Logical Negation of statements.

Question

1. Between every two distinct real numbers, there is a rational number

Answer: There is no rational numbers between two non-distinct real numbers.

2. For all natural numbers $n ∈ N, \sqrt n$ is either a natural number or an irrational number

Answer: For all natural numbers $n$, $\sqrt n$ is either not a natural number or not a irrational number.

3. Given any real number $x ∈ R$, there exists $n ∈ N$ satisfying $n>x$.

Answer: ??

Can someone tell me what is the general way to look at these things.

Answer 1

In order to negate:

1) "Between every two distinct real numbers, there is a rational number",

we have to assert that "There are two distinct real numbers such that there is no rational number between them".

It may help to formalize the statements with quantifiers:

$\forall r_1, r_2 \in \mathbb R \ (r_1 < r_2 \to \exists q \in \mathbb Q \ (r_1 < q < r_2))$.

The "procedure" to get the correct negation is simply to put the negation sign in front and then "move it inside" with the equivalences:

$\lnot \forall$ is equiv to $\exists \lnot$ and $\lnot \exists$ is equiv to$\forall \lnot$.

Thus, from:

$\lnot \ [\forall r_1, r_2 \in \mathbb R \ (r_1 < r_2 \to \exists q \in \mathbb Q \ (r_1 < q < r_2))],$

we get in the first step:

$\exists r_1, r_2 \in \mathbb R \ \lnot (r_1 < r_2 \to \exists q \in \mathbb Q \ (r_1 < q < r_2)).$

The next step is to use the propositional equivalence between $\lnot (p \to q)$ and $(p \land \lnot q)$, to get:

$\exists r_1, r_2 \in \mathbb R \ (r_1 < r_2 \land \lnot \exists q \in \mathbb Q \ (r_1 < q < r_2)).$

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For:

3) "Given any $x \in \mathbb R$, there exists $n \in \mathbb N$ satisfying $(n > x)$",

we have for the negated statement:

$\lnot \forall x \in \mathbb R \ \exists n \in \mathbb N \ (n > x)$.

Applying the above equivalences we get:

$\exists x \in \mathbb R \ \forall n \in \mathbb N \ (n \le x)$.

Answer 2

The general rule is that "For all" statements should be changed to "There exists" and "There exists" should be changed to "For all". Then apply this inward in the statement. It is often easier to rewrite statements in the syntactic logical way to get a better formal grip. (i.e. using $\exists x, \forall y $ etc.)

1. Say that "for all distinct ..." so we change this to a "there exists." This leads to the statement

   There exists two distinct real numbers such that there is no rational number between these.

Now it is not clear exactly how far one should push the negation, currently I let the negation stand on the "There exists" part, because this makes the formulation clear, but we could reformulate it as

There exists two distinct real numbers such that for any rational number, this number is not between the two distinct real numbers.

which pushes the negation as far as it can go. On the other hand a negation of the sentence is also

It is not the case that between every two distinct real numbers, there is a rational number

But this might not be what anyone wanted, still it is very clearly a negation of the sentence.

For sentence 2 and 3 there are similar discussions regarding the quantifiers which you need to change. Thus when the sentence starts with "For any real number x there exists a natural number n such that $n>x$ we will need to swap quantifiers to get

There is a real number $x$ such that for any natural number $n, n\not > x$