Given the following method, what is the result of getNumber(n)?
Public static double getNumber(double n)
{
Return Math.pow(Math.sqrt(n), 2) - n;
}
A) Always 0 for every argument n
B) Always 2 for every argument n
C) Close to 0, but not always 0
D) Compilation error

Question

Quizplus · Accepted Answer

The method first calculates the square root of n using the Math.sqrt() method, and then raises that result to the power of 2 using the Math.pow() method. This essentially just gives us n, so we can simplify the method to:

public static double getNumber(double n) {
   return n - n;
}

This further simplifies to:

public static double getNumber(double n) {
   return 0;
}

Therefore, the result of getNumber(n) will always be 0 for every argument n. However, it is worth noting that due to rounding errors in floating-point arithmetic, the result may not always be exactly 0, but will be very close to 0.

Given the Following Method, What Is the Result of GetNumber(n)