NCERT Solutions for Class 9 Maths Exercise 1.4 Question 1
Understanding the Question 🧐
In this problem, we need to classify a given set of numbers as either rational or irrational. Let’s quickly refresh these concepts. A rational number can be written in the form &&\frac{p}{q}&&, where &&p&& and &&q&& are integers and &&q \neq 0&&. An irrational number cannot be written this way and has a non-terminating, non-repeating decimal expansion. This exercise is a great way to test our understanding of these number types.
Classify the following numbers as rational or irrational:
(i) &&2 – \sqrt{5}&&
(ii) &&(3 + \sqrt{23}) – \sqrt{23}&&
(iii) &&\frac{2\sqrt{7}}{7\sqrt{7}}&&
(iv) &&\frac{1}{\sqrt{2}}&&
(v) &&2\pi&&
Part (i): Classifying &&2 – \sqrt{5}&& 📝
Step 1: Identify the type of each number in the expression.
- The number &&2&& is a rational number because it can be written as &&\frac{2}{1}&&.
- The number &&\sqrt{5}&& is an irrational number because &&5&& is a prime number and not a perfect square.
Step 2: Apply the property of operations between rational and irrational numbers.
- A key property states that the sum or difference of a rational number and an irrational number is always irrational.
Conclusion: Since we are subtracting an irrational number (&&\sqrt{5}&&) from a rational number (&&2&&), the result &&2 – \sqrt{5}&& is an irrational number.
Part (ii): Classifying &&(3 + \sqrt{23}) – \sqrt{23}&& 📝
Step 1: Simplify the given expression.
Before classifying, we should always simplify the expression if possible. Let’s remove the parentheses:
&&\Rightarrow (3 + \sqrt{23}) – \sqrt{23} = 3 + \sqrt{23} – \sqrt{23}&&
The terms &&+\sqrt{23}&& and &&-\sqrt{23}&& cancel each other out.
&&\Rightarrow 3 + 0 = 3&&
Step 2: Classify the simplified number.
- The simplified result is &&3&&.
- The number &&3&& can be written in the &&\frac{p}{q}&& form as &&\frac{3}{1}&&.
Conclusion: Therefore, the expression &&(3 + \sqrt{23}) – \sqrt{23}&& is a rational number.
Part (iii): Classifying &&\frac{2\sqrt{7}}{7\sqrt{7}}&& 📝
Step 1: Simplify the fraction.
Notice that the term &&\sqrt{7}&& appears in both the numerator and the denominator. We can cancel them out.
&&\Rightarrow \frac{2\sqrt{7}}{7\sqrt{7}} = \frac{2}{7}&&
Step 2: Classify the simplified fraction.
- The simplified number is &&\frac{2}{7}&&.
- This is clearly in the form &&\frac{p}{q}&&, where &&p=2&& and &&q=7&& (and &&q \neq 0&&).
Conclusion: Thus, the expression &&\frac{2\sqrt{7}}{7\sqrt{7}}&& is a rational number.
Part (iv): Classifying &&\frac{1}{\sqrt{2}}&& 📝
Step 1: Identify the type of numbers in the fraction.
- The numerator, &&1&&, is a rational number.
- The denominator, &&\sqrt{2}&&, is an irrational number (since &&2&& is a prime number and not a perfect square).
Step 2: Apply the property of division involving irrational numbers.
- The quotient of a non-zero rational number and an irrational number is always irrational.
Conclusion: Based on this property, &&\frac{1}{\sqrt{2}}&& is an irrational number.
Part (v): Classifying &&2\pi&& 📝
Step 1: Identify the type of each number being multiplied.
- The number &&2&& is a rational number.
- The number &&\pi&& (pi) is a famous mathematical constant and is known to be an irrational number. Its decimal representation goes on forever without repeating.
Step 2: Apply the property of multiplication involving irrational numbers.
- The product of a non-zero rational number and an irrational number is always irrational.
Conclusion: Therefore, &&2\pi&& is an irrational number.
Conclusion and Key Points ✅
Based on our step-by-step analysis, here is the final classification for each expression:
- &&2 – \sqrt{5}&& is Irrational.
- &&(3 + \sqrt{23}) – \sqrt{23}&& is Rational.
- &&\frac{2\sqrt{7}}{7\sqrt{7}}&& is Rational.
- &&\frac{1}{\sqrt{2}}&& is Irrational.
- &&2\pi&& is Irrational.
- Rational numbers can be written as &&\frac{p}{q}&& (&&q \neq 0&&).
- Irrational numbers have non-terminating, non-repeating decimal expansions.
- Rational &&\pm&& Irrational &&=&& Irrational.
- Rational &&\times&& Irrational &&=&& Irrational (if the rational number is not zero).
- Rational &&\div&& Irrational &&=&& Irrational (if the rational number is not zero).
- Constants like &&\pi&& and square roots of non-perfect squares (like &&\sqrt{2}, \sqrt{3}, \sqrt{5}&&) are irrational.
FAQ ❓
Q: Why is &&2 – \sqrt{5}&& irrational?
A: It’s irrational because you are subtracting an irrational number (&&\sqrt{5}&&) from a rational number (&&2&&). The result of this operation is always irrational.
Q: Is &&(3 + \sqrt{23}) – \sqrt{23}&& rational or irrational?
A: It is rational. When you simplify the expression, &&\sqrt{23} – \sqrt{23} = 0&&, which leaves you with just &&3&&. Since &&3&& can be written as &&\frac{3}{1}&&, it is a rational number.
Q: How do you classify the number &&\frac{2\sqrt{7}}{7\sqrt{7}}&&?
A: This number is rational. The &&\sqrt{7}&& in the numerator and denominator cancel each other, simplifying the expression to &&\frac{2}{7}&&, which is in the standard &&\frac{p}{q}&& form of a rational number.
Q: What makes &&\frac{1}{\sqrt{2}}&& an irrational number?
A: It’s the division of a rational number (&&1&&) by an irrational number (&&\sqrt{2}&&). The result of such a division is always an irrational number.
Q: Why is &&2\pi&& considered irrational if &&\pi&& is approximately &&\frac{22}{7}&&?
A: The value &&\frac{22}{7}&& is only a rational approximation of &&\pi&&; it is not its exact value. The number &&\pi&& itself is irrational. Multiplying an irrational number (&&\pi&&) by a non-zero rational number (&&2&&) results in an irrational number.
Q: Can the sum of two irrational numbers be rational?
A: Yes, it can. For example, &&(2 + \sqrt{3})&& and &&(2 – \sqrt{3})&& are both irrational, but their sum is &&(2 + \sqrt{3}) + (2 – \sqrt{3}) = 4&&, which is rational.
Further Reading 📖
For a deeper understanding of number systems, you can refer to the official NCERT textbook for Class 9 Maths. More resources are available on the NCERT website at https://ncert.nic.in/.
