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  1. When comparing a double to zero in Java, direct equality checks (== 0.0) can be unreliable due to floating-point precision errors. Instead, use an epsilon-based comparison to account for tiny rounding differences.

    Example:

    public class CompareDouble {
    public static void main(String[] args) {
    double discriminant = 0.00000000000000001236;
    double EPSILON = 1e-9; // tolerance level

    if (Math.abs(discriminant) < EPSILON) {
    System.out.println("Equal to zero");
    } else if (discriminant > 0) {
    System.out.println("Greater than zero");
    } else {
    System.out.println("Less than zero");
    }
    }
    }
    Copied!

    This approach ensures that values extremely close to zero are treated as zero, avoiding false negatives caused by floating-point representation issues.

    Why not use == directly? Floating-point numbers like 0.1 cannot be represented exactly in binary, so calculations may produce results like 0.00000000000000001236 instead of 0.0. Direct equality fails in such cases.

    Alternative Approaches:

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