Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Translate each of the following quantified formulas into an English sentence where the universal set is \(\mathbb{R}\). Label each as true or false. (a) \(\forall x(\exists y(x y=x))\) (b) \(\forall y(\exists x(x y=x))\) (c) \(\forall x(\exists y(x y=1))\) (d) \(\exists y(\forall x \neq 0(x y=1))\) (e) \(\exists x(\forall y(x y=x))\) (f) \((\forall x(x \neq 0 \rightarrow \exists y(x y=1))\)

Short Answer

Expert verified
(a) True, (b) False, (c) False, (d) False, (e) True, (f) True.

Step by step solution

01

Analyze the Formula (a)

The formula is \( \forall x(\exists y(x y = x)) \). This statement means "For every real number \( x \), there exists a real number \( y \) such that their product is equal to \( x \)." In simpler terms, this asks if every real number has a multiplicative identity. The answer is True because for any real number \( x \), \( y = 1 \) satisfies the equation \( x \cdot y = x \).
02

Analyze the Formula (b)

The formula is \( \forall y(\exists x(x y = x)) \). This statement means "For every real number \( y \), there exists a real number \( x \) such that their product is equal to \( x \)." This is False because when \( y eq 1 \), we cannot find such an \( x \) for all \( y \), since \( xy = x \) implies \( (1 - y)x = 0 \), which cannot always satisfy \( x eq 0 \) unless \( y = 1 \).
03

Analyze the Formula (c)

The formula is \( \forall x(\exists y(x y = 1)) \). This statement translates to "For every real number \( x \), there exists a real number \( y \) such that their product equals 1." The statement is False because for \( x = 0 \), \( y \) cannot be defined to satisfy \( x \cdot y = 1 \).
04

Analyze the Formula (d)

The formula is \( \exists y(\forall x eq 0(x y=1)) \). This translates to "There exists a real number \( y \) such that for all non-zero real numbers \( x \), their product equals 1." This is False, since there is no single \( y \) that, when multiplied with any non-zero \( x \), always equals 1. The only way \( x \cdot y = 1 \) for every non-zero \( x \) is if \( y \) itself varies as \( 1/x \), which violates the condition of a single \( y \).
05

Analyze the Formula (e)

The formula is \( \exists x(\forall y(x y=x)) \). This means "There exists a real number \( x \) such that for all real numbers \( y \), their product is \( x \)." This is True, as \( x = 0 \) satisfies the condition, since \( 0 \cdot y = 0 \) for any real \( y \).
06

Analyze the Formula (f)

The formula is \( (\forall x(x eq 0 \rightarrow \exists y(x y=1)) \). This translates to "For every non-zero real number \( x \), there exists a real number \( y \) such that their product equals 1." The statement is True, since for any non-zero \( x \), \( y = 1/x \) fulfills the requirement \( x \cdot y = 1 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Universal Set
In mathematics, the term "universal set" refers to a comprehensive set containing all the elements under consideration, which sets the context for the discussion. It acts as a boundary for what elements are possible in a given scenario.
When dealing with quantified formulas, the universal set specifies the domain of discourse for the variables involved. Essentially, it tells us which elements we are talking about.
In the provided exercise, the universal set is specified as the set of real numbers, denoted by \( \mathbb{R} \). This means that any quantified statements we are analyzing are referring to real numbers. Understanding this is crucial as it lays the groundwork for interpreting and assessing the truth of any mathematical statement or formula given. Without realizing which universal set we're operating under, interpretations could be incorrect since the variable constraints wouldn't be rightly applied.
Real Numbers
Real numbers are the backbone of many mathematical discussions and problems. They include all the numbers you can think of that are not imaginary, such as integers, fractions, and irrational numbers. This broad classification comes together under the notation \( \mathbb{R} \).
Real numbers are essential because they encompass natural numbers (like 1, 2, 3), whole numbers (including zero), and negative numbers. They also include fractions like 1/2, decimals like 3.14, and irrational numbers like \( \sqrt{2} \).
In the context of quantified formulas, being aware that the variables are confined to real numbers helps us understand the scope of any problem. For instance, in part (a) of the exercise, knowing that \( x \) and \( y \) are real numbers is key to correctly interpreting the statement \( \forall x(\exists y(xy=x)) \). This tells us there is a number \( y = 1 \) for each real number \( x \) that makes the equality hold true.
Quantifiers
Quantifiers are symbols used in logic and mathematics to specify the quantity of specimens in a statement. There are two main types: the universal quantifier, represented as \( \forall \), and the existential quantifier, represented as \( \exists \).
- **Universal Quantifier** (\( \forall \)): This conveys the idea of "for all" or "every." An expression like \( \forall x \) suggests that the following condition applies to every element within the universal set.
- **Existential Quantifier** (\( \exists \)): This signifies "there exists" or "there is at least one." An expression like \( \exists y \) implies that there is at least one instance of an element in the universal set that satisfies the given condition.
Understanding quantifiers is critical when evaluating mathematical formulas, as they shape the conditions under which such statements hold true or false. For example, in formula (f), \( \forall x(x eq 0 \rightarrow \exists y(xy=1)) \), it means that for every non-zero \( x \), there is some \( y \) such that their product is 1.
True or False Statements
Determining if a mathematical statement is true or false is like verifying its logical validity within the given universal set. It involves checking if the provided conditions and quantifiers hold consistently for all or some of the elements.
A statement's truth can depend on several factors:
  • The universal set under consideration – as this dictates the domain.
  • The specific quantifiers used – which set the scope of applicability.
  • The relationships and operations stipulated – such as equalities or inequalities.
When translating formulas into English sentences, as shown in the original exercise, one needs to assess these factors to declare if the statement is true or false. For instance, part (d) of the exercise, \( \exists y(\forall x eq 0(xy=1)) \), was concluded as false because there cannot be a single real number \( y \) that satisfies the equation for all non-zero real numbers \( x \). This systematic verification is part of the logical foundation that supports critical mathematical analysis.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the expression tree for the formula $$ ((\neg(p \wedge q)) \vee(\neg(q \wedge r))) \wedge((\neg(p \leftrightarrow(\neg(\neg s)))) \vee((r \wedge s) \vee(\neg q))) . $$ Evaluate the expression tree if proposition \(p\) is \(F\), proposition \(q\) is \(T\), proposition \(r\) is \(F\), and proposition \(s\) is \(T\).

Show that the following formulas from Table 2.5 are tautologies: (a) \((p \wedge p) \leftrightarrow p\) (b) \((p \wedge(p \rightarrow q)) \rightarrow q\) (c) \((p \rightarrow r) \leftrightarrow(\neg r \rightarrow \neg p)\)

Let \(\phi=(p \vee q) \rightarrow(r \wedge \neg s)\). For each of the following interpretations of \(p, q, r,\) and \(s,\) compute \(I(\phi)\) using the truth tables for \(\neg, \vee, \wedge, \rightarrow,\) and \(\leftrightarrow\) (a) \(I(p)=T, I(q)=T, I(r)=T,\) and \(I(s)=F\) (b) \(I(p)=T, I(q)=T, I(r)=F,\) and \(I(s)=F\) (c) \(I(p)=F, I(q)=T, I(r)=T,\) and \(I(s)=T\) (d) \(I(p)=F_{,} I(q)=F_{,} I(r)=T,\) and \(I(s)=T\)

Let \(U\) be the set of all problems on a comprehensive list of problems in science. Define four predicates over \(U\) by: \(P(x): x\) is a mathematics problem \(Q(x): x\) is difficult (according to some well-defined criterion: it does not matter for us What the criterion is) \(R(x): x\) is easy (according to some well-defined criterion) \(S(x): x\) is unsolvable (if you do not know what "unsolvable" means, do not worry about it here) Translate into English sentences each of the following formulas: (a) \(\forall x P(x)\) (b) \(\exists x Q(x)\) (c) \(\forall x(Q(x) \vee R(x))\) (d) \(\forall x(S(x) \rightarrow P(x))\) (e) \(\exists x(S(x) \wedge \neg P(x))\) (f) \(\neg(\forall x(\neg R(x) \vee S(x)))\) (g) \(\forall x(P(x) \rightarrow(Q(x) \leftrightarrow \neg R(x)))\) (h) \(\forall x \neg S(x)\) (i) \(\forall x(P(x) \rightarrow \neg S(x))\) (j) \(\forall x(P(x) \rightarrow(R(x) \vee S(x)))\) (k) \(\exists x(\neg Q(x) \wedge \neg R(x))\) (l) \(\exists x(R(x) \wedge S(x))\) (m) \( \forall x(Q(x) \leftrightarrow \neg R(x))\)

(a) Show that the following formula in CNF is unsatisfiable: $$ (p \vee q) \wedge(p \vee \neg q) \wedge(\neg p \vee q) \wedge(\neg p \vee \neg q) $$ (b) Show that the following formula in CNF is unsatisfiable: $$ \begin{array}{c} (p \vee q \vee r) \wedge(p \vee \neg q \vee r) \wedge(\neg p \vee q \vee r) \wedge(\neg p \vee \neg q \vee r) \\ \wedge(p \vee q \vee \neg r) \wedge(p \vee \neg q \vee \neg r) \wedge(\neg p \vee q \vee \neg r) \wedge(\neg p \vee \neg q \vee \neg r) \end{array} $$ Can you find an easier argument than just writing the entire truth table? (c) Generalize the above to some class of CNF formulas on an arbitrary number \(n \geq 1\) of proposition letters, and prove it by induction on \(n\).

See all solutions

Recommended explanations on Computer Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free