Prove that there is a positive integer that equals the sum of the positive integers not exceeding it. Is your proof constructive or nonconstructive?

Question

enlacamig · Accepted Answer

Assume n is a positive integer that equals the sum of the positive integers not exceeding it:  n=n+(n−1)+(n−2)+(n−3)+…+2+1  Using summation formula we get:  n=n(n+1)2⇒2n=n2+n⇒0=n2−n⇒0=n(n−1)  Therefore, n=0 or n=1. But n is a nonzero integer, so it must be n=1.  Result: We prove that there exists such positive integer and it is n=1, so our proof is constructive

nick1337 · Accepted Answer

Let n be a positive integer. We want to show that there exists a positive integer k such that k=∑i=1ni.By the formula for the sum of an arithmetic series, we have:∑i=1ni=n(n+1)2.Now, consider the positive integer k=n(n+1)2. We need to show that this value of k satisfies the required condition.Substituting k into the sum expression, we have:k=∑i=1ni=n(n+1)2.Hence, we have proven that there exists a positive integer k (specifically, k=n(n+1)2) that equals the sum of the positive integers not exceeding it.

Don Sumner · Accepted Answer

Step 1:To prove that there is a positive integer that equals the sum of the positive integers not exceeding it, we can use a nonconstructive proof.Let n be the positive integer we are considering. The sum of the positive integers not exceeding n can be expressed as 1+2+3+…+n, which is the sum of an arithmetic series. Using the formula for the sum of an arithmetic series, we have:n(n+1)2Step 2:Now, we need to show that there exists a positive integer n such that n(n+1)2=n. This can be rewritten as:n(n+1)=2nExpanding the equation, we get:n2+n=2nSimplifying further:n2−n=0Factoring out n, we have:n(n−1)=0This equation holds true for n=0 and n=1. Therefore, there exists a positive integer (n=1) that satisfies the equation.

Vasquez · Accepted Answer

1. Assume n is the positive integer we seek.2. The sum of positive integers not exceeding n can be expressed as ∑k=1nk.3. By using the formula for the sum of an arithmetic series, we can rewrite the sum as n(n+1)2.4. Now, we need to find an n that satisfies the equation n(n+1)2=n.5. Simplifying the equation, we have n2+n2=n.6. Multiplying both sides of the equation by 2, we get n2+n=2n.7. Rearranging the terms, we have n2−n=0.8. Factoring out n, we get n(n−1)=0.9. Therefore, the equation is satisfied when n=0 or n=1.10. However, since we are looking for a positive integer, we discard n=0.11. Thus, the positive integer that equals the sum of the positive integers not exceeding it is n=1.Therefore, the proof is constructive, as we have explicitly found the positive integer that satisfies the given condition, which is n=1.

Prove that there is a positive integer that equals the

Answered question

Answer & Explanation

New Questions in Other