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Q3:
Recall, $\pi$ is defined as the ratio of the circumference (say $c$) of a circle to its diameter (say $d$). That is, $\pi = \frac{c}{d}$. This seems to contradict the fact that $\pi$ is irrational. How will you resolve this contradiction?
Solution :
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 ($\mathbb{Z}$).
| Number Type | Form $\frac{p}{q}$ Condition | Decimal Expansion | Example |
|---|---|---|---|
| Rational ($\mathbb{Q}$) | $p, q \in \mathbb{Z}$, $q \neq 0$ | Terminating or Non-terminating repeating | $\frac{22}{7} \approx 3.142857...$ |
| Irrational ($\mathbb{R} \setminus \mathbb{Q}$) | Cannot be expressed with $p, q \in \mathbb{Z}$ | Non-terminating and Non-repeating | $\pi \approx 3.14159265...$ |
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.
More Questions from Class 9 Mathematics Number Systems EXERCISE 1.4
- Q1(i): Classify the following numbers as rational or irrational: (i) $2 - \sqrt{5}$
- Q1(ii): Classify the following numbers as rational or irrational: (ii) $(3 + \sqrt{23}) - \sqrt{23}$
- Q1(iii): Classify the following numbers as rational or irrational: (iii) $\frac{2\sqrt{7}}{7\sqrt{7}}$
- Q1(iv): Classify the following numbers as rational or irrational: (iv) $\frac{1}{\sqrt{2}}$
- Q1(v): Classify the following numbers as rational or irrational: (v) $2\pi$
- Q2(i): Simplify each of the following expressions: (i) $(3 + \sqrt{3})(2 + \sqrt{2})$
- Q2(ii): Simplify each of the following expressions: (ii) $(3 + \sqrt{3})(3 - \sqrt{3})$
- Q2(iii): Simplify each of the following expressions: (iii) $(\sqrt{5} + \sqrt{2})^2$
- Q2(iv): Simplify each of the following expressions: (iv) $(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})$
- Q4: Represent $\sqrt{9.3}$ on the number line.
- Q5(i): Rationalise the denominators of the following: (i) $\frac{1}{\sqrt{7}}$
- Q5(ii): Rationalise the denominators of the following: (ii) $\frac{1}{\sqrt{7} - \sqrt{6}}$
- Q5(iii): Rationalise the denominators of the following: (iii) $\frac{1}{\sqrt{5} + \sqrt{2}}$
- Q5(iv): Rationalise the denominators of the following: (iv) $\frac{1}{\sqrt{7} - 2}$
CBSE Solutions for Class 9 Mathematics Number Systems
Chapters in CBSE - Class 9 Mathematics
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