Help with negating proposition $P: \forall a\in\Bbb R, \forall b\in\Bbb R, f(a)=f(b)\rightarrow a=b$

Question

Let $\,f(x)=x^2-2x-8\,,\,$ $\forall x\in\Bbb R$.
Consider the proposition
$P: \forall a\in\Bbb R,\;\forall b\in\Bbb R,\, f(a)=f(b)\rightarrow a=b$

1. Find $f(x)=0$.
2. Negate the proposition $P$.
3. Find truth values for $P$.

So, I started by finding the values for $f(x)=0$, which the $x$ values are $4$ and $-2$. I started trying to negate it, but I wasn't sure how to. Do I negate the $\forall a$ ? And if I do, where does the negation symbol go? I’m not sure how to go about negation here.

Answer 1

Some rules of predicate logic:

1. $\neg\forall x\Phi(x)\equiv \exists x\neg\Phi(x)$
2. $A\to B\equiv \neg A\vee B$
3. $\neg(C\vee D)\equiv \neg C\wedge \neg D$

So $$\begin{align*}&\neg\left[\forall a\in\Bbb R\;\forall b\in\Bbb R\, \left(f(a)=f(b)\rightarrow a=b\right)\right] \\&\equiv\exists a\in\Bbb R\;\neg\left[\forall b\in\Bbb R\,\left(f(a)=f(b)\rightarrow a=b\right)\right]\\&\equiv \exists a\in\Bbb R\,\exists b\in\Bbb R\, \neg\left(f(a)=f(b)\rightarrow a=b\right)\\&\equiv\exists a\in\Bbb R\,\exists b\in\Bbb R\, \neg\left(f(a)\neq f(b)\vee a=b\right)\\&\equiv\exists a\in\Bbb R\,\exists b\in\Bbb R\, \left(f(a)=f(b)\wedge a\neq b\right) \end{align*}$$

Let's translate out of the formalism. The property $P$ says that if $f$ has the same value for arguments $a$ and $b$, then $a$ and $b$ are equal. In other words, different arguments $a$ and $b$ always result in different values $f(a)$ and $f(b)$. In other words, $f$ is injective.

The negation says injectivity fails for at least one pair of arguments $a$ and $b$. So we must have $a\neq b$, but nevertheless $f(a)=f(b)$.

Your answer to (3) shows that $f(4)=0$ and $f(-2)=0$, so we can set $a=4$ and $b=-2$ to show that the negation is true. (Logicians say that $a=4,b=-2$ "witnesses" the truth of $\neg P$.)

Answer 2

$\lnot P:\exists a\in\Bbb R,\,\exists b\in\Bbb R\,,\,f(a)=f(b)\,\land\,a\neq b\,.$

Let $\,P(a,b)\,$ be the proposition:
$(a,b)\in\Bbb R^2\,,\;f(a)=f(b)\rightarrow a=b\,.$

$P(a,b)\,$ is true if and only if $\;(a,b)\in\Bbb R^2\,$ and $\;a+b\neq2\;.$

Proof :

$P(a,b)\,$ is true $\;$ if and only if

$(a,b)\in\Bbb R^2\,,\; f(a)=f(b)\rightarrow a=b\;\;$ if and only if

$(a,b)\in\Bbb R^2\,,\;a^2\!-\!2a\!-\!8=b^2\!-\!2b\!-\!8\rightarrow a=b\;\;$ if and only if

$(a,b)\in\Bbb R^2\,,\;a^2-b^2+2b-2a=0\rightarrow a=b\;\;$ if and only if

$(a,b)\in\Bbb R^2,\;(a\!+\!b)(a\!-\!b)\!-\!2(a\!-\!b)\!=\!0\rightarrow a\!=\!b\;$ if and only if

$(a,b)\in\Bbb R^2\,,\;(a\!-\!b)(a\!+\!b\!-\!2)=0\rightarrow a\!-\!b=0\;\;$ if and only if

$(a,b)\in\Bbb R^2\,,\;a+b-2\neq0\;\;$ if and only if

$(a,b)\in\Bbb R^2\,,\;a+b\neq2\;.$

Answer 3

$II: (\forall a\in R) \land(\forall b \in R), f(a)=f(b)\implies a=b$

Might help to work backwards.

$p\implies q \equiv (q \lor \lnot p) \therefore \lnot(p \implies q)\equiv (p \land \lnot q)$

So negating the second part means $f(a)=f(b) $ and $a \ne b.$

So II can be rewritten as $\forall (a,b)\in R\times R, \lnot(p \land \lnot q)$

The negation of that is $\exists (a,b) \in R \times R, \lnot \lnot (p \land \lnot q)$

If for all given entities, statement S is true, the negation is there exist and entity for which it isn't true.

$\lnot II:\exists (a,b)\in R\times R , f(a)=f(b) \land a \ne b $

Answer 4

Let $\,f(x)=x^2-2x-8\,,\,$ $\forall x\in\Bbb R$.
Consider the proposition $P: \forall a\in\Bbb R,\;\forall b\in\Bbb R,\, f(a)=f(b)\rightarrow a=b$

Do I negate the $\forall a$ ? And if I do, where does the negation symbol go?

1. Negating means flipping the truth value, so not [every x satisfies Q(x)] means some x doesn't satisfy Q(x).

   Notice that the negation symbol does not need to be present; after all, every sentence is the negation of some sentence yet clearly not every sentence contains the negation symbol.

2. What does it mean to say that if A then B is false? (For example, is it possibly false if A is false?)

3. Can you see that proposition P refers to the horizontal-line test? What does it mean to pass or to fail this test?

4. Although ∀ does stand for 'for each' and ∃ for 'for some', it is bad practice to treat quantifier symbols as mere short forms whose positions convey no information: for example, instead of writing Q(x,y) ∃x ∀y (ambiguous), place the quantifiers at the beginning, like ∃x∀y Q(x,y) or ∀y∃x Q(x,y) (the former implies the latter, but not vice versa).