Step 1: Formal Definitions of the Number Sets

To evaluate the validity of the statement, we must first establish the rigorous mathematical definitions of the sets involved:

• Rational Numbers ($\mathbb{Q}$): A rational number is any number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers ($\mathbb{Z}$) and $q \neq 0$. Examples include $\frac{1}{2}$, $-\frac{3}{4}$, $5$ (since $5 = \frac{5}{1}$), and $0$.
• Whole Numbers ($\mathbb{W}$): The set of whole numbers consists of all non-negative integers. It is defined as $\mathbb{W} = \{0, 1, 2, 3, \dots\}$.

Step 2: Logical Analysis of the Proposition

The given statement asserts that "Every rational number is a whole number." In set theory notation, this proposition claims that the set of rational numbers is a subset of the set of whole numbers ($\mathbb{Q} \subseteq \mathbb{W}$).

[Per the axioms of set theory], for $\mathbb{Q} \subseteq \mathbb{W}$ to be true, every single element that exists in $\mathbb{Q}$ must also exist in $\mathbb{W}$. If we can find even one element in $\mathbb{Q}$ that is not in $\mathbb{W}$, the statement is proven false by counterexample.

Step 3: Disproving via Counterexamples

Let us test specific rational numbers against the definition of whole numbers to find a counterexample:

• Fractional Values: Consider the number $\frac{3}{5}$. It is a rational number because it is in the form $\frac{p}{q}$ where $p=3$, $q=5$, and $q \neq 0$. However, $\frac{3}{5} = 0.6$, which is a decimal value and not a non-negative integer. Therefore, $\frac{3}{5} \notin \mathbb{W}$.
• Negative Integers: Consider the number $-4$. It is a rational number because it can be written as $\frac{-4}{1}$. However, whole numbers strictly include only non-negative integers. Therefore, $-4 \notin \mathbb{W}$.

Step 4: Visualizing the Set Hierarchy

The relationship between these number systems is strictly hierarchical in the opposite direction: $\mathbb{W} \subset \mathbb{Q}$. Every whole number is a rational number, but the converse is not true. The Euler diagram below illustrates this structural relationship, proving that $\mathbb{Q}$ encompasses a much larger domain than $\mathbb{W}$.

Step 5: Final Conclusion

Because the set of rational numbers ($\mathbb{Q}$) contains fractions and negative integers—neither of which belong to the set of whole numbers ($\mathbb{W}$)—the proposition fails. The correct logical statement is that every whole number is a rational number, but not vice versa.

Final Solution: False. Rational numbers include fractions (e.g., $\frac{1}{2}$) and negative integers (e.g., $-3$), which are not whole numbers. Therefore, it is incorrect to state that every rational number is a whole number.

Given Expression & Theoretical Foundation

We are tasked with classifying the number $\sqrt{23}$ as either a rational or an irrational number. To do this rigorously, we must first establish the mathematical definitions of these two sets of numbers:

• Rational Number ($\mathbb{Q}$): Any number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers ($p, q \in \mathbb{Z}$) and $q \neq 0$. The decimal expansion of a rational number is always either terminating or non-terminating but repeating.
• Irrational Number ($\mathbb{I}$): Any real number that cannot be expressed as a simple fraction $\frac{p}{q}$. The decimal expansion of an irrational number is strictly non-terminating and non-repeating.

Step 1: Analyzing the Radicand

The given expression is $\sqrt{23}$. The number inside the square root, $23$, is called the radicand. We must first determine if the radicand is a perfect square.

Let us evaluate the squares of integers in the vicinity of $23$:

• $4^2 = 16$
• $5^2 = 25$

Since $16 < 23 < 25$, it follows that $\sqrt{16} < \sqrt{23} < \sqrt{25}$, which simplifies to $4 < \sqrt{23} < 5$. Because $23$ lies strictly between two consecutive perfect squares, it is not a perfect square. Its square root will not be an integer.

Step 2: Prime Factorization & Theorem Application

Next, we analyze the factors of the radicand $23$. The only positive divisors of $23$ are $1$ and $23$ itself. Therefore, $23$ is a prime number.

[Per the properties of real numbers], there is a fundamental theorem regarding the square roots of prime numbers:

Theorem: If $p$ is a prime number, then $\sqrt{p}$ is an irrational number.

Since $23$ is a prime number, applying this theorem directly classifies $\sqrt{23}$ as an irrational number.

Step 3: Rigorous Proof by Contradiction (Analytical Depth)

To provide a masterclass-level justification, we will prove that $\sqrt{23}$ is irrational using a proof by contradiction.

Assumption: Assume the contrary, that $\sqrt{23}$ is a rational number. Therefore, it can be written in the simplest fractional form:

$\sqrt{23} = \frac{a}{b}$

where $a$ and $b$ are coprime integers (meaning their greatest common divisor is $1$, $\gcd(a,b) = 1$), and $b \neq 0$.

Algebraic Manipulation:

1. Square both sides of the equation:
   $23 = \frac{a^2}{b^2}$
2. Multiply both sides by $b^2$:
   $a^2 = 23b^2$     (Equation 1)

Logical Deduction:

Equation 1 states that $a^2$ is a multiple of $23$. [Per Euclid's Lemma and the Fundamental Theorem of Arithmetic], if a prime number divides the square of an integer, it must also divide the integer itself. Therefore, $23$ divides $a$.

Let $a = 23k$, where $k$ is some integer. Substitute this back into Equation 1:

$(23k)^2 = 23b^2$

$529k^2 = 23b^2$

Divide both sides by $23$:

$23k^2 = b^2$

This new equation shows that $b^2$ is also a multiple of $23$. By the same logic applied previously, $23$ must divide $b$.

The Contradiction:

We have deduced that $23$ divides both $a$ and $b$. This means $a$ and $b$ share a common factor of $23$. However, this directly contradicts our initial assumption that $a$ and $b$ are coprime ($\gcd(a,b) = 1$). Because our assumption led to a logical contradiction, the assumption must be false.

Therefore, $\sqrt{23}$ cannot be expressed as a rational fraction.

Step 4: Visualizing on the Real Number Line

To further contextualize this irrational number, we can map its approximate decimal expansion ($\sqrt{23} \approx 4.795831523...$) onto a real number line. The non-terminating, non-repeating nature of this decimal confirms its irrationality.

Final Solution: The number $\sqrt{23}$ is an irrational number.

Initial Setup & Given Expression

We are tasked with evaluating the following exponential expression:

$32^{\frac{1}{5}}$

[Per the definition of rational exponents, an expression of the form $a^{\frac{1}{n}}$ represents the $n$-th principal root of the base $a$, denoted algebraically as $\sqrt[n]{a}$. Therefore, evaluating $32^{\frac{1}{5}}$ is mathematically equivalent to finding the 5th root of $32$.]

Step 1: Prime Factorization of the Base

To systematically simplify expressions involving fractional exponents, we first decompose the base ($32$) into its prime factors. This allows us to express the base as an exponential term, which can then be manipulated using the laws of exponents.

Dividing $32$ successively by the smallest prime number ($2$):

• $32 \div 2 = 16$
• $16 \div 2 = 8$
• $8 \div 2 = 4$
• $4 \div 2 = 2$
• $2 \div 2 = 1$

Thus, the prime factorization is $32 = 2 \times 2 \times 2 \times 2 \times 2 = 2^5$.

Step 2: Substitution into the Original Expression

We substitute the prime factored form of the base back into the original expression:

$32^{\frac{1}{5}} = (2^5)^{\frac{1}{5}}$

Step 3: Application of the Laws of Exponents

[Per the Power of a Power Property of Exponents], which states that $(a^m)^n = a^{m \times n}$, we multiply the inner exponent ($5$) by the outer fractional exponent ($\frac{1}{5}$).

$(2^5)^{\frac{1}{5}} = 2^{5 \times \frac{1}{5}}$

Step 4: Simplification and Final Calculation

We now perform the multiplication in the exponent:

$5 \times \frac{1}{5} = \frac{5}{5} = 1$

Substituting this simplified exponent back onto the base $2$ yields:

$2^1 = 2$

Final Solution: 2

Step 1: Analyzing the Components of the Expression

We are given the mathematical expression $2 - \sqrt{5}$. To classify this number, we must first analyze the individual terms that constitute the expression:

• The number $2$: This is an integer. By definition, any integer can be expressed in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$ (e.g., $\frac{2}{1}$). Therefore, $2$ is a rational number.
• The number $\sqrt{5}$: The number $5$ is a prime number and not a perfect square. The square root of any non-perfect square yields a non-terminating, non-repeating decimal. Therefore, $\sqrt{5}$ is an irrational number.

Step 2: Applying the Properties of Real Numbers

In the real number system, specific closure properties govern the arithmetic operations between rational and irrational numbers. The relevant theorem states:

[Theorem: The sum or difference of a non-zero rational number and an irrational number is always an irrational number.]

Since we are subtracting an irrational number ($\sqrt{5}$) from a rational number ($2$), the resulting difference must be irrational.

Step 3: Formal Proof by Contradiction

To rigorously prove that $2 - \sqrt{5}$ is irrational, we utilize a proof by contradiction.

Assumption: Let us assume the contrary, that $2 - \sqrt{5}$ is a rational number. Let this rational number be represented by $r$.

$2 - \sqrt{5} = r$

Rearranging the equation to isolate $\sqrt{5}$:

$2 - r = \sqrt{5}$

Logical Deduction:

• We know that $2$ is a rational number.
• By our assumption, $r$ is a rational number.
• [Per the Closure Property of Rational Numbers under Subtraction], the difference between two rational numbers ($2 - r$) must also be a rational number.
• This implies that $\sqrt{5}$ is equal to a rational number.

Contradiction: This contradicts the universally established mathematical fact that $\sqrt{5}$ is an irrational number. Therefore, our initial assumption that $2 - \sqrt{5}$ is rational must be false.

Step 4: Decimal Expansion Analysis (Verification)

We can also verify this classification by examining the decimal expansion of the expression:

The decimal expansion of $\sqrt{5}$ is non-terminating and non-repeating:

$\sqrt{5} \approx 2.236067977...$

Substituting this into our expression:

$2 - \sqrt{5} = 2 - 2.236067977...$

$2 - \sqrt{5} = -0.236067977...$

The resulting decimal is also non-terminating and non-repeating, which is the definitive characteristic of an irrational number.

Visual Representation: Number Line Mapping

The following geometric representation demonstrates the subtraction of the irrational length $\sqrt{5}$ from the rational coordinate $2$, landing on the irrational coordinate $2 - \sqrt{5}$.

Final Solution: The expression $2 - \sqrt{5}$ is an irrational number.