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sqrt5 contradiction MCQ Questions for Class 10

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2 questions tagged with sqrt5 contradiction.

Question 1/2 Easy Mathematics Real Numbers Proof of irrationality of √2, √3, √5 Class 10 Level 18

\(\sqrt{5}\) के प्रमाण में (p) और (q) दोनों (5) से विभाज्य मिलते हैं। यह किस बात के विरुद्ध है?

In the proof of \(\sqrt{5}\), both (p) and (q) are found divisible by (5). This is against what?

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Correct Answer

A. उनके सहअभाज्य होने केTheir being coprime

Step 1

Concept

At the beginning, (p) and (q) were assumed coprime.

Step 2

Why this answer is correct

If both are divisible by (5), then (5) becomes a common factor.

Step 3

Exam Tip

So this goes against their being coprime. चरण 1: (p) और (q) को शुरुआत में सहअभाज्य माना गया था। चरण 2: दोनों (5) से विभाज्य होने पर (5) साझा गुणनखंड बन जाता है। चरण 3: इसलिए यह सहअभाज्य होने के विरुद्ध है।

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Question 2/2 Easy Mathematics Real Numbers Proof of irrationality of √2, √3, √5 Class 10 Level 18

\(\sqrt{5}\) के प्रमाण में विरोधाभास किस बात से आता है?

What creates the contradiction in the proof of \(\sqrt{5}\)?

Explanation opens after your attempt
Correct Answer

B. (p) और (q) दोनों (5) से विभाज्य मिलते हैंBoth (p) and (q) are found divisible by (5)

Step 1

Concept

In the proof of \(\sqrt{5}\), \(p^2=5q^2\) makes (p) divisible by (5).

Step 2

Why this answer is correct

Then (q) is also found divisible by (5).

Step 3

Exam Tip

Having common factor (5) contradicts the coprime condition. चरण 1: \(\sqrt{5}\) के प्रमाण में \(p^2=5q^2\) से (p) (5) से विभाज्य मिलता है। चरण 2: फिर (q) भी (5) से विभाज्य मिलता है। चरण 3: दोनों में (5) साझा गुणनखंड होना सहअभाज्य शर्त से टकराता है।

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