Answer:
So for the first few small values of n, we have proven by demonstration that f(n) = n / (n+1).
Our task is to prove that if it works for any positive integer value of n, then it works for n + 1. This way, it must by induction work for all subsequent values of n.
Formally said, we need to prove that if for some positive integer n we can show that f(n) = n / (n+1), then we can conclude that f(n+1) = (n + 1) / (n + 2).
We begin the real "proof" by expanding f(n + 1):
f(n + 1) = f(n) + 1 / ((n+1)((n+1)+1)) because that's based on the construction.
= n / (n+1) + 1 / ((n+1)(n+2)) because f(n) = n / (n+1); this is called "using what you know from earlier".
= n(n+2) / ((n+1)(n+2)) + 1 / ((n+1)(n+2)) because we can multiply the left fraction by (n+2)/(n+2).
= (n2 + 2n + 1) / ((n+1)(n+2)) because we have a common denominator and can combine the numerators.
= (n+1)2 / ( (n+1)(n+2)) because we can factor the numerator now; it is a perfect square.
= (n+1) / (n+2) because we can cancel the common (n+1) factor from the numerator and denominator.
Q.E.D. (which means "that which was to be proven", in other words: "voilà")
Step-by-step explanation:
