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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5 Answers
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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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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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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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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}