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Q5(i):
Rationalise the denominators of the following:
(i) $\frac{1}{\sqrt{7}}$
Solution :
Initial Setup & Mathematical Objective
We are given the fractional expression:
$ \frac{1}{\sqrt{7}} $
The denominator of this fraction is $\sqrt{7}$, which is an irrational number. In mathematics, it is a standard convention to express fractions with a rational number in the denominator to facilitate easier addition, subtraction, and comparison of fractions. The process of converting an irrational denominator into a rational one without altering the overall value of the expression is known as rationalisation.
Step 1: Identifying the Rationalising Factor
To eliminate the square root from the denominator, we must multiply it by a value that results in a perfect square under the radical. [Per the fundamental property of radicals, $\sqrt{x} \cdot \sqrt{x} = \sqrt{x^2} = x$ for any positive real number $x$].
Given the denominator is $\sqrt{7}$, the smallest and most direct rationalising factor is $\sqrt{7}$ itself, because:
$ \sqrt{7} \times \sqrt{7} = \sqrt{49} = 7 $
Step 2: Applying the Multiplicative Identity Property
To ensure that the value of the original fraction remains unchanged, we must multiply the entire expression by $1$. We can express $1$ as a fraction where the numerator and the denominator are both equal to our rationalising factor, $\sqrt{7}$. [By the Multiplicative Identity Property, $a \cdot 1 = a$].
We set up the multiplication as follows:
$ \frac{1}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} $
Step 3: Algebraic Simplification
We now perform the multiplication across the numerators and the denominators respectively:
- Numerator Calculation: Multiply the numerators together.
$ 1 \times \sqrt{7} = \sqrt{7} $ - Denominator Calculation: Multiply the denominators together.
$ \sqrt{7} \times \sqrt{7} = \sqrt{7 \times 7} = \sqrt{49} = 7 $
Combining the simplified numerator and denominator yields the final rationalised expression:
$ \frac{\sqrt{7}}{7} $
Notice that the denominator is now $7$, which is a rational integer, fulfilling the objective of the problem.
Final Solution: The rationalised form of $\frac{1}{\sqrt{7}}$ is $\frac{\sqrt{7}}{7}$.
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(iii): Classify the following numbers as rational or irrational: (iii) $\frac{2\sqrt{7}}{7\sqrt{7}}$
- 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(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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