Write a
- 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:
- • 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.
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”.
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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Chapter 2 Solutions
Java: An Introduction to Problem Solving and Programming (8th Edition)
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Starting Out with Java: From Control Structures through Objects (7th Edition) (What's New in Computer Science)
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