# Thread: The negation of infinite

1. ## The negation of infinite

What is the correct negation of the statement

P: There are infinitely many integers $\displaystyle n$ such that $\displaystyle \sqrt{n}$ is irrational.

Is it correct to say $\displaystyle \sim P:$ There exist no integers $\displaystyle n$ such that $\displaystyle \sqrt{n}$ is irrational.

or

Is it correct to say $\displaystyle \sim P:$ There is a finite set of integers $\displaystyle n$ such that $\displaystyle \sqrt{n}$ is irrational.

2. Originally Posted by novice
What is the correct negation of the statement

P: There are infinitely many integers $\displaystyle n$ such that $\displaystyle \sqrt{n}$ is irrational.

Is it correct to say $\displaystyle \sim P:$ There exist no integers $\displaystyle n$ such that $\displaystyle \sqrt{n}$ is irrational.

or

Is it correct to say $\displaystyle \sim P:$ There is a finite set of integers $\displaystyle n$ such that $\displaystyle \sqrt{n}$ is irrational.
The second.

3. I disagree. Nether of those is the negation.
It is: $\displaystyle \left( {\exists j \in \mathbb{Z}^ + } \right)\left( {\forall n \in \mathbb{Z}^ + ,n > J \Rightarrow \sqrt n \in \mathbb{Q}} \right)$.
That is, there is at most a finite set of integers each of which has an irrational square root.
Or: “the set of integers having irrational square roots is finite."

The statement “There is a finite set of integers each having a irrational square root” is always true.

4. Originally Posted by Plato
I disagree. Nether of those is the negation.
It is: $\displaystyle \left( {\exists j \in \mathbb{Z}^ + } \right)\left( {\forall n \in \mathbb{Z}^ + ,n > J \Rightarrow \sqrt n \in \mathbb{\overline Q}} \right)$.
That is, there is at most a finite set of integers each of which has an irrational square root.
Or: “the set of integers having irrational square roots is finite."

The statement “There is a finite set of integers each having a irrational square root” is always true.
Plato,
I put a bar over your $\displaystyle \mathbb{Q}$ . I am just guessing because someone told me that there is no symbol for irrational numbers.
Be that as it may, what is n > J?

5. Originally Posted by novice
Plato,
I put a bar over your $\displaystyle \mathbb{Q}$ . I am just guessing because someone told me that there is no symbol for irrational numbers.
Be that as it may, what is n > J?