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Q1(i):
Classify the following numbers as rational or irrational:
(i) $2 - \sqrt{5}$
Solution :
Step 1: Analyzing the Components of the Expression
We are given the mathematical expression $2 - \sqrt{5}$. To classify this number, we must first analyze the individual terms that constitute the expression:
- The number $2$: This is an integer. By definition, any integer can be expressed in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$ (e.g., $\frac{2}{1}$). Therefore, $2$ is a rational number.
- The number $\sqrt{5}$: The number $5$ is a prime number and not a perfect square. The square root of any non-perfect square yields a non-terminating, non-repeating decimal. Therefore, $\sqrt{5}$ is an irrational number.
Step 2: Applying the Properties of Real Numbers
In the real number system, specific closure properties govern the arithmetic operations between rational and irrational numbers. The relevant theorem states:
[Theorem: The sum or difference of a non-zero rational number and an irrational number is always an irrational number.]
Since we are subtracting an irrational number ($\sqrt{5}$) from a rational number ($2$), the resulting difference must be irrational.
Step 3: Formal Proof by Contradiction
To rigorously prove that $2 - \sqrt{5}$ is irrational, we utilize a proof by contradiction.
Assumption: Let us assume the contrary, that $2 - \sqrt{5}$ is a rational number. Let this rational number be represented by $r$.
$2 - \sqrt{5} = r$
Rearranging the equation to isolate $\sqrt{5}$:
$2 - r = \sqrt{5}$
Logical Deduction:
- We know that $2$ is a rational number.
- By our assumption, $r$ is a rational number.
- [Per the Closure Property of Rational Numbers under Subtraction], the difference between two rational numbers ($2 - r$) must also be a rational number.
- This implies that $\sqrt{5}$ is equal to a rational number.
Contradiction: This contradicts the universally established mathematical fact that $\sqrt{5}$ is an irrational number. Therefore, our initial assumption that $2 - \sqrt{5}$ is rational must be false.
Step 4: Decimal Expansion Analysis (Verification)
We can also verify this classification by examining the decimal expansion of the expression:
The decimal expansion of $\sqrt{5}$ is non-terminating and non-repeating:
$\sqrt{5} \approx 2.236067977...$
Substituting this into our expression:
$2 - \sqrt{5} = 2 - 2.236067977...$
$2 - \sqrt{5} = -0.236067977...$
The resulting decimal is also non-terminating and non-repeating, which is the definitive characteristic of an irrational number.
Visual Representation: Number Line Mapping
The following geometric representation demonstrates the subtraction of the irrational length $\sqrt{5}$ from the rational coordinate $2$, landing on the irrational coordinate $2 - \sqrt{5}$.
Final Solution: The expression $2 - \sqrt{5}$ is an irrational number.
More Questions from Class 9 Mathematics Number Systems EXERCISE 1.4
- 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(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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