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Now that you’re familiar with direct proof and proof by contradiction, it’s time to discover a powerful technique of proof by induction.

Aside: do not confuse mathematical induction with inductive or deductive reasoning. Despite the name, mathematical induction is actually a form of deductive reasoning.

Let’s say, we want to prove that some statement \(P\) is true for all positive integers. In other words:

\(P(1)\) is true, \(P(2)\) is true, \(P(3)\) is true… etc.

We could try and prove each one directly or by contradiction, but the infinite number of positive integers makes this task rather grueling. Proof by induction is a sort of generalization that starts with the basis:

Basis: Prove that \(P(1)\) is true.

Then makes one generic step that can be applied indefinitely:

Induction step: Prove that for all \(n\geq1\), the following statement holds: If \(P(n)\) is true, then \(P(n+1)\) is also true.

See what we did there? We’ve devised another problem to solve, and it’s seemingly the same. But if the basis is true, then proving this inductive step will prove the theorem.

To do this, we chose an arbitrary \(n\geq1\) and assume that \(P(n)\) is true. This assumption is called the inductive hypothesis. The tricky part is this: we don’t prove the hypothesis directly, but prove the \(n+1\) version of it.

This is all rather amorphous, so let’s prove a real theorem.

Theorem 1. For all positive integers \(n\), the following is true:

\begin{equation} \label{eq:1} 1 + 2 + 3 + … + n = \frac{n(n+1)}{2} \end{equation}

Proof. Start with the basis when \(n\) is \(1\). Just calculate it:

\[ 1 = \frac{1(1+1)}{2}. \]

This is correct, so, the basis is proven. Now, assume that the theorem is true for any \(n\geq1\):

\begin{equation} \label{eq:2} 1 + 2 + 3 + … + n = \frac{n(n+1)}{2} \end{equation}

In the induction step we have to prove that it’s true for \(n+1\):

\begin{equation} \label{eq:3} 1 + 2 + 3 + … + (n+1) = \frac{(n+1)(n+2)}{2} \end{equation}

Having this equation, we should just try to expand it and prove directly. Since the last member on the left side is \(n+1\), the second last must be \(n\), so:

\[ 1 + 2 + 3 + … + (n + 1) = 1 + 2 + 3 + … + n + (n+1) \]

From our assumption, we know, that

\[ 1 + 2 + 3 + … + n = \frac{n(n+1)}{2}. \]

So, let’s replace it on the right hand side:

\[ 1 + 2 + 3 + … + (n + 1) = \frac{n(n+1)}{2} + (n+1) \]

And then make that addition so that the right hand side is a single fraction:

\[ 1 + 2 + 3 + … + (n + 1) = \frac{n(n+1)}{2} + \frac{2(n+1)}{2} \]

\[ = \frac{n(n+1) + 2(n+1)}{2} \]

\[ = \frac{(n+1)(n+2)}{2}. \]

Done, we have proven that the inductive step (\ref{eq:3}) is true.

There are two results:

  1. The theorem is true for \(n=1\).
  2. If the theorem is true for any \(n\), then it’s also true for \(n+1\).

Combining these two results we can conclude that the theorem is true for all positive integers \(n\).


I had troubles with this technique because for a long time I couldn’t for the life of me understand why is this enough and how is the basis helping?! The basis seemed redundant. We assume \(P(n)\) is true, then prove that \(P(n+1)\) is true given that \(P(n)\) is true, but so what? We didn’t prove the thing we assumed!

It clicked after I understood that we don’t have to prove \(P(n)\), we just take the concrete value from the basis and use it as \(n\). Since we have a proof of \(P(n+1)\) being true if \(P(n)\) is true, we conclude that if \(P(1)\) is true, then \(P(1+1)\) is true.

Well, if \(P(1+1)\) is true, then, using the same idea, \(P(1+1+1)\) is true, and so forth.

The basis was the cheat-code to kick-start the process by avoiding the need to prove the assumption \ref{eq:2}.