To prove 1/a(a+1) = 1/a -1/(a+1).
We start from the right side and show it is equal to the expression on the left.
RHS = 1/a -1/(a+1) .
The LCM of the denominators is a(a+1).
Therefore we convert each of the fractions 1/a and 1/(a+1) into equivalent fraction with LCM as the denominator.
So the first term 1/a = 1*(a+1)/a*(a+1) = (a+1)/a(a+1).
The second term 1/(a+1) = 1*a/(a+1)*a = a/(a(a+1).
Therefore 1/a -1/(a+1) = (a+1 -a)/a(a+1) = 1/a(a+1) = LHS.
Thus we proved 1/a -1/(a+1) = 1/a(a+1). Or 1/a(a+1) = 1/a -1/(a+1).
We notice that the denominator of the left side ratio is the least common denominator of 2 irreducible ratios.
We'll suppose that the ratio 1/a(a+1) is the result of addition or subtraction of 2 elementary fractions:
1/a(a+1) = A/a + B/(a+1) (1)
We'll multiply the ratio A/a by (a+1) and we'll multiply the ratio B/(a+1) by a.
1/a(a+1) = [A(a+1) + Ba]/a(a+1)
Since the denominators of both sides are matching, we'll write the numerators, only.
1 = A(a+1) + Ba
We'll remove the brackets:
1 = Aa + A + Ba
We'll factorize by a to the right side:
1 = a(A+B) + A
If the expressions from both sides are equivalent, the correspondent coefficients are equal.
A+B = 0
A = 1
1 + B = 0
B = -1
We'll substitute A and B into the expression (1):
1/a(a+1) = 1/a - 1/(a+1)
We remark that we've obtained the request from enunciation.
The identity has been proved.