The sum of a rational number and an even integer is rational. a) always true b) sometimes true c) usually true d) never true

I think that the sum will always be a rational number
let's prove that

any rational number can be represented as a/b where a and b are integers and b≠0

and an integer is the counting numbers plus their negatives and 0
so like -4,-3,-2,-1,0,1,2,3,4

so, 2 rational numbers can be represented as

a/b and c/d (where a,b,c,d are all integers and b≠0 and d≠0)

their sum is
a/b+c/d=
ad/bd+bc/bd=
(ad+bc)/bd

1. the numerator and denominator will be integers
2. that the denominator does not equal 0

alright
1.
we started with that they are all integers
ab+bc=?
if we multiply any 2 integers, we get an integer
like 3*4=12 or -3*4=-12 or -3*-4=12, etc.
even 0*4=0, that's an integer
the sum of any 2 integers is an integer
like 4+3=7, 3+(-4)=-1, 3+0=3, etc.
so we have established that the numerator is an integer

now the denominator
that is just a product of 2 integers so it is an integer

2. we originally defined that b≠0 and d≠0 so we're good

therefore, the sum of any 2 rational numbers will always be a rational number is the correct answer.

The answer is a good luck

Logical equivalent of the give question is its contrapositive