Which option gives the correct short order of the proof of √3?
Answer and explanation
Correct answer: Assume rational, then square, then obtain the contradiction that both are divisible by 3
The governing concept is the contradiction proof that √3 is irrational. Begin by assuming √3 is rational, so √3 = a/b where a and b are coprime integers and b is non-zero. Squaring gives 3b² = a². This shows that 3 divides a, so write a = 3k; substitution then shows that 3 also divides b. That contradicts the assumption that a and b have no common factor. Therefore the sequence in option B is the correct short order. A diagram, decimal approximation, zero assumption or subtraction does not establish irrationality rigorously.
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What is the correct answer to this question?
Assume rational, then square, then obtain the contradiction that both are divisible by 3
Why is this the correct answer?
The governing concept is the contradiction proof that √3 is irrational. Begin by assuming √3 is rational, so √3 = a/b where a and b are coprime integers and b is non-zero. Squaring gives 3b² = a². This shows that 3 divides a, so write a = 3k; substitution then shows that 3 also divides b. That contradicts the assumption that a and b have no common factor. Therefore the sequence in option B is the correct short order. A diagram, decimal approximation, zero assumption or subtraction does not establish irrationality rigorously.
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Number Systems. Topic: Proof of irrationality of square root 2 and square root 3.
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