Which option helps show that the claim (\sqrt{2}+\sqrt{3}+\sqrt{6}) is rational is false?
Answer and explanation
Correct answer: Roots of different non-perfect squares give independent irrational parts
Step 1: (\sqrt{2}), (\sqrt{3}), and (\sqrt{6}) are linked to different non-perfect squares. Step 2: Their irrational parts do not cancel through ordinary addition, so the sum is not rational. Step 3: Avoid false identities such as (\sqrt{a+b}=\sqrt{a}+\sqrt{b}).
Frequently asked questions
What is the correct answer to this question?
Roots of different non-perfect squares give independent irrational parts
Why is this the correct answer?
Step 1: (\sqrt{2}), (\sqrt{3}), and (\sqrt{6}) are linked to different non-perfect squares. Step 2: Their irrational parts do not cancel through ordinary addition, so the sum is not rational. Step 3: Avoid false identities such as (\sqrt{a+b}=\sqrt{a}+\sqrt{b}).
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Real Numbers. Topic: Irrational numbers.
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