The difference between an integer and its additive inverse is always even.
A. True
B. False

Hint: For solving this question you should know about the additive integer and the definition of integer. In this problem we will first find the definitions of integers and its additive inverse and then find the difference between both.

Complete step by step answer:
Integer: A number which comes under the category of real numbers which cannot have any fraction part (that is, we can write that number without any fraction or decimal) is called as an integer. For example: $-1,0,2,3$ are integers whereas $0.5,\sqrt{2},\dfrac{3}{4}$ are not integers.
Additive inverse: In mathematics the additive inverse of a number is the number that when added to the number yields 0 as the result. The number is also known as the opposite, sign change, and negation of a particular number. For a real number reversing the sign automatically generates the additive inverse of that number. For example: We have a real number. Let $p$ be a number which when added with $a$ yields 0.
$\begin{align}
  & \Rightarrow a+p=0 \\
 & \Rightarrow p=-a \\
\end{align}$
Then $p$ is the additive inverse of $a$ and its value is $-a$.
Now let us take an integer as $k$.
So the additive inverse for this will be $=-k$.
Now, by the subtraction of these,
$k-\left( -k \right)=k+k=2k$
And this is an even number.

So, the correct answer is “Option A”.

Note: Be careful while writing the definition of additive inverse. In simple words an additive inverse of any number only has an opposite sign by which adding with the given number we get 0 as result.