Q9(i):

Classify the following numbers as rational or irrational : (i) $\sqrt{23}$

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.