for all x vs there exists an x

[GeniusAreMade]

Can someone give me a mathematical proof to prove that the negation of "for all x" is "there exists an x"

 

[Steven G]

Suppose for x statement P is true.

The negation of for all x statement P is true is it is not the case that for all x that P is true. So there must exist an x such that P is not true.

 

[GeniusAreMade]

Ah okay, I got the logic behind it, thank you sir.

 

[Steven G]

I usually start off the negation of a statement P by saying that it is NOT the case statement P

 

[GeniusAreMade]

yes I learned that, it seems a fairly simple chapiter
I am spending my free time on The Collatz Conjecture, trying to connect the dots and prove the conjecture

 

[pka]

Can someone give me a mathematical proof to prove that the negation of "for all x" is "there exists an x"

The question you ask goes all the way back to Aristotle in about the second century BCE.
There four types of propositions: \(\mathcal{~A,~~E,~~I,~~O}\)
1) \(\mathcal{~A}\) are universal positive: \((\forall x)[P(x) \Rightarrow Q(x)\) All \(P\) IS \(Q\).
2) \(\mathcal{~E}\) are universal negative: \((\forall x)[P(x) \Rightarrow \neg Q(x)\) No \(P\) IS \(Q\)
3) \(\mathcal{~I}\) are existential positive: \((\exists x)[P(x) \wedge Q(x)\) SOME \(P\) IS \(Q\)
4) \(\mathcal{~O}\) are existential negative: \((\exists x)[P(x) \wedge \neg Q(x)\) SOME \(P\) IS NOT \(Q\).
Now this is known as the Square of Opposition
\(\begin{array}{*{20}{c}}
\mathcal{A}&{} {}&{}&{}&\mathcal{E} \\
{}& \nwarrow &{}&{}& \nearrow &{} \\
{}&{}& \nwarrow & \nearrow &{}&{} \\
{}&{}& \swarrow & \searrow &{}&{} \\
{}& \swarrow &{}&{}& \searrow &{} \\
\mathcal{I}&{}&{}&{}&{}&\mathcal{O} \end{array}\)

 

[GeniusAreMade]

wow this is a very interesting notion, it would be better if the university explains us the history of these notions because the whole chapiter will make more sense and easily to understand. Thanks for the culture pka.