negation of statement

**relyt** · March 28th 2009, 07:11 PM

Here is the statement about set of real numbers:
$(\forall x \in R) (x \neq 5 \rightarrow (\exists y \in R) (y < x))$

I need to find the negation of it.

This is what I have so far. Am I correct?

$(\exists x \in R) (x = 5 \wedge (\forall y \exists R) (y > x))$

**Jhevon** · March 28th 2009, 07:40 PM

Originally Posted by relyt

Here is the statement about set of real numbers:
$(\forall x \in R) (x \neq 5 \rightarrow (\exists y \in R) (y < x))$

I need to find the negation of it.

This is what I have so far. Am I correct?

$(\exists x \in R) (x = 5 \wedge (\forall y \exists R) (y > x))$

i think the $x \ne 5$ should stay

the negation of $P \implies Q$ is $P \wedge \sim Q$

so the antecedent does not change

**bmp05** · March 29th 2009, 12:01 AM

$(\forall x \in R) (x \neq 5 \rightarrow (\exists y \in R) (y < x))$

So, would you say this like:
"For the set of all real numbers x, if x does not equal 5 then their exists (at least) one y in the set of real numbers such that y is less than x."

A is "For the set of all real numbers x, x not equal to 5."
B is "There exists one y in the set of real numbers such that y is less than x"

A $\Rightarrow$ B $\equiv$ B $\vee \neg$ A.
The negation is A $\wedge \neg$ B

$\neg$ B $\equiv (\forall y \in R) (y > x)$
so,

$\equiv (\forall x \in R) (x \neq 5) \wedge (\forall y \in R) (y > x)$

And you would say, "For the set of all real numbers x, x not equal to 5 and the set of all real numbers y, y is greater than x."

So, can you simplify this to, "for the set of all real numbers, y is greater than x, with the exception of x equals 5, where x is not defined"?

**Plato** · March 29th 2009, 04:08 AM

Originally Posted by relyt

Here is the statement about set of real numbers:
$(\forall x \in R) (x \neq 5 \rightarrow (\exists y \in R) (y < x))$
I need to find the negation of it. This is what I have so far. Am I correct?
$(\exists x \in R) (x = 5 \wedge (\forall y {\color{red}\exists R}) (y {\color{red}>} x))$

Almost, two corrections.
$(\exists x \in R) (x \not= 5 \wedge (\forall y {\color{blue}\in R}) (y {\color{blue}\ge} x))$

**bmp05** · March 29th 2009, 05:08 AM

$A \Rightarrow B \equiv B \vee \neg A$

Are thes two statements equivalent?
$(\forall x \in \mathbb{R})(x \Rightarrow (\exists y \in \mathbb{R})(y)) \equiv (\exists y \in \mathbb{R})(y \vee (\exists x \in \mathbb{R})(x))$

**Plato** · March 29th 2009, 05:17 AM

Originally Posted by bmp05

$A \Rightarrow B \equiv B \vee \neg A$
Are thes two statements equivalent?
$(\forall x \in \mathbb{R})(x \Rightarrow (\exists y \in \mathbb{R})(y)) \equiv (\exists y \in \mathbb{R})(y \vee (\exists x \in \mathbb{R})(x))$

I have no idea what those statements could possibly mean.
But I do know that they are not equivalent.

**bmp05** · March 29th 2009, 05:34 AM

Originally Posted by Plato

I have no idea what those statements could possibly mean.
But I do know that they are not equivalent.

Yes, I just found this link, which might be helpful: An Elementary Introduction to Logic and Set Theory: Methods of Proof

And there's a section Rules of Replacement for Quantified Predicates with an example, which shows that what I wrote in my last post is definately wrong, but now, I wonder what the negation of (x) is?!
$\exists x \in \mathbb{R}: \neg (x)$ does that even make sense?

**Plato** · March 29th 2009, 06:14 AM

Originally Posted by bmp05

I wonder what the negation of (x) is?!
$\exists x \in \mathbb{R}: \neg (x)$ does that even make sense?

$\neg (x)$ is just not proper notation. What does not x mean?
Usually we have a propositional function.
Example: Let P(x) mean that x is a rational number.
Now $\neg P(x)$ reads “x is not a rational number.”
$\neg \left( {\forall x} \right)\left[ {P(x)} \right]$ reads: “It is false that every x is rational.
$\left( {\exists x} \right)\left[\neg {P(x)} \right]$ reads: “Some x is not rational.
Clearly those two statements are equivalent.

**bmp05** · March 29th 2009, 06:37 AM

Yes, this is really quite interesting, so you can negate a predicate, but is there so much different between P(x):= (x $\neq$ 5) and P(y):=(x)..?

So, the nagation of P $\forall x: x \neq 5 \equiv \exists x: x=5 \equiv x=5$ ? But then the negation of Q $\forall x: x \equiv \neg$ Q $\equiv$ Because, you can negate all x, that would be for any x, False.

Originally Posted by Plato

$\neg (x)$ is just not proper notation. What does not x mean?
Usually we have a propositional function.
Example: Let P(x) mean that x is a rational number.
Now $\neg P(x)$ reads “x is not a rational number.”
$\neg \left( {\forall x} \right)\left[ {P(x)} \right]$ reads: “It is false that every x is rational.
$\left( {\exists x} \right)\left[\neg {P(x)} \right]$ reads: “Some x is not rational.
Clearly those two statements are equivalent.

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