# Proof by Induction

## 2018 January 13

Unlike the natural sciences, mathematics is built on proofs, not experiments. To find truth, a mathematician would decide that some statements must be true. For example, Euclid's first axiom in his Elements is

Things which are equal to the same thing are also equal to one another.
i.e. a = c and b = c imply a = b.

If this sounds obvious, that's the point. Axioms are inherently boring, but they can be used to build theorems, which are very interesting, using mathematical proofs. Also, conjectures are statements which have not been proven.

There are many techniques for making a mathematical proof, one of which is a proof by induction. I'm reading the phenomenal Abstract Algebra by Judson, which starts by explaining a proof by induction. As always, I focus on intuition on my blog, not rigour, so you should read Judson for rigour.

Assume that you want to prove that some statement S(n) is true for all natural numbers n. Then you could prove that S(1) is true and that, if S(k) is true for some k, then S(k + 1) must be true, which is called the inductive step. The first step implies S(1), and the second implies S(1 + 1) (i.e. S(2)), and then we repeat to see that S(3) is true, and so forth. Clearly, we just proved that S(n) is true for all the natural numbers n, so we proved what was to be proved.

For example, here is a proof I did while reading.

Conjecture
Prove that
$$1^3 + 2^3 + \ldots + n^3 = \frac{n^2(n+1)^2}{4}$$
for all n ∈ ℕ

Proof

We see that the conjecture is true when n = 1, so now we need to prove that, if
$$1^3 + 2^3 + \ldots + n^3 = \frac{n^2(n+1)^2}{4}$$
is true, then so is
$$1^3 + 2^3 + \ldots + n^3 + (n+1)^3 = \frac{(n+1)^2(n+2)^2}{4}$$

We can just do some simple algebra to prove the inductive step.

Assume
$$1^3 + 2^3 + \ldots + n^3 = \frac{n^2(n+1)^2}{4}$$

$$1^3 + 2^3 + \ldots + n^3 + (n+1)^3 = \frac{n^2(n+1)^2}{4} + (n+1)^3$$

$$1^3 + 2^3 + \ldots + n^3 + (n+1)^3 = \frac{n^2(n+1)^2 + 4(n+1)^3}{4}$$

$$1^3 + 2^3 + \ldots + n^3 + (n+1)^3 = \frac{n^2(n+1)^2 + 4(n+1)(n+1)^2}{4}$$

$$1^3 + 2^3 + \ldots + n^3 + (n+1)^3 = \frac{(n^2 + 4(n+1))(n+1)^2}{4}$$

$$1^3 + 2^3 + \ldots + n^3 + (n+1)^3 = \frac{(n^2 + 4n + 4)(n+1)^2}{4}$$

$$1^3 + 2^3 + \ldots + n^3 + (n+1)^3 = \frac{(n + 2)^2(n+1)^2}{4}$$

$$1^3 + 2^3 + \ldots + n^3 + (n+1)^3 = \frac{(n+1)^2(n+2)^2}{4}$$

Which is the inductive step.

Now that we've finished both steps of the proof by induction, we've proven the conjecture for all natural n, so it's a theorem now.

This is obviously a very simple example, but it may be harder that it looks. If you want to try your hand at another, prove that, for all natural n ≥ 4,
n!>2n