Question

Use the method of direct proof to prove the following statements. If \(n \in \mathbb{Z},\) then \(5 n^{2}+3 n+7\) is odd. (Try cases.)

Short answer

Confirmed. The statement 'If \(n\) is an integer, then \(5n^{2}+3n+7\) is odd.' is true.

Step-by-step solution

Recognize cases for an integer

An integer can either be even or odd. An even integer can be expressed as \(2k\), where \(k\) is any integer, and an odd integer can be expressed as \(2m+1\), where \(m\) is any integer.

Case 1 - Assume n is even

Assume \(n\) is an even integer, i.e., \(n = 2k\). Substitute \(n = 2k\) into the expression \(5n^{2}+3n+7\) to obtain \(5(2k)^{2}+3(2k)+7 = 20k^{2}+6k+7\). No matter the values of \(k\), \(20k^{2}+6k\) will always yield an even integer. Adding an odd number (7) to an even number will always yield an odd integer.

Case 2 - Assume n is odd

Assume \(n\) is an odd integer, i.e., \(n = 2m+1\). Substitute \(n = 2m+1\) into the expression \(5n^{2}+3n+7\) to obtain \(5(2m+1)^{2}+3(2m+1)+7 = 20m^{2}+20m+5+6m+3+7 = 20m^{2}+26m+15\). Regardless of the values of \(m\), \(20m^{2}+26m\) will always yield an even number, and adding an odd number (15) to an even number will always yield an odd integer.

Conclusion of the proof

Since the expression \(5n^{2}+3n+7\) results in an odd integer no matter if \(n\) is even or odd (the only possibilities for an integer), it can be concluded that the statement 'If \(n\) is an integer, then \(5n^{2}+3n+7\) is odd.' is true.