Question 1 of 12

Question 1

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. An existentially quantified inequality $\exists x\ (2x^2\le x-1)$, a compound proposition in the three variables $p,q,r$, and two quantified implications built from predicates $P(x),Q(x)$.

Find. (a) the negation, written with the universal quantifier $\forall$. (b) the 8-row truth table of $(\neg r\lor q)\to(p\lor r)$. (c) whether the two quantified statements always have the same truth value.

Approach. (a) apply the quantifier-negation (De Morgan) law $\neg\exists x\,S(x)\equiv\forall x\,\neg S(x)$ and negate the inequality. (b) build the table over all $2^3=8$ assignments; a conditional is false only when its antecedent is true and its consequent false. (c) try to build a counterexample universe/predicate pair.

1. a) Negate $\exists x\ (2x^2\le x-1)$ using $\forall$. $\neg\exists x\,S(x)\equiv\forall x\,\neg S(x)$, and the negation of $\le$ is $>$: $$\boxed{\forall x\quad 2x^2>x-1}.$$ (Over the reals the negation is in fact true: $2x^2-x+1$ has discriminant $1-8=-7<0$ and positive leading coefficient, so $2x^2-x+1>0$ for every $x$; the original existential statement is therefore false.)
2. b) Truth table for $(\neg r\lor q)\to(p\lor r)$.

   Truth table

   $p$	$q$	$r$	$\neg r$	$\neg r\lor q$	$p\lor r$	$(\neg r\lor q)\to(p\lor r)$
   T	T	T	F	T	T	T
   T	T	F	T	T	T	T
   T	F	T	F	F	T	T
   T	F	F	T	T	T	T
   F	T	T	F	T	T	T
   F	T	F	T	T	F	F
   F	F	T	F	F	T	T
   F	F	F	T	T	F	F

   The consequent $p\lor r$ is false only when $p=r=F$, and then $\neg r=T$ makes the antecedent true, so $\boxed{\text{F exactly when } p=F,\ r=F\ (\text{either } q);\ \text{T in the other 6 rows}}$.
3. c) Compare $\forall x(P(x)\to Q(x))$ with $\forall x\,P(x)\to\forall x\,Q(x)$. (The second is read, by the usual precedence, as $(\forall x\,P(x))\to(\forall x\,Q(x))$.) They do NOT always agree. Take the universe $\{1,2\}$, $P(x):$ "$x=1$", $Q(x):$ "$x=2$". Then $P(1)\to Q(1)$ is $T\to F=F$, so $\forall x(P(x)\to Q(x))$ is $\boxed{F}$. On the right, $\forall x\,P(x)$ is false (fails at $x=2$), so the implication $(\forall x\,P(x))\to(\forall x\,Q(x))$ is vacuously $\boxed{T}$. Only the one-way implication $\forall x(P(x)\to Q(x))\Rightarrow(\forall x\,P(x)\to\forall x\,Q(x))$ holds in general.