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For the following predicates with universal set \(\mathbb{R}\), state the meaning of the predicate in a sentence. If it is false, give an example to show why. (Example: \(\forall x(\exists y(xz \wedge z>y))))\) (h) \(\forall x(\forall y(\forall z)(x>y \wedge y>z) \rightarrow x>z))\) )

Short Answer

Expert verified
(b) and (c) are false. Example for (b): y=1 and x=2 implies 2*1 ≠ 1. Example for (c): No x is greater than all real numbers.

Step by step solution

01

Understand part (a)

The predicate \( \forall x (\exists y (x eq 0 \rightarrow xy=a)) \) states: "For every real number \( x \), there exists a real number \( y \) such that if \( x eq 0 \), then the product of \( x \) and \( y \) equals \( a \)." This is true, as for any non-zero \( x \), \( y = \frac{a}{x} \) satisfies the equation \( xy = a \).
02

Analyze part (b)

The predicate \( \exists y (\forall x (x eq 0 \rightarrow xy=1)) \) states: "There exists a real number \( y \) such that for every real number \( x eq 0 \), the product of \( x \) and \( y \) equals 1." This is true when \( y = \frac{1}{x} \), meaning \( y \) can adjust based on \( x \), example being \( y = \frac{1}{x} \) for any non-zero \( x \). However, \( y \) is expected to be consistent for all \( x \), making it false; such a constant \( y \) cannot exist.
03

Evaluate part (c)

The predicate \( \exists x (\forall y (y \leq x)) \) means "There exists a real number \( x \) such that all real numbers \( y \) are less than or equal to \( x \)." This is false, because there is no greatest real number; any real number can have a larger number than it.
04

Examine part (d)

The predicate \( \forall x (\exists y (x + y = x)) \) translates to "For every real number \( x \), there exists a real number \( y \) such that \( x + y = x \)." This is true, and the number \( y \) that satisfies this is \( y = 0 \), making the equation hold for any real \( x \).
05

Analyze part (e)

The predicate \( \exists y (\forall x (x + y = x)) \) means "There exists a real number \( y \) such that for every real number \( x \), \( x + y = x \)." This is true when \( y = 0 \), as adding 0 to any \( x \) results in \( x \).
06

Investigate part (f)

The predicate \( \forall x (\forall y (\exists z (x < z \wedge z < y))) \) states, "For every pair of real numbers \( x \) and \( y \), there exists a real number \( z \) such that \( x < z < y \)." This is true; there is always a number between any two distinct real numbers \( x \) and \( y \).
07

Explore part (g)

The predicate \( \forall x (\forall y (x eq y \rightarrow \exists z ((xz \wedge z>y)))) \) means "For every pair of distinct real numbers \( x \) and \( y \), there exists a real number \( z \) such that \( z \) is strictly between \( x \) and \( y \)." This is true; there is always a value \( z \) that fits between two different numbers.
08

Examine part (h)

The predicate \( \forall x (\forall y (\forall z ( (x > y \wedge y > z) \rightarrow x > z ))) \) translates to "For any numbers \( x \), \( y \), and \( z \), if \( x > y \) and \( y > z \), then \( x > z \)." This statement is true; it exemplifies the transitive property of inequalities.

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Key Concepts

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

Universal Quantifiers
Universal quantifiers in predicate logic are expressed with the symbol \( \forall \). This symbol is used to indicate that the statement it precedes is true for all elements within a particular set or domain. For example, if we use the universal quantifier in the expression \( \forall x P(x) \), it means "for every element \( x \) in the set, the predicate \( P(x) \) is true."

These quantifiers are critical when formulating concepts that need to be applicable to every item in a group, like saying all real numbers satisfy a particular property. In simpler terms, when you see \( \forall \), think "no exceptions."

When dealing with real numbers, universal quantifiers help in expressing statements such as "for every real number \( x \), there is a real number \( y \) such that..." It's essential to check if the statement holds true universally, and even a single counterexample can invalidate the claim.
Existential Quantifiers
Existential quantifiers are represented by the symbol \( \exists \), which means "there exists." This quantifier is used to indicate that there is at least one element within a domain for which the given predicate or property holds. For instance, the statement \( \exists x P(x) \) means "there exists at least one \( x \) such that \( P(x) \) is true."

This contrasts with universal quantifiers as it looks for one or more specifics rather than applying to all. With real numbers, it might describe scenarios where "some number satisfies a condition," rather than all. For example, the predicate \( \exists y \forall x(x eq 0 \rightarrow xy=1) \) tries to find one value of \( y \) that satisfies all values of \( x \).

It is vital to carefully distinguish between these symbols to understand whether you are proving something about every instance (universal) or demonstrating the existence of at least one instance (existential).
Real Numbers
Real numbers are a fundamental part of mathematics encompassing all the numbers that can be found on the number line. This includes integers, rational numbers (like fractions), and irrational numbers (such as \( \sqrt{2} \)). Real numbers can be either positive, negative, or zero.

They form a continuous set of numbers without any gaps, unlike rational numbers, which can have fractional gaps. This continuity makes them crucial for describing quantities in the real world and helps in bridging gaps in logical statements in mathematics.

In predicate logic exercises, understanding real numbers is crucial because many of the expressions and predicates are evaluated over this set. For example, when using predicates that ask whether there exists some real number satisfying a condition, knowing that real numbers include such a wide range is key to analyzing the truth of the predicate.
Transitive Property
The transitive property is a fundamental concept in mathematics, describing a relation that holds between three elements. It states that if a relation holds between the first element and the second, and between the second and the third, then it must hold between the first and the third as well.

In logical terms, the transitive property is formalized as: if \( a > b \) and \( b > c \), then \( a > c \). This property is crucial when working with orderings and comparisons, as it allows the extension of relations over sequences of elements.

In the context of predicates involving comparisons, recognizing the transitive property can help identify when certain logical statements are inherently true. For instance, the transitive property explains why the predicate \( \forall x (\forall y (\forall z ((x > y \wedge y > z) \rightarrow x > z))) \) is always true, as it is a direct application of this property in inequalities.

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Most popular questions from this chapter

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 \(I(s)=T\)

Simplify the following boolean expressions: (a) \((x \wedge y) \vee(x \wedge \neg y) \vee(\neg x \wedge y) \vee(\neg x \wedge \neg y)\) (b) \((x \wedge y \wedge z) \vee(x \wedge \neg y \wedge z) \vee(\neg x \wedge y \wedge \neg z) \vee(\neg x \wedge \neg y \wedge z)\) (c) \((x \wedge y \wedge \neg z) \vee(x \wedge \neg y \wedge z) \vee(x \wedge \neg y \wedge \neg z)\)

Find formulas in CNF equivalent to each of the following formulas: (a) \(\neg(p \wedge T)\) (b) \(((p \rightarrow q) \rightarrow r) \rightarrow F\) (c) \(((p \rightarrow q) \rightarrow r) \rightarrow T\) (d) \((p \leftrightarrow q) \leftrightarrow r\) (e) \(\neg(p \leftrightarrow q) \leftrightarrow r\) (f) \(((p \vee q) \rightarrow r) \wedge(r \rightarrow \neg(p \vee q))\) (g) \((\neg r) \rightarrow(((p \vee q) \rightarrow r) \rightarrow \neg q)\)

Find a CNF for each of the following formulas, and prove that each formula is a tautology. (a) \((p \wedge p) \leftrightarrow p\) (b) \((p \wedge(p \rightarrow q)) \rightarrow q\) (c) \((p \rightarrow(r \rightarrow q)) \leftrightarrow((p \wedge r) \rightarrow q)\) (d) \((p \rightarrow r) \leftrightarrow(\neg r \rightarrow \neg p)\)

(a) Find the resolvant of \((p \vee q)\) and \((\neg p \vee r)\) on \(p\). (b) Find the resolvant of \((p \vee q \vee r \vee s)\) and \((\neg p \vee \neg q \vee t)\) on \(p\). (c) Find the resolvant of \((p \vee q)\) and \(\neg p\) on \(p\). (d) Find the resolvant of \((p)\) and \((\neg p)\) on \(p\). (e) Which resolvant above from parts (a) through (d) is a tautology? Which is tautologically false?

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