Find the best tutors and institutes for Class 10 Tuition
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
Top Tutors who teach Number Systems
10+ years of experience teaching Mathematics for Grades 10 to 12th • Consistent record of 100% academic results • Very friendly and supportive learning environment • Encourages students to ask questions freely and confidently • Strong focus on conceptual clarity and problem-solving skills • Helps students build confidence and excel in Mathematics • Fee Structure for One-to-One Classes: ₹700 per hour • Group Classes: ₹4,000 per month.
Sir clears all my doubts he makes difficult concepts easy to understand he takes Previous year questions which makes it easier to prepare
I have specifically guided Class 10 students for board exams, with a strong focus on NCERT syllabus, and topics like Real Numbers, Polynomials, Quadratic Equations, and Statistics. I provide regular tests, personalized feedback, and exam-oriented strategies to improve accuracy and speed. Many of my students have shown remarkable improvement and scored above 90% in their board exams.
It was an amazing experience learning and understanding maths with Saurav sir. He really helped me overcome my fear of maths and helped me score good marks in a subject that previously scared me. It really shows how dedicated he is as a teacher. Really good experience!
Dear Parents and students, I am Amit, We provide home-based tuition for subjects and we also do professional coaching for NEET/IIT-JEE and other competitive exams. We have experienced tutors in all the fields and subjects. We also provide demo class free of cost. We do online as well as offline mode of teaching...
I've taught at least 150 students of this Class till date ....online/offline together.
It was a great experience while taking the class tenth private tutions.
Fabulous experience learning with Indrajeet Sir, quickly made me grasp difficult concepts in preparation for my board examinations.
Find more Tutor for Number Systems in your City
- Bangalore Mathematics Tutors
- Delhi Mathematics Tutors
- Chennai Mathematics Tutors
- Gurgaon Mathematics Tutors
- Noida Mathematics Tutors
- Hyderabad Mathematics Tutors
- Mumbai Mathematics Tutors
- Ghaziabad Mathematics Tutors
- Chandigarh Mathematics Tutors
- Pune Mathematics Tutors
- Jaipur Mathematics Tutors
- Surat Mathematics Tutors
Download free CBSE - Class 9 Mathematics Number Systems EXERCISE 1.4 worksheets
Download Now