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Number systems — MATHS STD 9 — Question

Rajasthan BoardEnglish MediumSTD 9MATHSNumber systems1 Mark
Question
Consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:
  1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
  2. Both Assertion (A) and Reason (R) are true but Reason is not a correct explanation of Assertion (A).
  3. Assertion (A) is true and Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is true.
Assertion (A) Reason (R)
$\sqrt{3}$ is an irrational number. The sum of rational number and an irrational number is an irrational number.
The correct answer is: (a), (b), (c), (d).
✓

Answer

  1. Both Assertion (A) and Reason (R) are true but Reason is not a correct explanation of Assertion (A).
    Solution:
    If possible, let $\sqrt{3}$ be a rational number and its simplest form is $\frac{\text{a}}{\text{b}}.$
    Then, $\sqrt{3}=\frac{\text{a}}{\text{b}}\Rightarrow\frac{\text{a}^2}{\text{b}^2}=3\Rightarrow\frac{\text{a}^2}{\text{b}}=3\text{b}$
    Clearly, 3b is an integer and $\frac{\text{a}^2}{\text{b}}$ is not an integer since (a, b) = 1
    Thus, we arrive at a contradiction
    So, our supposition is wrong
    Hence, $\sqrt{3}$ is an irrational number
    So, the Assertion (A) is true.
    If possible, let the sum of a rational number a and an irrational number $\sqrt{\text{b}}$ be a rational number
    Then, $\text{a}+\sqrt{\text{b}}=\text{c}\Rightarrow\sqrt{\text{b}}=\text{c}-\text{a}$
    But, the difference of two irrational is a rational
    So, (c - a) is rational and thus, $\sqrt{\text{b}}$ is rational
    Thus, we arrive at a contradiction
    So, our supposition is wrong
    Hence, the sum of a rational and an irrational is irrational
    So, the reason (R) is true.
    Hence, the Assertion (A) and Reason (R) are true but Reason is not a correct explanation of Assertion (A).

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