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?

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