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Regard this statement $ x \ge 0$. According to my teacher, by negating this statement, it will become $ x < 0$. Why is this so; why does the $\ge$ morph into $<$, and not into $\le$?

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  • $\begingroup$ "Planning Algorithms" by LaValle sect. 4.3.2 defines a strangely different definition of interval negation where negating [-1,2] results in [-2,1] (which also redefines Minkowski differences). There is very little information on this, all the book says on it is "In some contexts... the Minkowski difference ... is defined differently." without saying what context the book is using. It also introduces new properties where A - B = A + (-B), which is not normally true. $\endgroup$
    – jrh
    Jul 30 '19 at 19:57
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There is no "morphing", and this is not just a game played arbitrarily with squiggles on the paper. The symbols mean things, and you can reason out their behaviors if you understand the meanings. $x\ge 0$ means that $x$ is equal to or greater than zero. Negating the statement means constructing a statement whose meaning is "$x$ is not equal to or greater than zero".

Which of $x<0$ and $x\le 0$ means "$x$ is not equal to or greater than zero"? It can't be $x\le 0$, because that means that $x$ is less than or equal to zero, and we are trying to say that it is not equal to zero.

$x<0$ is correct, because if $x$ is not greater than or equal to zero, then it must be less than zero, and that is exactly what $x<0$ means.

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logically, there is no "morphing" going on here; your teacher is simply restating the same thing meant by $\,(x \ngeq 0)\,$, but just put differently $\,(x \lt 0)\,$:

NOT $\geq$ means:

"is NOT (greater than OR equal to)", $\iff$ ("is NOT greater than" AND "is NOT equal to")*...

...which leaves us with ("is less than")

that is "NOT $\geq$" must mean "is less than"

Formally, this is an application of (*) DeMorgan's Law (recall: $\lnot(p \lor q) \equiv \lnot p \land \lnot q$), and of the Trichotomy Law: between any two real numbers $a, b$, one and only one of the following relations holds:

  • $a \lt b $
  • $a = b$
  • $a \gt b$

The negation of $\,x\geq0\,$ ($x\ngeq0$) rules out $x = 0$ and rules out $x \gt o$ leaving us with just $x < 0$

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Try with sets if you don't see it with inequalities directly.

The statement $x\geq 0$ is equivalent, by definition of intervals, to saying that $x$ belongs to $[0,+\infty)$ where $0$ is included.

Therefore the negation is: $x$ belongs to the complement of $[0,+\infty)$, namely $(-\infty,0)$, where $0$ is now excluded.

The last condition is by definition equivalent to $x<0$.

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Saying that $x \geq 0$ is the same as saying that $x$ is a number that is

  • greater than $0$ or
  • equal to zero.

If you wanted to negate that, you get that $x$ is not be a number greater than or equal to zero. And what are the numbers that are

  • not greater than $0$ and
  • not equal to $0$?

Those are exactly the negative numbers, so $x< 0$.

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the negation of $x \ge 0$ is $\neg x \ge 0$.

$\neg x \ge 0$ is equivalent to $x < 0$.

It's easy to see this if you consider the set of values of $x$ that make the formula hold true:

  • $x \ge 0$ holds for x in {0,1,2,3,4,5,...}
  • $\neg x \ge 0$ holds for x in {...,-4,-3,-2,-1} (that's everything not in the previous set)
  • $x < 0$ holds for x in {...,-4,-3,-2,-1}
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