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Chapter 2: Problem 6

Let \(p(x, y), q(x, y)\) denote the following open statements: $$ p(x, y): x^{2} \geq y \quad q(x, y): x+2

Short Answer

Expert verified
a) True b) False c) False d) True e) False f) False

Step by step solution

01

Solving for \(p(2,4)\)

Substitute \(x = 2\) and \(y = 4\) into the equation \(p(x,y): x^2 >= y\). Evaluating, we get \(2^2 >= 4\) or \(4 >= 4\). Since this statement is true, \(p(2,4)\) is true.
02

Solving for \(q(1, \pi)\)

Substitute \(x = 1\) and \(y = \pi\) into the equation \(q(x,y): x+2 < y\). Evaluating, we get \(1+2 < \pi\) or \(3 < \pi\). Since this statement is false (because \(\pi\) is approximately 3.14159, which is more than 3), \(q(1, \pi)\) is false.
03

Solving for \(p(-3,8) \wedge q(1,3)\)

First, find the truth value for \(p(-3,8)\). Substitute \(x = -3\) and \(y = 8\) into \(p(x,y): x^2 >= y\). Evaluating, we get \((-3)^2 >= 8\) or \(9 >= 8\), which is true. Next, find the truth value for \(q(1,3)\). Substitute \(x = 1\) and \(y = 3\) into \(q(x,y): x+2< y\). Evaluating, we get \(1+2 < 3\) or \(3 < 3\), which is false. The logical and (\(\wedge\)) of true and false is false, so \(p(-3,8) \wedge q(1,3)\) is false.
04

Solving for \(p\left(\frac{1}{2}, \frac{1}{3}\right) \vee \neg q(-2,-3)\)

First, get the truth value for \(p\left(\frac{1}{2}, \frac{1}{3}\right)\). Substitute \(x = \frac{1}{2}\) and \(y = \frac{1}{3}\) into \(p(x,y): x^2 >= y\). We get \(\left(\frac{1}{2}\right)^2 >= \frac{1}{3}\) or \(0.25 >= 0.33\), which is false. Next, find out the truth value for \( \neg q(-2,-3)\). Substitute \(x = -2\) and \(y = -3\) into \(q(x,y): x+2 < y\). We get \(-2+2 < -3\) or \(0 < -3\), which is false. The negation of false is true. The logical or (\(\vee\)) of false and true is true, so \(p\left(\frac{1}{2}, \frac{1}{3}\right) \vee \neg q(-2,-3)\) is true.
05

Solving for \(p(2,2) \rightarrow q(1,1)\)

Let's first find the truth value for \(p(2,2)\). Substitute \(x = 2\) and \(y = 2\) into \(p(x,y): x^2 >= y\). We get \(2^2 >= 2\) or \(4 >= 2\), which is true. Next, get the truth value for \(q(1,1)\). Substitute \(x = 1\) and \(y = 1\) into \(q(x,y): x+2 < y\). We get \(1+2 < 1\) or \(3 < 1\), which is false. With the implication operator (\(\rightarrow\)), true implies false is false, so \(p(2,2) \rightarrow q(1,1)\) is false.
06

Solving for \(p(1,2) \leftrightarrow \neg q(1,2)\)

First, get the truth value for \(p(1,2)\). Substitute \(x = 1\) and \(y = 2\) into \(p(x,y): x^2 >= y\). We get \(1^2 >= 2\) or \(1 >= 2\), which is false. Next, find out the truth value for \( \neg q(1,2)\). Substitute \(x = 1\) and \(y = 2\) into \(q(x,y): x+2 < y\). We get \(1+2 < 2\) or \(3 < 2\), which is false. The negation of false is true. The equivalence operator (\(\leftrightarrow\)) says that false is the same as true which is false, so \(p(1,2) \leftrightarrow \neg q(1,2)\) is false.

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Most popular questions from this chapter

\text { In the arithmetic of real numbers, there is a real number, namely } 0 \text {, called the identity of }addition because \(a+0=0+a=a\) for every real number \(a\). This may be expressed in symbolic form by $$ \exists z \forall a[a+z=z+a=a] \text {. } $$ (We agree that the universe comprises all real numbers.) a) In conjunction with the existence of an additive identity is the existence of additive inverses. Write a quantified statement that expresses "Every real number has an additive inverse." (Do not use the minus sign anywhere in your statement.) b) Write a quantified statement dealing with the existence of a multiplicative identity for the arithmetic of real numbers. c) Write a quantified statement covering the existence of multiplicative inverses for the nonzero real numbers. (Do not use the exponent \(-1\) anywhere in your statement.) d) Do the results in parts (b) and (c) change in any way when the universe is restricted to the integers?

Determine whether each of the following is true or false. Here \(p, q\), and \(r\) are arbitrary statements. a) An equivalent way to express the converse of " \(p\) is sufficient for \(q^{* \prime}\) is " \(p\) is necessary for \(q\), b) An equivalent way to express the inverse of " \(p\) is necessary for \(q\) " is " \(7 q\) is sufficient for \(\neg p . "\) c) An equivalent way to express the contrapositive of " \(p\) is necessary for \(q^{\prime \prime}\) is " \(\neg q\) is necessary for \(\neg p .^{31}\) d) An equivalent way to express the converse of \(p \rightarrow(q \rightarrow r)\) is \((\neg q \vee r) \rightarrow p\).

Let the universe for the variables in the following statements consist of all real numbers. In each case negate and simplify the given statement. a) \(\forall x \forall y[(x>y) \rightarrow(x-y>0)]\) b) \(\forall x \forall y\left[\left[(x>0) \wedge\left(y=\log _{10} x\right)\right] \rightarrow\left(x=10^{y}\right)\right]\) c) \(\forall x \forall y[(x0) \wedge(y>0))] \rightarrow[\exists z(x z>y)]\)

Suppose that \(p(x, y)\) is an open statement where the universe for each of \(x, y\) consists of only three integers: \(2,3,5\). Then the quantified statement \(\exists y p(2, y)\) is logically equivalent to \(p(2,2) \vee p(2,3) \vee p(2,5)\). The quantified statement \(\exists x \forall y p(x, y)\) is logically equivalent to \([p(2,2) \wedge p(2,3) \wedge p(2,5)] \vee[p(3,2) \wedge p(3,3) \wedge p(3,5)] \vee\) \([p(5,2) \wedge p(5,3) \wedge p(5,5)]\). Use conjunctions and/or disjunctions to express the following statements without quantifiers. a) \(\exists \mathrm{x} p(x, 5)\) b) \(\forall x p(x, 3)\) c) \(\forall y p(2, y)\) d) \(\exists x \exists y p(x, y)\) e) \(\forall x \forall y p(x, y)\) f) \(\forall y \exists x p(x, y)\)

For a prescribed universe and any open statements \(p(x), q(x)\) in the variable \(x\), prove that a) \(\exists x[p(x) \vee q(x)] \Leftrightarrow \exists x p(x) \vee \exists x q(x)\) b) \(\forall x[p(x) \wedge q(x)] \Leftrightarrow \forall x p(x) \wedge \forall x q(x)\)
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