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Q1(v):
Classify the following numbers as rational or irrational: (v) $2\pi$

Solution :

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.

O r = 1 Geometric Meaning of 2π Circumference (C) = 2πr If r = 1, C = 2π (An Irrational Length)

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.


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