Concept-wise Practice

sqrt2 contradiction MCQ Questions for Class 10

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Practice Questions

5 questions tagged with sqrt2 contradiction.

Question 1/5 Medium Mathematics Chapter 1: Real Numbers 6: Proof of irrationality of √2, √3, √5 Class 10 Level 18

\(\sqrt{2}\) की सिद्धि में (p) और (q) दोनों सम मिलना किस कथन को असत्य बनाता है?

In the proof of \(\sqrt{2}\), both (p) and (q) being even makes which statement false?

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Correct Answer

A. (p) और (q) सहअभाज्य हैं(p) and (q) are coprime

Step 1

Concept

If both are even, both have common factor (2).

Step 2

Why this answer is correct

Coprime numbers should not have any common factor other than (1).

Step 3

Exam Tip

So the statement that they are coprime becomes false. चरण 1: दोनों सम होने पर दोनों में (2) साझा गुणनखंड होगा। चरण 2: सहअभाज्य संख्याओं में (1) के अलावा कोई साझा गुणनखंड नहीं होना चाहिए। चरण 3: इसलिए सहअभाज्य होने की बात असत्य हो जाती है।

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

\(\sqrt{2}\) की सिद्धि में (p) और (q) दोनों सम मिलते हैं। इसे विरोधाभास क्यों कहा जाता है?

In the proof of \(\sqrt{2}\), both (p) and (q) are found even. Why is this called a contradiction?

Explanation opens after your attempt
Correct Answer

A. क्योंकि दोनों में (2) साझा गुणनखंड है, जबकि वे सहअभाज्य माने गए थेBecause both have common factor (2), while they were assumed coprime

Step 1

Concept

An even number is divisible by (2).

Step 2

Why this answer is correct

If both are even, (2) is a common factor.

Step 3

Exam Tip

Coprime numbers cannot have such a common factor. चरण 1: सम संख्या (2) से विभाज्य होती है। चरण 2: दोनों सम होने पर (2) साझा गुणनखंड है। चरण 3: सहअभाज्य संख्याओं में ऐसा साझा गुणनखंड नहीं हो सकता।

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

एक प्रमाण में \(p^2=2q^2\) से (p=2r) और फिर (q=2s) मिला। यह कौन सा विरोधाभास देता है?

In a proof, from \(p^2=2q^2\), we get (p=2r) and then (q=2s). What contradiction does this give?

Explanation opens after your attempt
Correct Answer

A. (p) और (q) सहअभाज्य नहीं हैं(p) and (q) are not coprime

Step 1

Concept

(p=2r) and (q=2s) mean both are divisible by (2).

Step 2

Why this answer is correct

So they cannot be coprime.

Step 3

Exam Tip

But they were assumed coprime at the start, which is the contradiction. चरण 1: (p=2r) और (q=2s) से दोनों (2) से विभाज्य हैं। चरण 2: इसलिए वे सहअभाज्य नहीं हो सकते। चरण 3: जबकि शुरुआत में उन्हें सहअभाज्य माना गया था, यही विरोधाभास है।

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Question 4/5 Medium Mathematics Chapter 1: Real Numbers 6: Proof of irrationality of √2, √3, √5 Class 10 Level 16

यदि \(\sqrt{2}=\frac{p}{q}\) और (p), (q) सहअभाज्य हैं, तो (p) और (q) दोनों सम सिद्ध होना किससे टकराता है?

If \(\sqrt{2}=\frac{p}{q}\) and (p), (q) are coprime, proving both (p) and (q) even contradicts what?

Explanation opens after your attempt
Correct Answer

A. (p) और (q) के सहअभाज्य होने सेThe coprime nature of (p) and (q)

Step 1

Concept

Coprime means there is no common factor except (1).

Step 2

Why this answer is correct

If both are even, (2) is a common factor.

Step 3

Exam Tip

Therefore it directly contradicts their being coprime. चरण 1: सहअभाज्य होने का अर्थ है कि साझा गुणनखंड (1) के अलावा नहीं है। चरण 2: दोनों सम होने पर (2) साझा गुणनखंड होगा। चरण 3: इसलिए यह सीधे सहअभाज्य होने से टकराता है।

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

कौन सा विकल्प \(\sqrt{2}\) के प्रमाण में विरोधाभास को सही भाषा में बताता है?

Which option states the contradiction in the proof of \(\sqrt{2}\) correctly?

Explanation opens after your attempt
Correct Answer

A. (a) और (b) सहअभाज्य माने गए थे, पर दोनों सम निकले(a) and (b) were assumed coprime, but both turned out even

Step 1

Concept

At the beginning, \(\frac{a}{b}\) is taken in lowest form.

Step 2

Why this answer is correct

The proof shows both (a) and (b) are even.

Step 3

Exam Tip

Being coprime and both even is impossible. चरण 1: शुरुआत में \(\frac{a}{b}\) को सरलतम रूप मानते हैं। चरण 2: प्रमाण में (a) और (b) दोनों सम मिलते हैं। चरण 3: सहअभाज्य होकर दोनों सम होना असंभव है।

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