Determine the logical value of the following proposition $\exists x \in\Bbb R: -23/4.$

Question

Since $-2<x<0$, then $x^2-2<x^2+x<x^2$. So we can take $x=-\frac{\sqrt{11}}{2} \in (-2,0)$. So we have, $\left(-\frac{\sqrt{11}}{2}\right)^2-2=3/4<x^2+x$. Is this reasoning correct?

Answer 1

Proposition $P : \exists x \in \Bbb R: (-2 \lt x \lt 0) \implies (x^2+x \gt 3/4)$

(A) With "Common" non-rigorous Interpretation : It is not true.

(B) With rigorous Interpretation : It is "Co-Incidentally" true , but not for the reason you think it is true.

Elaborating (A) :

When you intend to claim that "when-ever (1) $(-2 \lt x \lt 0)$ is true , (2) $(x^2+x \gt 3/4)$ must be true" , then we can check that $x=-1$ makes (1) true , but (2) is not. $P$ is thus not true.

Here , (1) does not Imply (2) through elementary transformation operations like addition & multiplication.

Proposition $X : \exists x \in \Bbb R: (-2 \lt x \lt 0) \land (x^2+x \gt 3/4)$
We can take $x=-1.9$ , which makes both (1) & (2) true. $X$ is thus true.

This $X$ is closer to what you might want to claim about Calculus & real number theory , rather than logical Implication.
If logical Implication is your focus , then the given Example is too complicated & might obfuscate what you want to highlight.

Proposition $Y : \exists x \in \Bbb R: (y \lt x) \implies (y^2 \gt x)$
We can take $x=0$ , which makes the Implication true.
We can take $x$ Negative too , which makes the Implication true.
Every time (3) $(y \lt x)$ is true , then (4) $(y^2 \gt x)$ is true too.
$Y$ is thus true.

That type of "truthness" is not there in $P$.

Elaborating (B) :

In logic , "non-true $\implies$ true" & "non-true $\implies$ not true".

Thus when we take some $x$ which makes (1) not true , then "Automatically" $P$ will become "vacuously" true without checking (2) which is redundant here. That "truthness" is not because of some Property of Calculus & real numbers.

Consider these 2 Statements in general template :
$\exists x : x \in Y \implies Z(x)$ : always "vacuously" true when we can take $x \not \in Y$ (focusing on logical Implication)
$\exists x : x \in Y \land Z(x)$ : we have to analyse to Prove it & figure out the validity (focusing on Calculus & real number theory)

When you want to make some such claim , while utilizing Properties of Calculus & real numbers , try something like this template :

Proposition $Z : \exists x \in \Bbb R: U(x,y) \implies V(x,y)$ , where you have $x$ value making $U$ & $V$ true.

Example :

Proposition $W : \exists x \in \Bbb R: (0 \lt y \lt x) \implies (y^2 \lt y)$
Here $x=1,0.7,0.05,0.002$ are values which make "true \implies true" , thus $W$ is true.

Property used here :
Squares of Natural numbers will not Decrease : $10^2 = 10 \gt 10$.
Considering real numbers , Squaring gives larger number , in general : $9.5^2 = 90.25 \gt 9.5$.
Weirdly , when we Square numbers between 0 & 1 , we reduce it : $0.6^2 = 0.36 \lt 0.6$.
Proposition $W$ utilizes that Property.

Answer 2

Your result is correct since you found an $x$ such that $P(x)$ and $Q(x)$ are satisfied, and then so is $P(x)→Q(x).$ But a much simpler solution would be to pick any $x$ such that $P(x)$ is not satisfied, and not bother about $Q(x).$ Here for instance $x=1.$