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gcd contradiction MCQ Questions for Class 10

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Practice Questions

2 questions tagged with gcd contradiction.

Question 1/2 Hard Mathematics Chapter 1: Real Numbers 6: Proof of irrationality of √2, √3, √5 Class 10 Level 17

\(\sqrt{2}\) की सिद्धि में यदि (a=2m) और (b=2n), तो यह किस बात के विरुद्ध है?

In the proof of \(\sqrt{2}\), if (a=2m) and (b=2n), what does this go against?

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Correct Answer

A. (\gcd(a,b)=1)

Step 1

Concept

(a=2m) and (b=2n) show common factor (2) in both.

Step 2

Why this answer is correct

So (\gcd(a,b)) cannot be (1).

Step 3

Exam Tip

This goes against the lowest-form condition. चरण 1: (a=2m) और (b=2n) से दोनों में (2) साझा गुणनखंड है। चरण 2: इसलिए (\gcd(a,b)) (1) नहीं हो सकता। चरण 3: यह सरलतम रूप की शर्त के विरुद्ध है।

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Question 2/2 Hard Mathematics Chapter 1: Real Numbers 6: Proof of irrationality of √2, √3, √5 Class 10 Level 16

यदि \(\sqrt{2}\) की सिद्धि में (p) और (q) दोनों सम मिलते हैं, तो यह किस कथन को खंडित करता है?

If both (p) and (q) are found even in the proof of \(\sqrt{2}\), which statement does it refute?

Explanation opens after your attempt
Correct Answer

A. (\gcd(p,q)=1)

Step 1

Concept

If both are even, both (p) and (q) are divisible by (2).

Step 2

Why this answer is correct

Then their greatest common divisor cannot remain (1).

Step 3

Exam Tip

Therefore the condition (\gcd(p,q)=1) is refuted. चरण 1: दोनों सम होने पर (p) और (q) दोनों (2) से विभाज्य हैं। चरण 2: तब उनका महत्तम समापवर्तक (1) नहीं रह सकता। चरण 3: इसलिए (\gcd(p,q)=1) की शर्त खंडित होती है।

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