Question 425253
you are trying to compute: ((a^2-1)/(a-1))-a for any number a in the field of Integers.

well, consider the inner bracket first; (a^2-1)/a-1: the numerator can be factored into: (a+1)(a-1) (expand and confirm this is equal to a^2-1), then our inner bracket can be rewritten as: ((a+1)(a-1))/(a-1). now there's a factor of (a-1) in the numerator and denominator, so they will cancel each other out leaving behind only (a+1). So our inner bracket is just really (a+1), then we have to subtract a from this since we had initially ignored it to handle the inner bracket, leaving behind 1.

In summary, ((a^2-1)/(a-1))-a = [{(a+1)(a-1)}/(a-1)]-a = [a+1]-a = 1 for any number a in the field of Integers