📘 Concept & Theory Important Theory Used ›
- A Rational Number can be written in the form: \[ \frac{p}{q},\quad q\neq0 \] where \(p\) and \(q\) are integers.
- An Irrational Number cannot be written in the form \[ \frac{p}{q} \] and its decimal expansion is non-terminating and non-repeating.
-
Important properties:
- Rational \(+\) Irrational \(=\) Irrational
- Rational \(-\) Irrational \(=\) Irrational
- Non-zero Rational \(\times\) Irrational \(=\) Irrational
- Non-zero Rational \(\div\) Irrational \(=\) Irrational
🗺️ Solution Roadmap Step-by-step Plan ›
- Identify whether each individual number is rational or irrational.
- Apply properties of rational and irrational numbers.
- Simplify the expression wherever possible.
- Write the final classification clearly.
📊 Graph / Figure Graph / Figure ›
✏️ Solution Complete Solution ›
-
\[
2-\sqrt{5}
\]
Solution:
We know that:
\[ 2 \text{ is a Rational Number} \]Also,
\[ \sqrt{5} \text{ is an Irrational Number} \]Using the property:
\[ \text{Rational Number} - \text{Irrational Number} = \text{Irrational Number} \]Therefore,
\[ \therefore\ 2-\sqrt{5} \text{ is an Irrational Number} \]Final Answer: \[ \boxed{2-\sqrt{5}\text{ is Irrational}} \] -
\[
(3+\sqrt{23})-\sqrt{23}
\]
Solution:
Simplify the expression step-by-step:
\[ (3+\sqrt{23})-\sqrt{23} \] \[ =3+\sqrt{23}-\sqrt{23} \] \[ =3+\left(\sqrt{23}-\sqrt{23}\right) \] \[ =3+0 \] \[ =3 \]Since \(3\) is an integer and every integer is a rational number,
\[ \therefore\ 3\text{ is a Rational Number} \]Final Answer: \[ \boxed{(3+\sqrt{23})-\sqrt{23}\text{ is Rational}} \] -
\[
\frac{2\sqrt7}{7\sqrt7}
\]
Solution:
Simplify numerator and denominator carefully:
\[ \frac{2\sqrt7}{7\sqrt7} \]Since \(\sqrt7\) is common in numerator and denominator, we cancel it:
\[ =\frac{2\cancel{\sqrt7}}{7\cancel{\sqrt7}} \] \[ =\frac{2}{7} \]Here,
\[ \frac{2}{7} \]is of the form
\[ \frac{p}{q},\quad q\neq0 \]Therefore,
\[ \therefore\ \frac{2}{7} \text{ is a Rational Number} \]Final Answer: \[ \boxed{\frac{2\sqrt7}{7\sqrt7}\text{ is Rational}} \] -
\[
\frac{1}{\sqrt2}
\]
Solution:
We know that:
\[ \sqrt2 \text{ is an Irrational Number} \]Also, \(1\) is a rational number.
Using the property:
\[ \frac{\text{Rational Number}} {\text{Irrational Number}} = \text{Irrational Number} \]Hence,
\[ \therefore\ \frac{1}{\sqrt2} \text{ is an Irrational Number} \]Final Answer: \[ \boxed{\frac{1}{\sqrt2}\text{ is Irrational}} \] -
\[
2\pi
\]
Solution:
We know that:
\[ \pi \text{ is an Irrational Number} \]Also,
\[ 2 \text{ is a Rational Number} \]Using the property:
\[ \text{Rational Number}\times\text{Irrational Number} = \text{Irrational Number} \]Therefore,
\[ \therefore\ 2\pi \text{ is an Irrational Number} \]Final Answer: \[ \boxed{2\pi\text{ is Irrational}} \]
🎯 Exam Significance Exam Significance ›
- Very important for understanding operations on irrational numbers.
- Frequently asked in CBSE Board examinations.
- Concepts are directly used in Olympiads and entrance examinations.
- Builds foundation for algebraic simplification and surds.