The negation of infinite

**novice** · March 1st 2010, 07:12 PM

What is the correct negation of the statement

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

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

or

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

**Drexel28** · March 1st 2010, 07:13 PM

Originally Posted by novice

What is the correct negation of the statement

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

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

or

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

The second.

**Plato** · March 2nd 2010, 07:14 AM

I disagree. Nether of those is the negation.
It is: $\left( {\exists j \in \mathbb{Z}^ + } \right)\left( {\forall n \in \mathbb{Z}^ + ,n > J \Rightarrow \sqrt n \in \mathbb{Q}} \right)$ .
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.

**novice** · March 2nd 2010, 07:42 AM

Originally Posted by Plato

I disagree. Nether of those is the negation.
It is: $\left( {\exists j \in \mathbb{Z}^ + } \right)\left( {\forall n \in \mathbb{Z}^ + ,n > J \Rightarrow \sqrt n \in \mathbb{\overline Q}} \right)$ .
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 $\mathbb{Q}$ . 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?

**Plato** · March 2nd 2010, 07:51 AM

Originally Posted by novice

Plato,
I put a bar over your $\mathbb{Q}$ . 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?

Why add an overline?
That says that every integer greater than J has a rational square root.
That means there are only finitely many integers having irrational square roots.

**Drexel28** · March 2nd 2010, 03:29 PM

I'm sorry. I guess I didn't read the actual responses close enough.

**novice** · March 2nd 2010, 05:25 PM

Originally Posted by Drexel28

I'm sorry. I guess I didn't read the actual responses close enough.

It's perfectly alright. I am very grateful that you always help me.

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