Which option identifies an invalid shortcut in the proof of √2?
Answer and explanation
Correct answer: c² = 2d² implies c = 2d
The governing concept is the parity argument used in the proof that √2 is irrational. Starting from c²=2d², the right side is even, so c² is even; an integer whose square is even must itself be even. Writing c=2u and substituting gives 4u²=2d², hence d²=2u², so d is also even. If c/d was assumed to be in lowest terms, both numerator and denominator being even contradicts that assumption. However, c²=2d² does not imply c=2d. Equality of squares with a factor of 2 does not permit taking such a direct linear conclusion. Therefore option C is the invalid shortcut; A, B, and D represent valid steps or conclusions in the proof.
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What is the correct answer to this question?
c² = 2d² implies c = 2d
Why is this the correct answer?
The governing concept is the parity argument used in the proof that √2 is irrational. Starting from c²=2d², the right side is even, so c² is even; an integer whose square is even must itself be even. Writing c=2u and substituting gives 4u²=2d², hence d²=2u², so d is also even. If c/d was assumed to be in lowest terms, both numerator and denominator being even contradicts that assumption. However, c²=2d² does not imply c=2d. Equality of squares with a factor of 2 does not permit taking such a direct linear conclusion. Therefore option C is the invalid shortcut; A, B, and D represent valid steps or conclusions in the proof.
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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