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>
First you need to know some rules 2 numbers that are the same with coefficients divided : you differentiate those coeff and multyplied :you sum up them
The difference between 79 and 22 is 57. 79 is THE number. All you have to do is add 22 and 57 because the answer you get would ALWAYS have one of the numbers added as the difference.