Question:
Assertion (A) : (\(\sqrt{3}\) + \(\sqrt{5}\)) is an irrational number.
Reason (R) : Sum of the any two irrational numbers is always irrational.
Assertion (A) : (\(\sqrt{3}\) + \(\sqrt{5}\)) is an irrational number.
Reason (R) : Sum of the any two irrational numbers is always irrational.
Reason (R) : Sum of the any two irrational numbers is always irrational.
Show Hint
The sum, difference, product, or quotient of two irrational numbers is not always irrational.
Keep simple counterexamples like \((2 + \sqrt{3})\) and \((2 - \sqrt{3})\) in mind to quickly verify these properties during exams!
Keep simple counterexamples like \((2 + \sqrt{3})\) and \((2 - \sqrt{3})\) in mind to quickly verify these properties during exams!
Updated On: Jul 22, 2026
- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
The topic of this question is Real Numbers.
We need to evaluate the mathematical truth of two statements: the Assertion (A) about the irrationality of the sum of two square roots of prime numbers, and the Reason (R) about a general property of operations on irrational numbers.
Step 2: Key Formula or Approach:
- An irrational number is a real number that cannot be expressed as a simple fraction of two integers.
- We can prove the irrationality of \(\sqrt{3} + \sqrt{5}\) using a proof by contradiction.
- To evaluate the truth of the Reason, we will search for a counterexample that shows the sum of two irrational numbers can be rational.
Step 3: Detailed Explanation:
• Let us analyze Assertion (A):
Assume for contradiction that \(\sqrt{3} + \sqrt{5} = x\) is a rational number.
Rearrange the equation:
\[ x - \sqrt{3} = \sqrt{5} \]
Square both sides:
\[ (x - \sqrt{3})^2 = (\sqrt{5})^2 \]
\[ x^2 + 3 - 2x\sqrt{3} = 5 \]
Rearrange terms to isolate \(\sqrt{3}\):
\[ x^2 - 2 = 2x\sqrt{3} \]
\[ \sqrt{3} = \frac{x^2 - 2}{2x} \]
Since \(x\) is rational, the right hand side \(\frac{x^2 - 2}{2x}\) must be a rational number.
However, we know that \(\sqrt{3}\) is irrational.
This is a contradiction, which means our initial assumption is false.
Therefore, \(\sqrt{3} + \sqrt{5}\) is irrational. Assertion (A) is true.
• Let us analyze Reason (R):
The statement claims that the "Sum of any two irrational numbers is always irrational."
Let us test this claim with a counterexample:
Let the first irrational number be \(a = \sqrt{3}\).
Let the second irrational number be \(b = -\sqrt{3}\).
Their sum is:
\[ a + b = \sqrt{3} + (-\sqrt{3}) = 0 \]
Since 0 is a rational number, the sum of these two irrational numbers is rational.
Thus, Reason (R) is false.
Step 4: Final Answer:
Assertion (A) is true, but Reason (R) is false.
Therefore, the correct option is (C).
The topic of this question is Real Numbers.
We need to evaluate the mathematical truth of two statements: the Assertion (A) about the irrationality of the sum of two square roots of prime numbers, and the Reason (R) about a general property of operations on irrational numbers.
Step 2: Key Formula or Approach:
- An irrational number is a real number that cannot be expressed as a simple fraction of two integers.
- We can prove the irrationality of \(\sqrt{3} + \sqrt{5}\) using a proof by contradiction.
- To evaluate the truth of the Reason, we will search for a counterexample that shows the sum of two irrational numbers can be rational.
Step 3: Detailed Explanation:
• Let us analyze Assertion (A):
Assume for contradiction that \(\sqrt{3} + \sqrt{5} = x\) is a rational number.
Rearrange the equation:
\[ x - \sqrt{3} = \sqrt{5} \]
Square both sides:
\[ (x - \sqrt{3})^2 = (\sqrt{5})^2 \]
\[ x^2 + 3 - 2x\sqrt{3} = 5 \]
Rearrange terms to isolate \(\sqrt{3}\):
\[ x^2 - 2 = 2x\sqrt{3} \]
\[ \sqrt{3} = \frac{x^2 - 2}{2x} \]
Since \(x\) is rational, the right hand side \(\frac{x^2 - 2}{2x}\) must be a rational number.
However, we know that \(\sqrt{3}\) is irrational.
This is a contradiction, which means our initial assumption is false.
Therefore, \(\sqrt{3} + \sqrt{5}\) is irrational. Assertion (A) is true.
• Let us analyze Reason (R):
The statement claims that the "Sum of any two irrational numbers is always irrational."
Let us test this claim with a counterexample:
Let the first irrational number be \(a = \sqrt{3}\).
Let the second irrational number be \(b = -\sqrt{3}\).
Their sum is:
\[ a + b = \sqrt{3} + (-\sqrt{3}) = 0 \]
Since 0 is a rational number, the sum of these two irrational numbers is rational.
Thus, Reason (R) is false.
Step 4: Final Answer:
Assertion (A) is true, but Reason (R) is false.
Therefore, the correct option is (C).
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