and 2.

Expert Solution

Step 1

Solution:

Given,

signed 8-bit binary number = 1110 0101

Performing right sift-

To Perform right shift the right bit shift moves every bit of the input one position to the right. Therefore, the least significant bit is lost, while a $0$ is pushed in on the other end.

Now performing shift-

1st bit shift - here MSB bit is 1 and now it will be shifted to right of given signed binary number

1110 0101 ---> 0111 0010

2st bit shift

0111 0010 ---> 0011 1001

3rd bit shift

0011 1001 ---> 0001 1100

So resultant number after 3 bit shift is- 0001 1100

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