Asked by anna
It has been demonstrated that square root of, 6, end square root
6
is between 2.44 and 2.45. If we were to continue this process, it would
because square root of, 6, end square root
6
is
All Answers 1
Answered by
GPT-5 mini
AI
Complete sentence: "If we were to continue this process, it would go on forever because √6 is irrational."
Short proof: Suppose √6 = a/b in lowest terms (integers a,b with gcd(a,b)=1). Squaring gives 6b^2 = a^2. Then 3 divides a^2, so 3 divides a; write a = 3k. Substituting: 6b^2 = 9k^2, so 2b^2 = 3k^2. Hence 3 divides the left side, so 3 divides b^2 and therefore b. Thus both a and b are divisible by 3, contradicting gcd(a,b)=1. So √6 is not rational; its decimal expansion never terminates or repeats, and the narrowing process can continue without end.
Short proof: Suppose √6 = a/b in lowest terms (integers a,b with gcd(a,b)=1). Squaring gives 6b^2 = a^2. Then 3 divides a^2, so 3 divides a; write a = 3k. Substituting: 6b^2 = 9k^2, so 2b^2 = 3k^2. Hence 3 divides the left side, so 3 divides b^2 and therefore b. Thus both a and b are divisible by 3, contradicting gcd(a,b)=1. So √6 is not rational; its decimal expansion never terminates or repeats, and the narrowing process can continue without end.
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.