What problem appears when a² = 2b² is viewed through prime factors in the proof of √2?
Answer and explanation
Correct answer: The left side is a perfect square but the right side can have an odd exponent of 2
The correct answer is A. In the prime factorisation of any perfect square, every prime occurs with an even exponent. Thus the left side a² must have an even exponent for the prime 2, as well as for every other prime. On the right side, 2b² contains the factor 2 outside the square b². If the exponent of 2 in b² is 2r, then its exponent in 2b² is 2r + 1, which is odd. The same integer cannot simultaneously have an even and an odd exponent of 2 in its unique prime factorisation. This contradiction is another way to see why the assumed rational representation of √2 is impossible. The other options make false claims about zero, negativity, or decimals.
Frequently asked questions
What is the correct answer to this question?
The left side is a perfect square but the right side can have an odd exponent of 2
Why is this the correct answer?
The correct answer is A. In the prime factorisation of any perfect square, every prime occurs with an even exponent. Thus the left side a² must have an even exponent for the prime 2, as well as for every other prime. On the right side, 2b² contains the factor 2 outside the square b². If the exponent of 2 in b² is 2r, then its exponent in 2b² is 2r + 1, which is odd. The same integer cannot simultaneously have an even and an odd exponent of 2 in its unique prime factorisation. This contradiction is another way to see why the assumed rational representation of √2 is impossible. The other options make false claims about zero, negativity, or decimals.
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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