Prove: if $x$ is even, then $x + 5$ is odd. I am trying to prove or disprove that if $x$ is even, then $x + 5$ is odd.
This is what I have thus far, but I am stuck:


*

*Assume that the chose variable (x) are in the domain:
(x) is an integer

*Assume the IF part of the statement:
(x) is even

*Prove the THEN part: 
Let a be a dummy variable that we assume is an integer.
$x=2a+1$ (Definition of an Odd Integer)
Now find $x+5$.
$x+5=2a+1+5$ (Because we added the same value to both sides of the equation)
$x+5=2a+6$ 
$x+5=2(a+3)$ (Because of the Distributive Property of Multiplication over Addition)
Let $z=a+3$;
$x + 5 = 2z,$ but this is the definition of an even integer; where did I mess up?
 A: Your first line of proof does not agree with your assumption that x is even:
Assume the IF part of the statement: (x) is even
x=2a+1 (Definition of an Odd Integer)

A: given $x$ is even so $x=2k$ where you took $x=2k+1$ 
A: In general when faced with "even" or "odd" assumptions in proof for some variable $x \in \mathbb{Z}$, it's helpful to express $x$ in terms of some other integer $k \in \mathbb{Z}$.
In your case, if $x$ is even, then there must exist some integer- we'll call it $k$- such that $x=2k$. We don't know exactly what value $k$ is, and frankly, we don't need to know. 
Since $x=2k$, and we want to show that $x+5$ is odd, simply add 5 to the first value:
$$x+5= 2k+5$$
$$=2k+4+1$$
$$=2(k+2)+1$$
In general, we know that an odd number $m$ is one where for some integer $n$, $m=2n+1.$
Note that regardless of what $k$ is, $k+2$ will maintain its evenness or oddness (depending on $k$'s value); however, $2(k+2)$ will always be even. 
And thus, $2(k+2)+1$ is odd.
A: Notice, since $x$ is even then it can be rewritten as $x=2m$ 
Where, $m$ is some integer 
Now, we have $$x+5=2k+5=2k+4+1$$$$=2(k+2)+1=\color{blue}{2n+1}$$
Where, $n=k+2$ is an integer 
Thus, we get $\color{red}{x+5}=\color{blue}{2n+1}$ which is an odd integer i.e. $\color{red}{x+5}$ is not even.
