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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.
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.
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) $\frac{1}{\sqrt{2}}$
- 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) $(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})$
- 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?
- 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) $\frac{1}{\sqrt{7} - 2}$
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