I think that the sum will always be a rational number let's prove that
<span>any rational number can be represented as a/b where a and b are integers and b≠0
</span>and an integer is the counting numbers plus their negatives and 0 so like -4,-3,-2,-1,0,1,2,3,4....
<span>so, 2 rational numbers can be represented as
</span>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 <span>like 3*4=12 or -3*4=-12 or -3*-4=12, etc. </span>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
<span>2. we originally defined that b≠0 and d≠0 so we're good
</span>therefore, the sum of any 2 rational numbers will always be a rational number <span>is the correct answer.</span>
You can simply make 200/5 = 40 5 minute calls per month.
This works because every call is a 5-minute call. So then we just have to find how many times we can make a 5-minute call up to 200 minutes. We already have an operation for this: Division. This acts as the opposite of multiplication and speeds our process up.