doubt reg NaN (from JLS)

[Rajeshwari Natarajan]

In JLS section 4.2.3, the following points are given.

NaN is unordered, so the numerical comparison operators <, <=, >, and >= return false if either or both operands are NaN (�15.20.1). The equality operator == returns false if either operand is NaN, and the inequality operator != returns true if either operand is NaN (�15.21.1). In particular, x!=x is true if and only if x is NaN, and (x<y) == !(x>=y) will be false if x or y is NaN.

I don't get this. can anyone explain?

[Valentin Crettaz]

In clear, it means that NaN cannot be compared with numbers, that is,
NaN > 3 yields false
3 > NaN yields false
NaN >= 3 yields false
3 >= NaN yields false
NaN < 3 yields false
3 < NaN yields false
NaN <= 3 yields false
3 <= NaN yields false
3 == NaN yields false
3 != NaN yields true
NaN != NaN yields true (!!!)
(NaN < 3) == !(NaN >= 3) yields false
(3 < NaN) == !(3 >= NaN) yields false

[Rajeshwari Natarajan]

thanx a lot..that was a very clear and easy to understand explanation