Answer:
A proof can be as follows:
Step-by-step explanation:
Remember that an odd interger is of the form [tex]2p+1[/tex] where [tex]p[/tex] is a integer and remember that two consecutive integer are two numbers of the form [tex]p, p+1[/tex]
[tex](\Rightarrow)[/tex] Suppose the [tex]n[/tex] is an odd integer.
Then [tex]n-1[/tex] must be an even integer and hence divisible by 2. Then we define
[tex]p=\dfrac{n-1}{2}\\q=\dfrac{n-1}{2}+1[/tex]
Then we have that
[tex]p+q=\dfrac{n-1}{2}+\dfrac{n-1}{2}+1=\frac{(n-1)+(n-1)}{2}+1=\frac{2(n-1)}{2}+1=n-1+1=n[/tex]
The converse is as follows:
[tex](\Leftarrow)[/tex] Let [tex]p[/tex] an integer, then[tex]p,p+1[/tex] are two consecutive integers. Then
[tex]n=p+(p+1)=2p+1[/tex] is an odd integer.
