Problem 53 - Discrete Mathematics with Applications

Question

Let UD \(=\) set of real numbers and \(\mathrm{P}(x, y) : y^{2} < x .\) Determine the truth value of each proposition. $$(\forall y)(\exists x) \mathrm{P}(x, y)$$

Step-by-Step Solution

Answer

The proposition \((\forall y)(\exists x) \mathrm{P}(x, y)\) is true for the set of real numbers, as we can find a value of x that satisfies the inequality \(y^2 < x\) for each case of positive, negative, and zero values of y. One such choice of x is \(1 + y^2\), which satisfies the inequality for all real values of y.

Step 1: Understanding the proposition

The proposition \((\forall y)(\exists x) \mathrm{P}(x, y)\) reads as "For all real numbers y, there exists a real number x such that \(y^2 < x\)." We want to see if we can find a real number x that satisfies the inequality \(y^2 < x\) for every real number y.

Step 2: Check for positive values of y

Let's begin by considering positive values of y, i.e., \(y > 0\). To satisfy the inequality, we need to find an x such that \(y^2 < x\). For example, choose x as \(1 + y^2\). It's evident that for any positive value of y, we have \(y^2 < x\), because \(x = 1 + y^2 > y^2\).

Step 3: Check for negative values of y

Now let's consider negative values of y, i.e., \(y < 0\). Since \(y^2 > 0\), we need to find an x such that \(y^2 < x\). For instance, choose x as \(1 + y^2\). It is clear that for any negative value of y, we have \(y^2 < x\), because \(1 + y^2 > y^2\).

Step 4: Check for y = 0

Lastly, let's consider y = 0. The inequality becomes \(0^2 < x\), which simplifies to \(0 < x\). We can choose any positive x, for example, x = 1, and the inequality will be satisfied.

Step 5: Determine the truth value

Because we've found a value of x that satisfies the inequality for each case of positive, negative, and zero values of y, we can conclude that the proposition \((\forall y)(\exists x) \mathrm{P}(x, y)\) is true for the set of real numbers.

Key Concepts

Propositional Logic

Propositional logic forms the foundation of reasoning and argument with statements that hold a truth value—either true or false. In this exercise, we are tasked to evaluate the truth of a complex proposition:

• It states, "For every real number \(y\), there exists a real number \(x\) such that \(y^2 < x\)."
• This means we're using logical symbols to express and check if there is such an \(x\) for any given \(y\) in the real numbers.

Breaking down the notation, \(\forall y\) is a universal quantifier meaning "for all \(y\)," and follows with \(\exists x\), an existential quantifier meaning "there exists an \(x\)." Together, they frame our search for a condition that holds under all instances of \(y\).
To resolve this proposition, the problem is boiled down logically to finding any real number \(x\) for every \(y\), which makes \(y^2 < x\). If successful in every instance, the proposition is said to be true.

Inequality

An inequality expresses a relation where one value is not equal and appears smaller or larger than another. Here, the inequality \(y^2 < x\) represents our desired condition that must be satisfied by values \(x\) and \(y\).

• This inequality states that \(x\) must be greater than the square of \(y\), noted as \(y^2\).
• For each real number \(y\), it's necessary to find an \(x\) surpassing \(y^2\) to deem our original proposition truthful.

The solution walks us through demonstrating this with positive, negative, and zero values of \(y\):

• For positive \(y\), choose \(x = 1 + y^2\); this ensures \(x > y^2\).
• For negative \(y\), \(x = 1 + y^2\) still suffices as \(x > y^2\).
• For \(y = 0\), any \(x\) where \(x > 0\) works, commonly choosing \(x = 1\).

These choices of \(x\) validate the inequality for any \(y\), thus proving the inequality's satisfaction across the real numbers.

Universal Quantifier

Universal quantifiers express that a condition holds for all elements within a designated set. The symbol \(\forall\) effectively frames propositions by stating that a subsequent condition applies universally to its variable—in this case, \(y\).

• It suggests that for every real number \(y\), a condition needs to be met by some number \(x\).
• It is crucial to find such \(x\) that always fulfills the condition set by \(P(x, y)\) being true.

To affirm the truth of \((\forall y)(\exists x) \mathrm{P}(x, y)\), we examined every possibility within the universe of real numbers:

• Positive \(y\), where choosing an \(x\) larger than \(y^2\) works.
• Negative \(y\) applies the same logic, finding an \(x\) exceeding \(y^2\).
• \(y = 0\) has \(x = 1\) or any positive \(x\) to ensure \(0 < x\).

In all cases, we confirmed that such \(x\) could indeed be identified for each \(y\), making the universal statement true. This exercise emphasizes the importance of covering every instance within \(y\)'s realm to substantiate the inherent condition universally for a valid proposition.