RBSE Solutions Class 9 Maths Chapter 2 Number System Exercise 2.2

Question 1. Classify the following numbers as rational or irrational.
(i) \( 2 - \sqrt{5} \)
(ii) \( (3 + \sqrt{23}) - \sqrt{23} \)
(iii) \( \frac{2\sqrt{11}}{7\sqrt{11}} \)
(iv) \( \frac{1}{\sqrt{3}} \)
(v) \( 2\pi \)
Answer:
(i) \( 2 - \sqrt{5} \) is an irrational number because it is the difference of a rational number and an irrational number, which always results in an irrational number. The square root of 5 is a non-repeating, non-terminating decimal.
(ii) \( (3 + \sqrt{23}) - \sqrt{23} = 3 + \sqrt{23} - \sqrt{23} = 3 \). Since 3 can be written as \( \frac{3}{1} \), it is a rational number. This shows that combining rational and irrational numbers can sometimes lead to a rational result.
(iii) \( \frac{2\sqrt{11}}{7\sqrt{11}} = \frac{2}{7} \). This is a rational number because it is in the form \( \frac{p}{q} \) where p and q are integers and q is not zero. The \( \sqrt{11} \) terms cancel each other out.
(iv) \( \frac{1}{\sqrt{3}} \) is an irrational number. When you divide a rational number (1) by an irrational number (\( \sqrt{3} \)), the result is always irrational.
(v) \( 2\pi \) is an irrational number. The value of \( \pi \) is an irrational number, and multiplying an irrational number by a non-zero rational number always yields an irrational number.
In simple words: A rational number can be written as a fraction, while an irrational number cannot. When you add, subtract, multiply, or divide a rational number with an irrational number, the answer is usually irrational, unless the irrational parts cancel out.

🎯 Exam Tip: Remember the basic rules for combining rational and irrational numbers: (rational ± irrational = irrational), (rational × irrational = irrational, if rational ≠ 0), (irrational × irrational can be rational or irrational).

Question 2. Rationalize the denominator of the following:
(i) \( \frac{1}{5+3\sqrt{7}} \)
(ii) \( \frac{1}{\sqrt{2} + \sqrt{3}} \)
(iii) \( \frac{1}{\sqrt{7}-2} \)
Answer:
(i) To rationalize \( \frac{1}{5+3\sqrt{7}} \), we multiply the numerator and denominator by the conjugate of the denominator, which is \( 5-3\sqrt{7} \).
\[ \frac{1}{5+3\sqrt{7}} \times \frac{5-3\sqrt{7}}{5-3\sqrt{7}} \]
\[ = \frac{5-3\sqrt{7}}{(5)^2 - (3\sqrt{7})^2} \]
\[ = \frac{5-3\sqrt{7}}{25 - (9 \times 7)} \]
\[ = \frac{5-3\sqrt{7}}{25 - 63} \]
\[ = \frac{5-3\sqrt{7}}{-38} \]
\[ = \frac{-(5-3\sqrt{7})}{38} \]
\[ = \frac{-5+3\sqrt{7}}{38} \]
So, the rationalized form is \( \frac{3\sqrt{7}-5}{38} \).
(ii) To rationalize \( \frac{1}{\sqrt{2} + \sqrt{3}} \), we multiply the numerator and denominator by the conjugate of the denominator, which is \( \sqrt{2}-\sqrt{3} \).
\[ \frac{1}{\sqrt{2} + \sqrt{3}} \times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}} \]
\[ = \frac{\sqrt{2}-\sqrt{3}}{(\sqrt{2})^2 - (\sqrt{3})^2} \]
\[ = \frac{\sqrt{2}-\sqrt{3}}{2 - 3} \]
\[ = \frac{\sqrt{2}-\sqrt{3}}{-1} \]
\[ = -(\sqrt{2}-\sqrt{3}) \]
\[ = \sqrt{3}-\sqrt{2} \]
Thus, the rationalized form is \( \sqrt{3}-\sqrt{2} \).
(iii) To rationalize \( \frac{1}{\sqrt{7}-2} \), we multiply the numerator and denominator by the conjugate of the denominator, which is \( \sqrt{7}+2 \).
\[ \frac{1}{\sqrt{7}-2} \times \frac{\sqrt{7}+2}{\sqrt{7}+2} \]
\[ = \frac{\sqrt{7}+2}{(\sqrt{7})^2 - (2)^2} \]
\[ = \frac{\sqrt{7}+2}{7 - 4} \]
\[ = \frac{\sqrt{7}+2}{3} \]
Therefore, the rationalized form is \( \frac{\sqrt{7}+2}{3} \).
In simple words: To get rid of a square root in the bottom part of a fraction, you multiply both the top and bottom by a special number. If the bottom is like "A + B", you multiply by "A - B". If it's "A - B", you multiply by "A + B". This uses a math trick that makes the square roots disappear from the denominator.

🎯 Exam Tip: Always remember to multiply by the conjugate of the denominator to use the identity \( (a+b)(a-b) = a^2 - b^2 \), which eliminates the square roots from the denominator.

Question 3. If \( \frac {3+2\sqrt { 2 } }{ 3-\sqrt {2}} =a+b\sqrt {2} \), where a and b are rational, then find the values of a and b.
Answer: We need to rationalize the left side of the equation to match the form \( a+b\sqrt{2} \).
The given equation is \( \frac {3+2\sqrt { 2 } }{ 3-\sqrt {2}} =a+b\sqrt {2} \).
First, rationalize the denominator of the left side by multiplying the numerator and denominator by the conjugate of \( 3-\sqrt{2} \), which is \( 3+\sqrt{2} \).
\[ \frac{3+2\sqrt{2}}{3-\sqrt{2}} \times \frac{3+\sqrt{2}}{3+\sqrt{2}} \]
Now, multiply the terms in the numerator and the denominator.
For the numerator: \( (3+2\sqrt{2})(3+\sqrt{2}) = 3(3) + 3(\sqrt{2}) + 2\sqrt{2}(3) + 2\sqrt{2}(\sqrt{2}) \)
\[ = 9 + 3\sqrt{2} + 6\sqrt{2} + 2(2) \]
\[ = 9 + 9\sqrt{2} + 4 \]
\[ = 13 + 9\sqrt{2} \]
For the denominator: \( (3-\sqrt{2})(3+\sqrt{2}) = (3)^2 - (\sqrt{2})^2 \)
\[ = 9 - 2 \]
\[ = 7 \]
So, the left side becomes: \( \frac{13+9\sqrt{2}}{7} \)
We can write this as: \( \frac{13}{7} + \frac{9\sqrt{2}}{7} \)
Now, we equate this to \( a+b\sqrt{2} \):
\( \frac{13}{7} + \frac{9}{7}\sqrt{2} = a+b\sqrt{2} \)

\( \implies \) By comparing the rational and irrational parts, we get:
\( a = \frac{13}{7} \)
\( b = \frac{9}{7} \)
In simple words: To find 'a' and 'b', we first make the bottom of the fraction simple by removing the square root. After doing that, we get a new fraction with two parts: one part without a square root and one part with a square root. We then match these parts with 'a' and 'b' to find their values.

🎯 Exam Tip: When comparing expressions like \( x + y\sqrt{k} = a + b\sqrt{k} \), remember that you can equate the rational parts (\( x=a \)) and the coefficients of the irrational parts (\( y=b \)) only if \( \sqrt{k} \) is irrational.