Copyright (C) 2005-2008, David H. Hovemeyer
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Proof by induction is a very useful technique for proving that a property is true for integers 1, 2, ..., n, regardless of how big n is. Since n can be arbitrarily large, we can't try prove each case 1..n individually. Instead of requiring us to prove an arbitrary number of cases, proof by induction lets us prove every case with just two easy steps!
The basis step. Prove that the property holds for some small n, usually 1.
The induction step. Prove that if the property is true for n, then it also holds for n + 1.
It is easy to see how these two steps can be used to "cover" all values of n starting from n=1, up to any arbitrary n.
Example: let's prove that the sum of integers 1..n is (n(n+1)) / 2. In the proof, we will refer to this formula as f(n).
Basis step. Obviously, the sum of the sequence of integers consisting of the integer 1 by itself is 1. f(1) = (1(1+1)) / 2 = 2/2 = 1. So, the formula holds for n=1.
Induction Step. We assume that the formula holds for f(n), and based on this assumption, prove that it also holds for f(n+1). We start out by noting that f(n+1) = ((n+1)(n+2))/2. This is the result we expect for f(n+1).
We apply the induction step by adding (n+1) to f(n). Because f(n) is the sum of the integers 1..n, then adding (n+1) to this sum will obviously result in the sum of the integers 1..n+1.
f(n+1) = f(n) + (n+1)
= (n(n+1))/2 + (n+1) expand f(n)
= (n(n+1))/2 + 2(n+1)/2 multiply (n+1) by 2/2
= (n2+n)/2 + (2n+2)/2 expand terms
= (n2+3n+2)/2 combine terms
= ((n+1)(n+2))/2 factor polynomial
By "plugging in" the formula for f(n), we arrived at the expected result for f(n+1). This proves the induction step.