Mathematical Induction

Quite often in mathematics we find ourselves wanting to prove a statement that we think is true for every natural number $n$.

For example, you may have met the formula $\frac{1}{6}n(n+1)(2n+1)$ for the sum $$ \sum_{i=1}^n i^2=1^2+2^2+\ldots+n^2. $$ We can try some values of $n$, and see that the formula seems to be right: \begin{eqnarray} n=1: & & \sum_{i=1}^n i^2 = 1^2 = 1;\\ & & \frac{1}{6}n(n+1)(2n+1) = \frac{1}{6}\times 1\times 2\times 3 = 1\\ \\ n=2: & & \sum_{i=1}^n i^2 = 1^2 + 2^2 = 1+4 = 5;\\ & & \frac{1}{6}n(n+1)(2n+1) = \frac{1}{6}\times 2\times 3\times 5 = 5\\ \\ n=3: & & \sum_{i=1}^n i^2 = 1^2 + 2^2 + 3^2 = 1+4+9 = 14;\\ & & \frac{1}{6}n(n+1)(2n+1) = \frac{1}{6}\times 3\times 4\times 7 = 14.\\ \end{eqnarray} But we want to prove that this is true for all positive integers $n$, and it's going to be impossible if we try to do this by putting in all possible values of $n$! Instead, we're going to have to be a bit more cunning$\ldots$

Let's go back to our example from above, about sums of squares, and use induction to prove the result. We know that $P(1)$ is true (we did that before!). Now let's see what would happen if we knew that $P(n)$ is true. Then we'd know that $\sum_{i=1}^n i^2 = \frac{1}{6}n(n+1)(2n+1)$. So what can we say about $\sum_{i=1}^{n+1} i^2$? Well, \begin{eqnarray} \sum_{i=1}^{n+1} i^2 & = & (n+1)^2 + \sum_{i=1}^n i^2\\ & = & (n+1)^2 + \frac{1}{6}n(n+1)(2n+1)\quad\textrm{(by the inductive hypothesis)}\\ & = & \frac{1}{6}(n+1)\left[6(n+1)+n(2n+1)\right]\\ & = & \frac{1}{6}(n+1)\left[6n+6+2n^2+n\right]\\ & = & \frac{1}{6}(n+1)\left[2n^2+7n+6\right]\\ & = & \frac{1}{6}(n+1)(n+2)(2n+3). \end{eqnarray} This looks familiar; what is $P(n+1)$? Well, let's substitute $n+1$ in place of $n$ in our statement of $P(n)$. So we're aiming for $$ \sum_{i=1}^{n+1} i^2 = \frac{1}{6}(n+1)(n+2)(2(n+1)+1)=\frac{1}{6}(n+1)(n+2)(2n+3), $$ which is exactly what we've just got! (Working out what you're aiming for can often give you an idea of how to manipulate the algebra.) So we've shown that if $P(n)$ is true, then $P(n+1)$ is true. Since we also know that $P(1)$ is true, we know that $P(2)$ is true, so $P(3)$ is true, so $P(4)$ is true, so $\ldots$ In other words, we've shown that $P(n)$ is true for all $n\geq 1$, by mathematical induction.

Warning: When encountering induction for the first time, people usually remember to show ``if $P(n)$ is true then $P(n+1)$ is true'', because that's what induction is really about, but a common mistake is to forget to show that $P(1)$ is true. Proof by induction is a two-stage process, even if one stage is usually very easy. The dominoes won't fall over unless you knock over the first one!

Don't forget that your first domino doesn't have to be $P(1)$. It could be $P(2)$, or $P(19)$, or $P(1000000)$. For example, we can use induction to show $3^n> n^3$ for $n\geq 4$ (see the exercises below)

One last thing: induction is only a method of proof . For example, if you're trying to sum a list of numbers and have a guess for the answer, then you may be able to use induction to prove it. But you can't use induction to find the answer in the first place. Also, there are often other methods of proof: I've given some examples below of things that you might like to try to prove by induction, but several of them can be proved at least as easily by other methods (indeed, you've probably seen some proved by other methods).

Don't forget that if you get stuck with these, you can ask for help on Ask NRICH. You can also consult the Hints if you need to.

Vicky finished a degree in Maths at Cambridge last summer and is now doing a fourth year course studying Combinatorics, Number Theory and Algebra, still at Trinity College, Cambridge.