verdelite 写了： 2022年 12月 23日 21:22
谁有时间可以看看openAI证明得对不对？
me: Let S be the set of all positive integers that can be expressed as the sum of two or more consecutive positive integers. For example, 15 can be expressed as the sum of 3 + 4 + 5, so 15 is an element of S. Prove that a natural number n is not in S if and only if n = 2^i for some non-negative integer i.
chatGPT: First, let's prove that if n is in S, then n cannot be equal to 2^i for any non-negative integer i. Suppose for the sake of contradiction that n is in S and n = 2^i for some non-negative integer i. This means that n can be expressed as the sum of two or more consecutive positive integers. Let's consider the smallest such sum, which must have at least two terms.
Without loss of generality, let the smallest term in the sum be x, and let the second smallest term be x+1. Then the sum of the consecutive integers from x to x+k for some positive integer k is equal to n. This can be expressed as:
x + (x+1) + (x+2) + ... + (x+k) = n
If we simplify this equation, we get:
kx + (1+2+...+k) = n
kx + k(k+1)/2 = n
If we solve for k in this equation, we get:
k = [2(n - (k(k+1)/2))]/k
Since k is a positive integer, it follows that 2(n - (k(k+1)/2)) is a multiple of k. In other words, n - (k(k+1)/2) is a multiple of k/2. However, since n is equal to 2^i for some non-negative integer i, n is an even number, so n - (k(k+1)/2) must also be an even number. Therefore, k/2 must be an integer. But since k is a positive integer, it follows that k must be even.
Now, let's consider the sum x + (x+1) + ... + (x+k). This sum can be expressed as:
(x + (x+k)) + (x+1 + (x+k-1)) + ... + (x+k/2 + (x+k/2))
This simplifies to:
(2x + k) + (2(x+1) + k-2) + ... + (2(x+k/2-1) + 2) + (2(x+k/2) + 0)
If we simplify this further, we get:
(k+1)x + k(k+1)/2 = n
But we already know that n = kx + k(k+1)/2, so we have arrived at a contradiction. Therefore, if n is in S, then n cannot be equal to 2^i for any non-negative integer i.
Now, let's prove the converse: if n is not equal to 2^i for any non-negative integer i, then n is not in S. Suppose for the sake of contradiction that n is not equal to 2^i for any non-negative integer i, but n is in S. Then n can be expressed as the sum of two or more consecutive positive integers.
Without loss of generality, let the smallest term in the sum be x, and let the second smallest term be x+1. Then the sum of the consecutive integers from x to x+k for some positive integer k is equal to n. This can be expressed as:
x + (x+1) + (x+2) + ... + (x+k) = n
If we simplify this equation, we get:
kx + (1+2+...+
me: it is not done yet. please continue.
chatGPT: Sure, I apologize for the incomplete proof. Here is the rest of the proof:
kx + (1+2+...+k) = n
kx + k(k+1)/2 = n
If we solve for k in this equation, we get:
k = [2(n - (k(k+1)/2))]/k
Since k is a positive integer, it follows that 2(n - (k(k+1)/2)) is a multiple of k. In other words, n - (k(k+1)/2) is a multiple of k/2.
We know that n is not equal to 2^i for any non-negative integer i, so n cannot be an even number. Therefore, n - (k(k+1)/2) must also be an odd number. However, since k/2 is an integer, it follows that k must be even. This means that k/2 is also an integer.
Now, let's consider the sum x + (x+1) + ... + (x+k). This sum can be expressed as:
(x + (x+k)) + (x+1 + (x+k-1)) + ... + (x+k/2 + (x+k/2))
This simplifies to:
(2x + k) + (2(x+1) + k-2) + ... + (2(x+k/2-1) + 2) + (2(x+k/2) + 0)
If we simplify this further, we get:
(k+1)x + k(k+1)/2 = n
But we already know that n = kx + k(k+1)/2, so we have arrived at a contradiction. Therefore, if n is not equal to 2^i for any non-negative integer i, then n is not in S.
This completes the proof that a natural number n is not in S if and only if n = 2^i for some non-negative integer i.