What is the negation of this statement and which statement is true, R or ¬R?

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Discussion Overview

The discussion revolves around the logical statement R concerning prime numbers, even numbers, and divisibility, specifically focusing on the formulation of its negation and the truth value of R versus ¬R. Participants explore the logical implications and equivalences involved in the negation process.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Participants define R as (∀x ∈ Z)(P(x) ⇒ ((∃y ∈ Z)(E(y) ∧ D(x, y)))) and attempt to derive its negation ¬R.
  • Some participants propose that ¬R can be expressed as (∃x ∈ Z)(P(x) ∧ (∀y ∈ Z)((¬E(y)) ∨ (¬D(x, y)))) while others question this formulation.
  • There is a discussion about the logical equivalence of (P(x) ⇒ Q(x)) and (P(x) ∨ ¬Q(x)), with some participants asserting that ¬(P(x) ⇒ Q(x)) simplifies to (P(x) ∧ (¬Q(x))).
  • One participant expresses uncertainty about the correctness of their negation and the implications of the logical equivalences discussed.
  • Another participant suggests that R may be true by providing an example where y=2x, but acknowledges potential confusion regarding the definitions involved.
  • There is a recognition of conflicting interpretations of the logical statements, with some participants correcting each other on the equivalences.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correctness of the negation or the truth value of R versus ¬R. Multiple competing views and interpretations of the logical statements remain present throughout the discussion.

Contextual Notes

There are unresolved issues regarding the definitions of the logical constructs and the implications of the negation process. Some participants express confusion over the logical equivalences, indicating a need for clarity in the foundational concepts being discussed.

rayne1
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Let P(x) be assertion “x is prime”, E(x) be “x is even”, and D(x, y) be “x divides y” (i.e., y/x is an integer). Consider the following statement:
R = (∀x ∈ Z)(P(x) ⇒ ((∃y ∈ Z)(E(y) ∧ D(x, y))))

Write the negation of R, and determine which statement is true, R or ¬R.

I tried, but I'm not sure if I got the correct answer:
¬R = (∃x ∈ Z)(P(x) ∧ (∀y ∈ Z)((¬E(y)) ∨ (¬D(x, y)))

It seems that ¬R is true.
 
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rayne said:
Let P(x) be assertion “x is prime”, E(x) be “x is even”, and D(x, y) be “x divides y” (i.e., y/x is an integer). Consider the following statement:
R = (∀x ∈ Z)(P(x) ⇒ ((∃y ∈ Z)(E(y) ∧ D(x, y))))

Write the negation of R, and determine which statement is true, R or ¬R.

I tried, but I'm not sure if I got the correct answer:
¬R = (∃x ∈ Z)(P(x) ∧ (∀y ∈ Z)((¬E(y)) ∨ (¬D(x, y)))

It seems that ¬R is true.

Hi rayne!

Let's define:
Q(x) = (∃y ∈ Z)(E(y) ∧ D(x, y))

Then we have:
R = (∀x ∈ Z)(P(x) ⇒Q(x))

So:
¬R = (∃x ∈ Z)(¬(P(x) ⇒Q(x)))

Did you know that:
(P(x) ⇒Q(x)) = (P(x) ∨ ¬Q(x))
(Wondering)

So what does that make ¬(P(x) ⇒Q(x))?
 
I like Serena said:
Hi rayne!

Let's define:
Q(x) = (∃y ∈ Z)(E(y) ∧ D(x, y))

Then we have:
R = (∀x ∈ Z)(P(x) ⇒Q(x))

So:
¬R = (∃x ∈ Z)(¬(P(x) ⇒Q(x)))

Did you know that:
(P(x) ⇒Q(x)) = (P(x) ∨ ¬Q(x))
(Wondering)

So what does that make ¬(P(x) ⇒Q(x))?
P(x)∧(¬Q(x))
 
rayne said:
P(x)∧(¬Q(x))

How did you get that?
 
I like Serena said:
How did you get that?
I know that they're logically equivalent
 
rayne said:
I know that they're logically equivalent

So you're saying that:
¬(P(x) ⇒Q(x)) = (P(x)∧(¬Q(x)))

Let's take another look.
Substitute (P(x) ⇒Q(x)) = (P(x) ∨ ¬Q(x)) to get:
¬(P(x) ⇒Q(x)) = ¬(P(x) ∨ ¬Q(x))

... but wait! Isn't that the negation of what you are saying? :eek:
 
I like Serena said:
So you're saying that:
¬(P(x) ⇒Q(x)) = (P(x)∧(¬Q(x)))

Let's take another look.
Substitute (P(x) ⇒Q(x)) = (P(x) ∨ ¬Q(x)) to get:
¬(P(x) ⇒Q(x)) = ¬(P(x) ∨ ¬Q(x))

... but wait! Isn't that the negation of what you are saying? :eek:

I guess so.
 
I have to go right now, so can't reply in the next 24 hours.
 
rayne said:
Let P(x) be assertion “x is prime”, E(x) be “x is even”, and D(x, y) be “x divides y” (i.e., y/x is an integer). Consider the following statement:
R = (∀x ∈ Z)(P(x) ⇒ ((∃y ∈ Z)(E(y) ∧ D(x, y))))

Write the negation of R, and determine which statement is true, R or ¬R.

I tried, but I'm not sure if I got the correct answer:
¬R = (∃x ∈ Z)(P(x) ∧ (∀y ∈ Z)((¬E(y)) ∨ (¬D(x, y)))

It seems that ¬R is true.
Rayne, you are right on both counts.

I like Serena said:
Did you know that:
(P(x) ⇒Q(x)) = (P(x) ∨ ¬Q(x))
No, (P(x) ⇒ Q(x)) ⇔ (¬P(x) ∨ Q(x)).
 
  • #10
Evgeny.Makarov said:
Rayne, you are right on both counts.

Let's read what R says:
For every x in Z, if x is prime, then there is an y such that y is even and x divides y.
I believe this is true. Pick y=2x.

No, (P(x) ⇒ Q(x)) ⇔ (¬P(x) ∨ Q(x)).

Ah. My bad.
 
  • #11
I like Serena said:
Let's read what R says:
For every x in Z, if x is prime, then there is an y such that y is even and x divides y.
I believe this is true. Pick y=2x.
Yes, of course. I must have read $D(x,y)$ the other way. The negation is still constructed correctly, I believe.

Well, this thread has some unfortunate responses, doesn't it?
 

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