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Q1(iv):
Classify the following numbers as rational or irrational:
(iv) $\frac{1}{\sqrt{2}}$
Solution :
Initial Setup & Theoretical Foundation
We are tasked with classifying the real number $\frac{1}{\sqrt{2}}$ as either rational or irrational. To do this rigorously, we must rely on the fundamental definitions and theorems governing the real number system.
- Rational Number: A number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.
- Irrational Number: A real number that cannot be expressed as a simple fraction of two integers. Its decimal expansion is non-terminating and non-repeating.
Step 1: Analyzing the Components of the Expression
The given expression is a fraction where:
- The numerator is $1$, which is a non-zero rational number (since it can be written as $\frac{1}{1}$).
- The denominator is $\sqrt{2}$. [Per the fundamental theorem of arithmetic and the properties of square roots, the square root of any prime number is an irrational number. Since $2$ is a prime number, $\sqrt{2}$ is irrational.]
Step 2: Applying the Quotient Theorem of Real Numbers
We apply the established theorem regarding the arithmetic operations between rational and irrational numbers:
Theorem: The quotient of a non-zero rational number and an irrational number is always an irrational number.
Let $r = 1$ (a non-zero rational number) and $s = \sqrt{2}$ (an irrational number). The quotient $\frac{r}{s} = \frac{1}{\sqrt{2}}$ must, by definition, be irrational.
Step 3: Alternative Proof via Rationalization
To provide exhaustive proof, we can also manipulate the expression algebraically by rationalizing the denominator. This transforms the expression into a product, allowing us to apply the product theorem.
Multiply both the numerator and the denominator by $\sqrt{2}$:
$ \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2} $
This can be rewritten as a product:
$ \frac{\sqrt{2}}{2} = \frac{1}{2} \times \sqrt{2} $
Here, we have the product of $\frac{1}{2}$ (a non-zero rational number) and $\sqrt{2}$ (an irrational number). [Per the Product Theorem of Real Numbers: The product of a non-zero rational number and an irrational number is always irrational.] Therefore, the result is irrational.
Geometric Visualization of $\frac{1}{\sqrt{2}}$
To understand this number spatially, we can construct a right-angled isosceles triangle where the hypotenuse is exactly $1$ unit in length. By the Pythagorean theorem ($a^2 + b^2 = c^2$), the lengths of the two equal legs will be exactly $\frac{1}{\sqrt{2}}$. Because the hypotenuse is a rational integer ($1$), the legs represent an incommensurable (irrational) magnitude.
Final Conclusion
Whether analyzed through the quotient of a rational and irrational number, or by rationalizing the denominator to form a product, the mathematical logic strictly dictates that the resulting value cannot be expressed as a simple integer fraction.
Final Solution: The number $\frac{1}{\sqrt{2}}$ is an irrational 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(iii): Classify the following numbers as rational or irrational: (iii) $\frac{2\sqrt{7}}{7\sqrt{7}}$
- 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
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