Initial Setup & Objective

We are given the following fractional expression containing irrational terms in the denominator:

$ \frac{1}{\sqrt{5} + \sqrt{2}} $

The mathematical objective is to rationalise the denominator. Rationalisation is the process of eliminating irrational numbers (such as surds or square roots) from the denominator of an algebraic fraction. This is achieved by multiplying the numerator and the denominator by a carefully chosen multiplier known as the conjugate.

Step 1: Identifying the Conjugate (Rationalising Factor)

To eliminate the square roots in a binomial denominator of the form $\sqrt{a} + \sqrt{b}$, we utilize its conjugate pair, which is $\sqrt{a} - \sqrt{b}$. [Per the fundamental properties of surds, multiplying a binomial by its conjugate leverages the difference of squares identity to yield a rational integer].

For our specific denominator, $\sqrt{5} + \sqrt{2}$, the conjugate is:

$ \sqrt{5} - \sqrt{2} $

Step 2: Multiplying Numerator and Denominator

To maintain the equivalence of the fraction [Per the Multiplicative Identity Property, multiplying by $1$ does not change the value], we multiply both the numerator and the denominator by the conjugate $\sqrt{5} - \sqrt{2}$:

$ \frac{1}{\sqrt{5} + \sqrt{2}} \times \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}} $

Step 3: Applying the Difference of Squares Identity

We now expand the numerator and the denominator. The numerator is simply multiplied by $1$:

$ \text{Numerator} = 1 \times (\sqrt{5} - \sqrt{2}) = \sqrt{5} - \sqrt{2} $

For the denominator, we apply the algebraic identity for the difference of squares: $(x + y)(x - y) = x^2 - y^2$.

Let $x = \sqrt{5}$ and $y = \sqrt{2}$:

$ \text{Denominator} = (\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2}) = (\sqrt{5})^2 - (\sqrt{2})^2 $

Step 4: Evaluating the Squares and Final Simplification

Evaluating the squares in the denominator [Per the definition of a square root, $(\sqrt{n})^2 = n$ for any non-negative real number $n$]:

• $(\sqrt{5})^2 = 5$
• $(\sqrt{2})^2 = 2$

Substitute these values back into the denominator's expression:

$ \text{Denominator} = 5 - 2 = 3 $

Now, recombine the evaluated numerator and the rationalised denominator to form the final simplified fraction:

$ \frac{\sqrt{5} - \sqrt{2}}{3} $

The denominator is now $3$, which is a rational integer. The rationalisation process is complete.

Final Solution: $ \frac{\sqrt{5} - \sqrt{2}}{3} $

Initial Setup & Given Expression

We are tasked with classifying the following mathematical expression as either a rational or an irrational number:

$\frac{2\sqrt{7}}{7\sqrt{7}}$

Step 1: Algebraic Simplification

Before classifying any real number, it must be reduced to its simplest form. We begin by analyzing the factors in both the numerator and the denominator.

• Numerator: $2\sqrt{7}$ (which is the product of the integer $2$ and the irrational number $\sqrt{7}$)
• Denominator: $7\sqrt{7}$ (which is the product of the integer $7$ and the irrational number $\sqrt{7}$)

[Per the fundamental properties of fractions and real numbers, any non-zero common factor present in both the numerator and the denominator can be divided out without changing the value of the expression]. Since $\sqrt{7} \approx 2.645$ and strictly $\sqrt{7} \neq 0$, we can safely cancel this common factor:

$\frac{2\sqrt{7}}{7\sqrt{7}} = \frac{2 \cdot \sqrt{7}}{7 \cdot \sqrt{7}} = \frac{2}{7} \cdot \frac{\sqrt{7}}{\sqrt{7}}$

$\frac{2}{7} \cdot 1 = \frac{2}{7}$

Step 2: Theoretical Classification

We must now evaluate the simplified expression, $\frac{2}{7}$, against the formal axioms of the real number system.

[By definition, a rational number ($\mathbb{Q}$) is any number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers ($\mathbb{Z}$), and the denominator $q$ is not equal to zero ($q \neq 0$)].

Analyzing our simplified fraction $\frac{2}{7}$:

• $p = 2$, which is an integer ($2 \in \mathbb{Z}$).
• $q = 7$, which is an integer ($7 \in \mathbb{Z}$) and strictly non-zero ($7 \neq 0$).

Because the simplified form perfectly satisfies the necessary and sufficient conditions of a rational number, the original expression is inherently rational.

Visual Representation of the Number System Classification

Final Solution: The given expression $\frac{2\sqrt{7}}{7\sqrt{7}}$ simplifies algebraically to $\frac{2}{7}$. Because it can be expressed as the ratio of two integers where the denominator is not zero, it is a rational number.

Initial Setup & Theoretical Foundation

We are tasked with classifying the real number $\frac{1}{\sqrt{2}}$ as either rational or irrational. To do this rigorously, we must rely on the fundamental definitions and theorems governing the real number system.

• Rational Number: A number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.
• Irrational Number: A real number that cannot be expressed as a simple fraction of two integers. Its decimal expansion is non-terminating and non-repeating.

Step 1: Analyzing the Components of the Expression

The given expression is a fraction where:

• The numerator is $1$, which is a non-zero rational number (since it can be written as $\frac{1}{1}$).
• The denominator is $\sqrt{2}$. [Per the fundamental theorem of arithmetic and the properties of square roots, the square root of any prime number is an irrational number. Since $2$ is a prime number, $\sqrt{2}$ is irrational.]

Step 2: Applying the Quotient Theorem of Real Numbers

We apply the established theorem regarding the arithmetic operations between rational and irrational numbers:

Theorem: The quotient of a non-zero rational number and an irrational number is always an irrational number.

Let $r = 1$ (a non-zero rational number) and $s = \sqrt{2}$ (an irrational number). The quotient $\frac{r}{s} = \frac{1}{\sqrt{2}}$ must, by definition, be irrational.

Step 3: Alternative Proof via Rationalization

To provide exhaustive proof, we can also manipulate the expression algebraically by rationalizing the denominator. This transforms the expression into a product, allowing us to apply the product theorem.

Multiply both the numerator and the denominator by $\sqrt{2}$:

$ \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2} $

This can be rewritten as a product:

$ \frac{\sqrt{2}}{2} = \frac{1}{2} \times \sqrt{2} $

Here, we have the product of $\frac{1}{2}$ (a non-zero rational number) and $\sqrt{2}$ (an irrational number). [Per the Product Theorem of Real Numbers: The product of a non-zero rational number and an irrational number is always irrational.] Therefore, the result is irrational.

Geometric Visualization of $\frac{1}{\sqrt{2}}$

To understand this number spatially, we can construct a right-angled isosceles triangle where the hypotenuse is exactly $1$ unit in length. By the Pythagorean theorem ($a^2 + b^2 = c^2$), the lengths of the two equal legs will be exactly $\frac{1}{\sqrt{2}}$. Because the hypotenuse is a rational integer ($1$), the legs represent an incommensurable (irrational) magnitude.

Final Conclusion

Whether analyzed through the quotient of a rational and irrational number, or by rationalizing the denominator to form a product, the mathematical logic strictly dictates that the resulting value cannot be expressed as a simple integer fraction.

Final Solution: The number $\frac{1}{\sqrt{2}}$ is an irrational number.

Initial Setup & Given Expression

We are tasked with rationalizing the denominator of the following irrational fraction:

$E = \frac{1}{\sqrt{7} - 2}$

Objective: To rationalize the denominator means to transform the expression such that the denominator contains only rational numbers, without altering the overall mathematical value of the expression.

Step 1: Identifying the Rationalizing Factor (The Conjugate)

The denominator is a binomial of the form $a - b$, specifically $\sqrt{7} - 2$. To eliminate the square root, we utilize the algebraic identity for the difference of squares:

$(a - b)(a + b) = a^2 - b^2$

[Per the Difference of Squares Theorem], multiplying the binomial by its conjugate will square both terms, thereby neutralizing the square root. The conjugate of $\sqrt{7} - 2$ is obtained by changing the sign between the terms, resulting in $\sqrt{7} + 2$.

Step 2: Multiplying by the Multiplicative Identity

To maintain the equivalence of the fraction, we must multiply both the numerator and the denominator by the conjugate. This is equivalent to multiplying the entire expression by $1$ [Per the Multiplicative Identity Property, $\frac{x}{x} = 1$].

$E = \frac{1}{\sqrt{7} - 2} \times \frac{\sqrt{7} + 2}{\sqrt{7} + 2}$

Step 3: Expanding the Numerator and Denominator

Now, we perform the algebraic multiplication for both the top and bottom of the fraction.

• Numerator: $1 \cdot (\sqrt{7} + 2) = \sqrt{7} + 2$
• Denominator: $(\sqrt{7} - 2)(\sqrt{7} + 2)$

Applying the difference of squares identity $(a-b)(a+b) = a^2 - b^2$ to the denominator, where $a = \sqrt{7}$ and $b = 2$:

$\text{Denominator} = (\sqrt{7})^2 - (2)^2$

Step 4: Simplifying the Expression

Evaluate the squares in the denominator:

• $(\sqrt{7})^2 = 7$ [Since squaring a square root yields the original radicand]
• $(2)^2 = 4$

Subtract the squared values:

$\text{Denominator} = 7 - 4 = 3$

Substitute the simplified denominator back into the fraction:

$E = \frac{\sqrt{7} + 2}{3}$

The denominator is now $3$, which is a rational integer. The rationalization process is complete.

Final Solution: The rationalized form of the expression is $\frac{\sqrt{7} + 2}{3}$.

Step 1: Analysis of the Given Expression

We are tasked with classifying the number $2\pi$ as either rational or irrational. To do this rigorously, we must decompose the expression into its constituent mathematical factors:

• Factor 1: The number $2$. This is an integer. By definition, any integer $z$ can be expressed as the quotient of two integers $\frac{z}{1}$. Therefore, $2 = \frac{2}{1}$, making it a rational number.
• Factor 2: The number $\pi$ (pi). $\pi$ is a mathematical constant defined as the ratio of a circle's circumference to its diameter. Its decimal expansion ($3.14159265...$) is non-terminating and non-repeating. Therefore, $\pi$ cannot be expressed as a simple fraction $\frac{p}{q}$ (where $p, q \in \mathbb{Z}$ and $q \neq 0$). This makes $\pi$ a fundamental irrational number.

Step 2: Geometric Visualization of $2\pi$

Geometrically, the value $2\pi$ is not merely an abstract algebraic product; it represents the exact circumference of a unit circle (a circle with a radius of $1$ unit). Because the circumference is a continuous length mapped by an irrational multiplier, the total length itself cannot be measured as an exact rational fraction of the radius.

Step 3: Theoretical Justification via Number System Properties

To classify the product of a rational number and an irrational number, we rely on a foundational theorem of real numbers:

Theorem: The product of a non-zero rational number and an irrational number is always an irrational number.

Since $2$ is a non-zero rational number and $\pi$ is an irrational number, their product, $2 \times \pi = 2\pi$, must inherently be irrational [Per the Closure Properties of Real Numbers].

Step 4: Algebraic Proof by Contradiction

To provide absolute mathematical rigor, we can prove this classification using the method of contradiction.

• Assumption: Let us assume the opposite of our expected conclusion. Assume that $2\pi$ is a rational number.
• Definition Application: If $2\pi$ is rational, it can be expressed as the ratio of two coprime integers $p$ and $q$ (where $q \neq 0$).
  
  $2\pi = \frac{p}{q}$
• Algebraic Manipulation: Isolate $\pi$ by dividing both sides of the equation by $2$:
  
  $\pi = \frac{p}{2q}$
• Logical Deduction: Since $p$ and $q$ are integers, the product $2q$ is also an integer. Therefore, the fraction $\frac{p}{2q}$ represents the quotient of two integers, which by definition is a rational number.
• The Contradiction: The equation $\pi = \frac{p}{2q}$ implies that $\pi$ is a rational number. However, it is a universally proven mathematical fact that $\pi$ is irrational. This creates a logical contradiction.
• Conclusion of Proof: Because our initial assumption led to a false statement, the initial assumption must be false. Therefore, $2\pi$ cannot be rational.

Final Solution: The number $2\pi$ is an irrational number.

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.

Initial Setup & Mathematical Objective

We are given the fractional expression:

$ \frac{1}{\sqrt{7}} $

The denominator of this fraction is $\sqrt{7}$, which is an irrational number. In mathematics, it is a standard convention to express fractions with a rational number in the denominator to facilitate easier addition, subtraction, and comparison of fractions. The process of converting an irrational denominator into a rational one without altering the overall value of the expression is known as rationalisation.

Step 1: Identifying the Rationalising Factor

To eliminate the square root from the denominator, we must multiply it by a value that results in a perfect square under the radical. [Per the fundamental property of radicals, $\sqrt{x} \cdot \sqrt{x} = \sqrt{x^2} = x$ for any positive real number $x$].

Given the denominator is $\sqrt{7}$, the smallest and most direct rationalising factor is $\sqrt{7}$ itself, because:

$ \sqrt{7} \times \sqrt{7} = \sqrt{49} = 7 $

Step 2: Applying the Multiplicative Identity Property

To ensure that the value of the original fraction remains unchanged, we must multiply the entire expression by $1$. We can express $1$ as a fraction where the numerator and the denominator are both equal to our rationalising factor, $\sqrt{7}$. [By the Multiplicative Identity Property, $a \cdot 1 = a$].

We set up the multiplication as follows:

$ \frac{1}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} $

Step 3: Algebraic Simplification

We now perform the multiplication across the numerators and the denominators respectively:

• Numerator Calculation: Multiply the numerators together.
  $ 1 \times \sqrt{7} = \sqrt{7} $
• Denominator Calculation: Multiply the denominators together.
  $ \sqrt{7} \times \sqrt{7} = \sqrt{7 \times 7} = \sqrt{49} = 7 $

Combining the simplified numerator and denominator yields the final rationalised expression:

$ \frac{\sqrt{7}}{7} $

Notice that the denominator is now $7$, which is a rational integer, fulfilling the objective of the problem.

Final Solution: The rationalised form of $\frac{1}{\sqrt{7}}$ is $\frac{\sqrt{7}}{7}$.

Initial Setup & Mathematical Objective

We are given the following fractional expression featuring an irrational denominator:

$ E = \frac{1}{\sqrt{7} - \sqrt{6}} $

The mathematical objective is to rationalize the denominator. Rationalization is the process of eliminating irrational numbers (such as surds or roots) from the denominator of an algebraic fraction. This is achieved by multiplying the numerator and the denominator by a suitable rationalizing factor, ensuring the fraction's overall value remains unchanged [Per the Multiplicative Identity Property].

Step 1: Identifying the Conjugate Surd

To eliminate the square roots in a binomial denominator of the form $a - b$, we utilize its conjugate. The conjugate of a binomial $a - b$ is $a + b$.

• The given denominator is: $\sqrt{7} - \sqrt{6}$
• The conjugate of this denominator is: $\sqrt{7} + \sqrt{6}$

Multiplying a binomial surd by its conjugate leverages the difference of squares identity, which squares both terms and effectively removes the square root radicals.

Step 2: Multiplying Numerator and Denominator by the Conjugate

We multiply both the numerator and the denominator of the fraction by the conjugate $\sqrt{7} + \sqrt{6}$. This is equivalent to multiplying the entire expression by $1$, thus preserving its fundamental value.

$ E = \frac{1}{\sqrt{7} - \sqrt{6}} \times \frac{\sqrt{7} + \sqrt{6}}{\sqrt{7} + \sqrt{6}} $

Step 3: Applying Algebraic Identities

Combine the numerators and the denominators:

$ E = \frac{1 \cdot (\sqrt{7} + \sqrt{6})}{(\sqrt{7} - \sqrt{6})(\sqrt{7} + \sqrt{6})} $

For the denominator, we apply the foundational algebraic identity for the difference of two squares [$(a - b)(a + b) = a^2 - b^2$].

• Let $a = \sqrt{7}$
• Let $b = \sqrt{6}$

Substituting these into the identity yields:

$ (\sqrt{7} - \sqrt{6})(\sqrt{7} + \sqrt{6}) = (\sqrt{7})^2 - (\sqrt{6})^2 $

Step 4: Simplifying the Expression

Now, we evaluate the squares in the denominator. By definition, the square of a square root returns the original radicand [$(\sqrt{x})^2 = x$ for $x \ge 0$].

• $(\sqrt{7})^2 = 7$
• $(\sqrt{6})^2 = 6$

Substitute these values back into the denominator:

$ \text{Denominator} = 7 - 6 = 1 $

The numerator remains unchanged as it was multiplied by $1$:

$ \text{Numerator} = \sqrt{7} + \sqrt{6} $

Reassembling the fraction gives:

$ E = \frac{\sqrt{7} + \sqrt{6}}{1} $

Since any number divided by $1$ is the number itself, the expression simplifies completely to a linear binomial surd.

Final Solution: \sqrt{7} + \sqrt{6}

Initial Setup & Mathematical Objective

We are tasked with simplifying the product of two binomial expressions containing irrational numbers (surds). The given expression is:

$(3 + \sqrt{3})(2 + \sqrt{2})$

Step 1: Applying the Distributive Property

To multiply two binomials, we utilize the distributive property of multiplication over addition, commonly referred to as the FOIL method (First, Outer, Inner, Last) in elementary algebra. The general algebraic identity is defined as:

$(a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd$

[Per the Axioms of Real Numbers, multiplication distributes over addition].

Mapping our given values to this identity:

• $a = 3$
• $b = \sqrt{3}$
• $c = 2$
• $d = \sqrt{2}$

Step 2: Term-by-Term Multiplication

We will now systematically multiply each term:

• First terms ($ac$): $3 \times 2 = 6$
• Outer terms ($ad$): $3 \times \sqrt{2} = 3\sqrt{2}$
• Inner terms ($bc$): $\sqrt{3} \times 2 = 2\sqrt{3}$
• Last terms ($bd$): $\sqrt{3} \times \sqrt{2} = \sqrt{3 \times 2} = \sqrt{6}$
  [Per the Product Rule for Radicals: $\sqrt{x} \cdot \sqrt{y} = \sqrt{xy}$ for all real numbers $x, y \ge 0$].

Step 3: Aggregating the Expanded Terms

Combining the results from Step 2, we construct the expanded expression:

$6 + 3\sqrt{2} + 2\sqrt{3} + \sqrt{6}$

Step 4: Analyzing for Like Terms

In radical expressions, terms can only be added or subtracted if they are "like surds" (i.e., they share the exact same irrational radicand). Let us evaluate our terms:

• $6$ is a rational integer.
• $3\sqrt{2}$ contains the irrational root $\sqrt{2}$.
• $2\sqrt{3}$ contains the irrational root $\sqrt{3}$.
• $\sqrt{6}$ contains the irrational root $\sqrt{6}$.

Because $\sqrt{2}$, $\sqrt{3}$, and $\sqrt{6}$ are distinct, mutually prime radicands, none of these terms are like terms. Therefore, no further algebraic simplification or combination is mathematically permissible.

Geometric Visualization: Area Model

The multiplication of these binomials can be geometrically represented as calculating the total area of a rectangle with side lengths $(3 + \sqrt{3})$ and $(2 + \sqrt{2})$. The total area is the sum of the four smaller rectangular regions.

Final Solution: $6 + 3\sqrt{2} + 2\sqrt{3} + \sqrt{6}$

Initial Setup & Given Expression

We are tasked with simplifying the following irrational algebraic expression:

$(3 + \sqrt{3})(3 - \sqrt{3})$

Step 1: Identification of the Algebraic Identity

The given expression is in the form of the product of the sum and difference of two binomial terms. We can optimize the simplification process by applying the fundamental algebraic identity for the difference of two squares.

[Per the algebraic identity for the difference of two squares]:

$(a + b)(a - b) = a^2 - b^2$

Step 2: Variable Mapping and Substitution

By comparing our specific expression $(3 + \sqrt{3})(3 - \sqrt{3})$ with the standard identity $(a + b)(a - b)$, we can map the variables as follows:

• $a = 3$
• $b = \sqrt{3}$

Substituting these values into the right-hand side of the identity ($a^2 - b^2$), we obtain:

$(3)^2 - (\sqrt{3})^2$

Step 3: Simplification of Individual Terms

We now evaluate the square of each term independently.

First Term ($a^2$):

$(3)^2 = 3 \times 3 = 9$

Second Term ($b^2$):

[Per the definition of a principal square root, the square of a square root of a non-negative real number $x$ returns the number itself: $(\sqrt{x})^2 = x$]

$(\sqrt{3})^2 = 3$

Step 4: Final Arithmetic Operation

Substitute the simplified values back into the expression from Step 2:

$9 - 3 = 6$

Final Solution: The simplified value of the expression $(3 + \sqrt{3})(3 - \sqrt{3})$ is 6.

Given Expression & Algebraic Foundation

We are tasked with simplifying the following mathematical expression involving irrational numbers:

$(\sqrt{5} + \sqrt{2})^2$

To expand this expression systematically, we utilize the standard binomial square identity [Derived from the distributive property of multiplication over addition]:

$(a + b)^2 = a^2 + 2ab + b^2$

Step 1: Variable Substitution

By mapping the given expression to the binomial identity, we establish our variables:

• Let $a = \sqrt{5}$
• Let $b = \sqrt{2}$

Substituting these values into the identity yields:

$(\sqrt{5} + \sqrt{2})^2 = (\sqrt{5})^2 + 2(\sqrt{5})(\sqrt{2}) + (\sqrt{2})^2$

Step 2: Simplifying the Squared Terms

We evaluate the first and third terms of the expansion. [Per the fundamental definition of a principal square root, $(\sqrt{x})^2 = x$ for any non-negative real number $x \ge 0$].

• First term: $(\sqrt{5})^2 = 5$
• Third term: $(\sqrt{2})^2 = 2$

Step 3: Simplifying the Cross-Product Term

Next, we evaluate the middle term, $2(\sqrt{5})(\sqrt{2})$. [Per the Product Property of Radicals, $\sqrt{x} \cdot \sqrt{y} = \sqrt{xy}$ for $x, y \ge 0$].

$2(\sqrt{5})(\sqrt{2}) = 2(\sqrt{5 \cdot 2}) = 2\sqrt{10}$

Step 4: Combining Like Terms

Substituting the simplified components back into the expanded equation, we get:

$5 + 2\sqrt{10} + 2$

We group the rational numbers together and leave the irrational term distinct [Per the commutative property of addition]:

$(5 + 2) + 2\sqrt{10} = 7 + 2\sqrt{10}$

Geometric Visualization of the Expansion

The algebraic identity $(a+b)^2 = a^2 + 2ab + b^2$ can be visualized as the area of a square with side length $(a+b)$, partitioned into four distinct rectangular regions. Below is the geometric proof applied specifically to $a = \sqrt{5}$ and $b = \sqrt{2}$.

Summing the areas of the four regions confirms our algebraic derivation:

$\text{Total Area} = 5 + \sqrt{10} + \sqrt{10} + 2 = 7 + 2\sqrt{10}$

Final Solution: $7 + 2\sqrt{10}$

Initial Setup & Given Expression

We are tasked with simplifying the following product of two binomials involving irrational numbers (surds):

$ (\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2}) $

Step 1: Identification of the Algebraic Identity

The given expression perfectly matches the structural form of the Difference of Squares identity. [Per the fundamental theorems of polynomial algebra], the product of the difference and the sum of the same two terms is equal to the square of the first term minus the square of the second term.

The identity is formally stated as:

$ (a - b)(a + b) = a^2 - b^2 $

Step 2: Geometric Verification of the Identity

To rigorously understand why this identity holds true, we can visualize the area of a square of side $a$ from which a smaller square of side $b$ is removed. The remaining area, $a^2 - b^2$, can be geometrically rearranged into a rectangle with dimensions $(a - b)$ and $(a + b)$.

Step 3: Variable Substitution

By mapping the variables from our specific problem to the algebraic identity, we establish the following equivalences:

• Let $ a = \sqrt{5} $
• Let $ b = \sqrt{2} $

Substituting these values into the right side of the identity ($a^2 - b^2$), we get:

$ (\sqrt{5})^2 - (\sqrt{2})^2 $

Step 4: Application of Exponent Rules on Surds

Next, we evaluate the squares of the square roots. [Per the definition of principal square roots and rational exponents], for any non-negative real number $x$, the operation of squaring its square root returns the original radicand. Mathematically, this is expressed as:

$ (\sqrt{x})^2 = (x^{\frac{1}{2}})^2 = x^{\frac{1}{2} \times 2} = x^1 = x \quad \text{for } x \ge 0 $

Applying this axiom to our specific terms:

• $ (\sqrt{5})^2 = 5 $
• $ (\sqrt{2})^2 = 2 $

Step 5: Final Arithmetic Evaluation

Substitute the simplified rational numbers back into the expression from Step 3:

$ 5 - 2 = 3 $

The irrational components have been completely rationalized through the algebraic expansion, resulting in a pure integer.

Final Solution: 3

Given Variables & Initial Setup

We are presented with the classical geometric definition of the mathematical constant $\pi$:

$\pi = \frac{c}{d}$

Where:

• $c$ represents the circumference of a circle (the total boundary length).
• $d$ represents the diameter of the same circle (the straight-line distance passing through the center).

The apparent contradiction arises from the definition of rational and irrational numbers. A rational number is defined as any number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Because $\pi$ is written as a ratio ($\frac{c}{d}$), it superficially resembles a rational number. However, it is a proven mathematical fact that $\pi$ is an irrational number (its decimal expansion is non-terminating and non-repeating).

Step 1: Analyzing the Definition of Rational Numbers

To resolve this contradiction, we must rigorously apply the definition of a rational number. [Per the Fundamental Axioms of the Real Number System], for a fraction $\frac{x}{y}$ to be strictly classified as a rational number, both the numerator $x$ and the denominator $y$ must be integers ($\mathbb{Z}$).

Step 2: The Limitation of Physical Measurement

When we measure the circumference ($c$) or the diameter ($d$) of a physical circle using a scale, tape, or any measuring device, we are limited by the precision of the instrument. We only ever obtain an approximate rational value.

Because of this physical limitation, we might mistakenly assume that both $c$ and $d$ are rational numbers (e.g., measuring $d = 7\text{ cm}$ and $c \approx 22\text{ cm}$). However, exact mathematical truth dictates that we can never measure an irrational length with absolute precision in the physical world.

Step 3: Theoretical Resolution of the Ratio

In pure mathematics, if we construct a circle where the diameter $d$ is a perfect rational number (e.g., exactly $2\text{ units}$), the circumference $c$ will mathematically evaluate to an irrational number ($c = 2\pi$). Conversely, if we force the circumference $c$ to be a rational number, the diameter $d$ must be irrational ($d = \frac{c}{\pi}$).

[Per the Properties of Irrational Numbers], the quotient of a non-zero rational number and an irrational number (or vice versa) is always an irrational number. Therefore:

• If $c$ is irrational and $d$ is rational: $\frac{\text{Irrational}}{\text{Rational}} = \text{Irrational}$
• If $c$ is rational and $d$ is irrational: $\frac{\text{Rational}}{\text{Irrational}} = \text{Irrational}$
• If both $c$ and $d$ are irrational: The ratio can be irrational (as is the case with $\pi$).

Step 4: Conclusion of the Logical Proof

The equation $\pi = \frac{c}{d}$ does not satisfy the condition $\frac{p}{q}$ where $p, q \in \mathbb{Z}$. At least one of the parameters, $c$ or $d$ (or both), is an irrational number. Because the numerator and denominator are not both integers, the ratio $\frac{c}{d}$ does not meet the definition of a rational number.

Final Solution: There is no contradiction. The ratio $\pi = \frac{c}{d}$ only yields a rational number if both $c$ and $d$ are integers. In reality, it is mathematically impossible for both the circumference and the diameter of the same circle to be integers simultaneously. At least one of them is always irrational, which ensures that their ratio, $\pi$, remains strictly irrational.

Initial Expression & Setup

We are tasked with classifying the following mathematical expression as either a rational or an irrational number:

$ (3 + \sqrt{23}) - \sqrt{23} $

To determine its classification, we must first simplify the expression to its most fundamental form.

Step 1: Algebraic Simplification

We begin by removing the parentheses. [Per the Associative Property of Addition, grouping symbols can be removed when only addition and subtraction are involved].

$ (3 + \sqrt{23}) - \sqrt{23} = 3 + \sqrt{23} - \sqrt{23} $

Step 2: Cancellation of Irrational Terms

Next, we group the like terms. The expression contains two terms involving the square root of 23: $+\sqrt{23}$ and $-\sqrt{23}$. These terms are additive inverses of each other. [Per the Additive Inverse Property, any number added to its negative equals zero, i.e., $x - x = 0$].

$ 3 + (\sqrt{23} - \sqrt{23}) = 3 + 0 $

$ = 3 $

Step 3: Theoretical Classification of the Result

The simplified result is the integer $3$. We must now evaluate this result against the formal definitions of rational and irrational numbers.

• Rational Number: A number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers, and $q \neq 0$.
• Irrational Number: A number that cannot be expressed as a simple fraction (its decimal expansion is non-terminating and non-repeating).

The integer $3$ can be rewritten as a fraction by placing it over a denominator of $1$:

$ 3 = \frac{3}{1} $

Here, $p = 3$ (an integer) and $q = 1$ (an integer where $1 \neq 0$). Because it strictly satisfies the condition $\frac{p}{q}$, the number is rational.

Visual Simplification Flowchart

Final Solution: The expression $(3 + \sqrt{23}) - \sqrt{23}$ simplifies to $3$, which is a rational number.

Step 1: Initial Geometric Setup & Line Segment Definition

To represent the irrational number $\sqrt{9.3}$ on a number line, we utilize a geometric construction based on the properties of right-angled triangles and semicircles. We begin by translating the numerical value into a physical line segment.

• Draw a horizontal line and mark a starting point $A$.
• From point $A$, measure a distance of exactly $9.3$ units and mark this point as $B$. Thus, the length of the line segment $\overline{AB} = 9.3$ units.
• From point $B$, extend the line segment further by exactly $1$ unit and mark the new point as $C$. Thus, the length of $\overline{BC} = 1$ unit.

[By the Segment Addition Postulate], the total length of the line segment $\overline{AC}$ is:

$AC = AB + BC = 9.3 + 1 = 10.3 \text{ units}$

Step 2: Locating the Midpoint and Constructing the Semicircle

Next, we must find the geometric center of the segment $\overline{AC}$ to serve as the origin for a semicircle.

• Bisect the line segment $\overline{AC}$ to locate its midpoint, $O$.
• The distance from the midpoint $O$ to either endpoint ($A$ or $C$) represents the radius $r$ of the semicircle.

$r = OA = OC = \frac{AC}{2} = \frac{10.3}{2} = 5.15 \text{ units}$

With $O$ as the center and $r = 5.15$ units as the radius, construct a semicircle passing through points $A$ and $C$.

Step 3: Erecting the Perpendicular to Determine $\sqrt{9.3}$

At point $B$ (the boundary between the $9.3$ unit segment and the $1$ unit segment), construct a perpendicular line segment intersecting the semicircle.

• Draw a line perpendicular to $\overline{AC}$ at point $B$ ($\angle OBD = 90^\circ$).
• Let the point where this perpendicular intersects the semicircle be $D$.

The length of the segment $\overline{BD}$ is exactly $\sqrt{9.3}$ units.

Step 4: Rigorous Mathematical Proof of the Construction

To rigorously prove that $BD = \sqrt{9.3}$, we analyze the right-angled triangle $\triangle OBD$.

• The hypotenuse $\overline{OD}$ is a radius of the semicircle. Therefore, $OD = 5.15$ units.
• The base $\overline{OB}$ can be found by subtracting $\overline{BC}$ from the radius $\overline{OC}$: $OB = OC - BC = 5.15 - 1 = 4.15 \text{ units}$

[Per the Pythagorean Theorem], in right $\triangle OBD$:

$OD^2 = OB^2 + BD^2$

$BD^2 = OD^2 - OB^2$

Substitute the known values into the equation:

$BD^2 = (5.15)^2 - (4.15)^2$

Using the difference of squares identity, $a^2 - b^2 = (a - b)(a + b)$:

$BD^2 = (5.15 - 4.15)(5.15 + 4.15)$

$BD^2 = (1.00)(9.30) = 9.3$

$BD = \sqrt{9.3} \text{ units}$

Step 5: Mapping the Magnitude onto the Number Line

Now that we have a physical segment $\overline{BD}$ of length $\sqrt{9.3}$, we must map it onto a standard number line.

• Define point $B$ as the origin ($0$) of the number line.
• Consequently, point $C$ represents the number $1$ (since $BC = 1$ unit).
• Place the compass point at $B$ and the pencil point at $D$. Draw an arc downwards to intersect the extended number line at a new point, $E$.

Since the arc represents a circle with center $B$ and radius $BD$, the distance $BE$ is equal to $BD$. Therefore, point $E$ represents the exact location of $\sqrt{9.3}$ on the number line.

Final Solution: By treating point $B$ as the origin ($0$) and drawing an arc with radius $BD = \sqrt{9.3}$ units, the intersection of the arc with the number line at point $E$ accurately represents the irrational number $\sqrt{9.3}$.