Write a program that demonstrates the approximate nature of floating-point values by performing the following tasks: Use scanner to read a floating-point value x . Compute 1.0 I x and store the result in y. Display x, y, and the product of x and y. Subtract 1 from the product of x and y and display the result. Try your program with values of x that range from 2e−11 to 2ell. What can you conclude?

Question

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Answer 1

Textbook Question

Chapter 2, Problem 1E

Write a program that demonstrates the approximate nature of floating-point values by performing the following tasks:

Use scanner to read a floating-point value x.
Compute 1.0 I x and store the result in y.
Display x, y, and the product of x and y.
Subtract 1 from the product of x and y and display the result.

Try your program with values of x that range from 2e−11 to 2ell. What can you conclude?

Expert Solution & Answer

Program Plan Intro

Program Plan:

• Include the required header files.
• Define main function.
- ○ Create an object for scanner.
- ○ Get the floating-point value from the user.
- ○ Calculate the “y” value.
- ○ Display “x”, “y” and product of “x” and “y”.
- ○ Subtract “1” from product of “x” and “y” and then display the result.

Program Description Answer

The program is used to read a floating point value then compute “y” and display “x”, “y”, product of “x” and “y” and also subtract “1” from product of “x” and “y” then display the result as follows:

Explanation of Solution

Program:

//include required header files

import java.util.Scanner;

//definition of "Approximation" class

public class Approximation

{

//definition of main method

public static void main(String[] args)

{

//create an object for scanner

Scanner keyboard = new Scanner(System.in);

//declare the required variable

float x = 0, y;

//get the input from the user

System.out.println("Enter a floating point value.");

x = keyboard.nextFloat();

//calculate "y" value

y = (float) (1.0/x);

//display "x" and "y" value

System.out.println("x value is: "+ x);

System.out.println("y value is: "+ y);

//display product of "x" and "y"

System.out.println("Product of x and y is: "+ (x*y));

/*subtract 1 from product of "x" and "y" and display the result*/

System.out.println("The difference of x*y and 1 is " + ((x*y) - 1.0));

}

Conclusion:

The “x” depends on user but the product of “x” and “y” will always give “1” and difference of “x * y” and “1” is always “0” because, “1” is divided by “y” and then “y” is multiplied with the “x”. So there is no change in product of “x” and “y” and difference of “x * y” and “1”.

Sample Output

Output:

Enter a floating point value.

2.5

x value is: 2.5

y value is: 0.4

Product of x and y is: 1.0

The difference of x*y and 1 is 0.0

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04:31

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Answer 2

Textbook Question

Chapter 2, Problem 1E

Write a program that demonstrates the approximate nature of floating-point values by performing the following tasks:

Use scanner to read a floating-point value x.
Compute 1.0 I x and store the result in y.
Display x, y, and the product of x and y.
Subtract 1 from the product of x and y and display the result.

Try your program with values of x that range from 2e−11 to 2ell. What can you conclude?

Expert Solution & Answer

Program Plan Intro

Program Plan:

• Include the required header files.
• Define main function.
- ○ Create an object for scanner.
- ○ Get the floating-point value from the user.
- ○ Calculate the “y” value.
- ○ Display “x”, “y” and product of “x” and “y”.
- ○ Subtract “1” from product of “x” and “y” and then display the result.

Program Description Answer

The program is used to read a floating point value then compute “y” and display “x”, “y”, product of “x” and “y” and also subtract “1” from product of “x” and “y” then display the result as follows:

Explanation of Solution

Program:

//include required header files

import java.util.Scanner;

//definition of "Approximation" class

public class Approximation

{

//definition of main method

public static void main(String[] args)

{

//create an object for scanner

Scanner keyboard = new Scanner(System.in);

//declare the required variable

float x = 0, y;

//get the input from the user

System.out.println("Enter a floating point value.");

x = keyboard.nextFloat();

//calculate "y" value

y = (float) (1.0/x);

//display "x" and "y" value

System.out.println("x value is: "+ x);

System.out.println("y value is: "+ y);

//display product of "x" and "y"

System.out.println("Product of x and y is: "+ (x*y));

/*subtract 1 from product of "x" and "y" and display the result*/

System.out.println("The difference of x*y and 1 is " + ((x*y) - 1.0));

}

Conclusion:

The “x” depends on user but the product of “x” and “y” will always give “1” and difference of “x * y” and “1” is always “0” because, “1” is divided by “y” and then “y” is multiplied with the “x”. So there is no change in product of “x” and “y” and difference of “x * y” and “1”.

Sample Output

Output:

Enter a floating point value.

2.5

x value is: 2.5

y value is: 0.4

Product of x and y is: 1.0

The difference of x*y and 1 is 0.0

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04:31

Answer 3

Textbook Question

Chapter 2, Problem 1E

Write a program that demonstrates the approximate nature of floating-point values by performing the following tasks:

Use scanner to read a floating-point value x.
Compute 1.0 I x and store the result in y.
Display x, y, and the product of x and y.
Subtract 1 from the product of x and y and display the result.

Try your program with values of x that range from 2e−11 to 2ell. What can you conclude?

Expert Solution & Answer

Program Plan Intro

Program Plan:

• Include the required header files.
• Define main function.
- ○ Create an object for scanner.
- ○ Get the floating-point value from the user.
- ○ Calculate the “y” value.
- ○ Display “x”, “y” and product of “x” and “y”.
- ○ Subtract “1” from product of “x” and “y” and then display the result.

Program Description Answer

The program is used to read a floating point value then compute “y” and display “x”, “y”, product of “x” and “y” and also subtract “1” from product of “x” and “y” then display the result as follows:

Explanation of Solution

Program:

//include required header files

import java.util.Scanner;

//definition of "Approximation" class

public class Approximation

{

//definition of main method

public static void main(String[] args)

{

//create an object for scanner

Scanner keyboard = new Scanner(System.in);

//declare the required variable

float x = 0, y;

//get the input from the user

System.out.println("Enter a floating point value.");

x = keyboard.nextFloat();

//calculate "y" value

y = (float) (1.0/x);

//display "x" and "y" value

System.out.println("x value is: "+ x);

System.out.println("y value is: "+ y);

//display product of "x" and "y"

System.out.println("Product of x and y is: "+ (x*y));

/*subtract 1 from product of "x" and "y" and display the result*/

System.out.println("The difference of x*y and 1 is " + ((x*y) - 1.0));

}

Conclusion:

The “x” depends on user but the product of “x” and “y” will always give “1” and difference of “x * y” and “1” is always “0” because, “1” is divided by “y” and then “y” is multiplied with the “x”. So there is no change in product of “x” and “y” and difference of “x * y” and “1”.

Sample Output

Output:

Enter a floating point value.

2.5

x value is: 2.5

y value is: 0.4

Product of x and y is: 1.0

The difference of x*y and 1 is 0.0

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

04:31

Answer 4

Textbook Question

Chapter 2, Problem 1E

Write a program that demonstrates the approximate nature of floating-point values by performing the following tasks:

Use scanner to read a floating-point value x.
Compute 1.0 I x and store the result in y.
Display x, y, and the product of x and y.
Subtract 1 from the product of x and y and display the result.

Try your program with values of x that range from 2e−11 to 2ell. What can you conclude?

Expert Solution & Answer

Program Plan Intro

Program Plan:

• Include the required header files.
• Define main function.
- ○ Create an object for scanner.
- ○ Get the floating-point value from the user.
- ○ Calculate the “y” value.
- ○ Display “x”, “y” and product of “x” and “y”.
- ○ Subtract “1” from product of “x” and “y” and then display the result.

Program Description Answer

The program is used to read a floating point value then compute “y” and display “x”, “y”, product of “x” and “y” and also subtract “1” from product of “x” and “y” then display the result as follows:

Explanation of Solution

Program:

//include required header files

import java.util.Scanner;

//definition of "Approximation" class

public class Approximation

{

//definition of main method

public static void main(String[] args)

{

//create an object for scanner

Scanner keyboard = new Scanner(System.in);

//declare the required variable

float x = 0, y;

//get the input from the user

System.out.println("Enter a floating point value.");

x = keyboard.nextFloat();

//calculate "y" value

y = (float) (1.0/x);

//display "x" and "y" value

System.out.println("x value is: "+ x);

System.out.println("y value is: "+ y);

//display product of "x" and "y"

System.out.println("Product of x and y is: "+ (x*y));

/*subtract 1 from product of "x" and "y" and display the result*/

System.out.println("The difference of x*y and 1 is " + ((x*y) - 1.0));

}

Conclusion:

The “x” depends on user but the product of “x” and “y” will always give “1” and difference of “x * y” and “1” is always “0” because, “1” is divided by “y” and then “y” is multiplied with the “x”. So there is no change in product of “x” and “y” and difference of “x * y” and “1”.

Sample Output

Output:

Enter a floating point value.

2.5

x value is: 2.5

y value is: 0.4

Product of x and y is: 1.0

The difference of x*y and 1 is 0.0

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04:31

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

Program Plan Intro

Program Plan:

• Include the required header files.
• Define main function.
- ○ Create an object for scanner.
- ○ Get the floating-point value from the user.
- ○ Calculate the “y” value.
- ○ Display “x”, “y” and product of “x” and “y”.
- ○ Subtract “1” from product of “x” and “y” and then display the result.

Program Description Answer

The program is used to read a floating point value then compute “y” and display “x”, “y”, product of “x” and “y” and also subtract “1” from product of “x” and “y” then display the result as follows:

Explanation of Solution

Program:

//include required header files

import java.util.Scanner;

//definition of "Approximation" class

public class Approximation

{

//definition of main method

public static void main(String[] args)

{

//create an object for scanner

Scanner keyboard = new Scanner(System.in);

//declare the required variable

float x = 0, y;

//get the input from the user

System.out.println("Enter a floating point value.");

x = keyboard.nextFloat();

//calculate "y" value

y = (float) (1.0/x);

//display "x" and "y" value

System.out.println("x value is: "+ x);

System.out.println("y value is: "+ y);

//display product of "x" and "y"

System.out.println("Product of x and y is: "+ (x*y));

/*subtract 1 from product of "x" and "y" and display the result*/

System.out.println("The difference of x*y and 1 is " + ((x*y) - 1.0));

}

Conclusion:

The “x” depends on user but the product of “x” and “y” will always give “1” and difference of “x * y” and “1” is always “0” because, “1” is divided by “y” and then “y” is multiplied with the “x”. So there is no change in product of “x” and “y” and difference of “x * y” and “1”.

Sample Output

Output:

Enter a floating point value.

2.5

x value is: 2.5

y value is: 0.4

Product of x and y is: 1.0

The difference of x*y and 1 is 0.0

Answer 8

Expert Solution & Answer

Program Plan Intro

Program Plan:

• Include the required header files.
• Define main function.
- ○ Create an object for scanner.
- ○ Get the floating-point value from the user.
- ○ Calculate the “y” value.
- ○ Display “x”, “y” and product of “x” and “y”.
- ○ Subtract “1” from product of “x” and “y” and then display the result.

Program Description Answer

The program is used to read a floating point value then compute “y” and display “x”, “y”, product of “x” and “y” and also subtract “1” from product of “x” and “y” then display the result as follows:

Explanation of Solution

Program:

//include required header files

import java.util.Scanner;

//definition of "Approximation" class

public class Approximation

{

//definition of main method

public static void main(String[] args)

{

//create an object for scanner

Scanner keyboard = new Scanner(System.in);

//declare the required variable

float x = 0, y;

//get the input from the user

System.out.println("Enter a floating point value.");

x = keyboard.nextFloat();

//calculate "y" value

y = (float) (1.0/x);

//display "x" and "y" value

System.out.println("x value is: "+ x);

System.out.println("y value is: "+ y);

//display product of "x" and "y"

System.out.println("Product of x and y is: "+ (x*y));

/*subtract 1 from product of "x" and "y" and display the result*/

System.out.println("The difference of x*y and 1 is " + ((x*y) - 1.0));

}

Conclusion:

The “x” depends on user but the product of “x” and “y” will always give “1” and difference of “x * y” and “1” is always “0” because, “1” is divided by “y” and then “y” is multiplied with the “x”. So there is no change in product of “x” and “y” and difference of “x * y” and “1”.

Sample Output

Output:

Enter a floating point value.

2.5

x value is: 2.5

y value is: 0.4

Product of x and y is: 1.0

The difference of x*y and 1 is 0.0

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Program Plan Intro

Program Plan:

• Include the required header files.
• Define main function.
- ○ Create an object for scanner.
- ○ Get the floating-point value from the user.
- ○ Calculate the “y” value.
- ○ Display “x”, “y” and product of “x” and “y”.
- ○ Subtract “1” from product of “x” and “y” and then display the result.

Answer 12

Program Description Answer

The program is used to read a floating point value then compute “y” and display “x”, “y”, product of “x” and “y” and also subtract “1” from product of “x” and “y” then display the result as follows:

Answer 13

Explanation of Solution

Program:

//include required header files

import java.util.Scanner;

//definition of "Approximation" class

public class Approximation

{

//definition of main method

public static void main(String[] args)

{

//create an object for scanner

Scanner keyboard = new Scanner(System.in);

//declare the required variable

float x = 0, y;

//get the input from the user

System.out.println("Enter a floating point value.");

x = keyboard.nextFloat();

//calculate "y" value

y = (float) (1.0/x);

//display "x" and "y" value

System.out.println("x value is: "+ x);

System.out.println("y value is: "+ y);

//display product of "x" and "y"

System.out.println("Product of x and y is: "+ (x*y));

/*subtract 1 from product of "x" and "y" and display the result*/

System.out.println("The difference of x*y and 1 is " + ((x*y) - 1.0));

}

Conclusion:

The “x” depends on user but the product of “x” and “y” will always give “1” and difference of “x * y” and “1” is always “0” because, “1” is divided by “y” and then “y” is multiplied with the “x”. So there is no change in product of “x” and “y” and difference of “x * y” and “1”.