**Title: Proof by Contraposition****Text:**Using a proof by contraposition, prove that for any integer \( n \), if \( n^2 \) is even then \( n \) is even.**Explanation:**To prove this statement by contraposition, we need to show that the contrapositive is true. The original statement is: "If \( n^2 \) is even, then \( n \) is even." The contrapositive of this statement is: "If \( n \) is odd, then \( n^2 \) is odd."1. Assume \( n \) is odd. This means \( n = 2k + 1 \) for some integer \( k \), where \( 2k + 1 \) is the standard form of an odd number.2. Calculate \( n^2 \): \[ n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1 \]3. Since \( n^2 = 2(2k^2 + 2k) + 1 \), it is of the form \( 2m + 1 \) for some integer \( m \), which is the standard form of an odd number.4. Therefore, if \( n \) is odd, then \( n^2 \) is odd.Since the contrapositive is proven to be true, the original statement is also true by contraposition.

Name: **Title: Proof by Contraposition****Text:**Using a proof by contrapos
Uploaded: 2024-03-20T06:38:44.000Z
Channel: Bartleby
Description: **Title: Proof by Contraposition****Text:**Using a proof by contraposition, prove that for any integer \( n \), if \( n^2 \) is even then \( n \) is even.**Explanation:**To prove this statement by contraposition, we need to show that the contraposit