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Q1(iii):
Classify the following numbers as rational or irrational:
(iii) $\frac{2\sqrt{7}}{7\sqrt{7}}$
Solution :
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.
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(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) $(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})$
- 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) $\frac{1}{\sqrt{7} - 2}$
CBSE Solutions for Class 9 Mathematics Number Systems
Chapters in CBSE - Class 9 Mathematics
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