In what way do relational operators not obey the compareTo contract with floating point values?

From the javadoc:

 public int compareTo(Double anotherDouble)

Compares two Double objects numerically. There are two ways in which comparisons performed by this method differ from those performed by the Java language numerical comparison operators (<, <=, ==, >=, >) when applied to primitive double values: Double.NaN is considered by this method to be equal to itself and greater than all other double values (including Double.POSITIVE_INFINITY). 0.0d is considered by this method to be greater than -0.0d. This ensures that the natural ordering of Double objects imposed by this method is consistent with equals.

From JavaDoc for Double::compareTo

Compares two Double objects numerically. There are two ways in which comparisons performed by this method differ from those performed by the Java language numerical comparison operators (<, <=, ==, >= >) when applied to primitive double values:

• Double.NaN is considered by this method to be equal to itself and greater than all other double values (including Double.POSITIVE_INFINITY).

• 0.0d is considered by this method to be greater than -0.0d.

This ensures that Double.compareTo(Object) (which forwards its behavior to this method) obeys the general contract for Comparable.compareTo, and that the natural order on Doubles is consistent with equals.

    double d1 =Double.NaN;
    double d2 = Double.NaN;

    System.out.println(Double.valueOf(d1).equals(d2));    ---> true
    System.out.println(Double.valueOf(d1).compareTo(d2));  ---> 0
    System.out.println(d1 == d2);                          --->false