# Mathematical Proof

As mentioned in a previous post, I find mathematical proof truly beautiful. To me, the timelessness of a theorem once it has been proved is truly unique to mathematics; work done by mathematicians who lived thousands of years ago still holds true today and is a vital component to modern day mathematics.

There are many ways to go about mathematical proof, and I would like to highlight three: proof by contradiction, proof by induction and direct proof.

To prove something by contradiction, we assume that what we want to prove is not true, and then show that the consequences of this assumption are not possible. This method of proof was favoured and skilfully used by Greek mathematicians.

One well-known use of this method is the proof that the square root of 2 is irrational:

Source

Proof by Induction

I only learnt this form of proof when I studied Further Pure Mathematics 1 in AS Levels last year. However, I had come across the term when I read Simon Singh’s book ‘Fermat’s Last Theorem’, as it was using this technique that Andrew Wiles proved this theorem.

So how do we prove by induction?

Suppose we have a statement P(n) about natural numbers, that may be true for some natural numbers n.

1. Firstly, we must show that the first case (normally P(0) or P(1)) is true.
2. We then suppose that P(n) is true.
3. If we can then show that P(n+1) is true, P(n) must be true for all n greater or equal to the first case.

This works as, say we know that P(1) is true and P(n+1) is true, this signifies that P(1+1) = P(2) is true, which means P(2+1) = P(3) is true and so on until infinity.

Example:

•  Show the statement holds true for P(0)

LHS = RHS, therefore true for P(0).

• Show that if P(k) holds, then also P(k + 1) holds. This can be done as follows.

Assume P(k) is true

Using this assumption, the left-hand side can be rewritten to:

Thus,

Showing that P(k+1) is true, if P(k) is true.

Source

Direct Proof

To prove a theorem directly we start with something known to be true and then proceed by making small logical steps, which are clearly correct, until arriving at the desired result. So, because the starting point was true and each small step clearly correct, we know the result to be true. This is perhaps the most classic way to prove a theorem, however this method can be quite challenging, as breaking down a mathematical argument into small steps requires patience and clear thinking. Personally, I find this the hardest form of proof, as I never know where to start!

An example of this method being employed is in the proof of the formula for the roots of a quadratic equation:

Source

Let me know your favourite proofs in the comments below! M x

Collection of proofs

Article by Economist: ‘Proof and Beauty’

1. youroundingup says:

This is a nice post, but I have a small suggestion about your induction example.

After “Assume P(k) is true”, you have written down what is to be shown.

I would start by writing 0 + 1 + 2 + …+ k + (k+1) = k(k+1)/2 +(k+1) (by assumption).

Then you can rewrite the right hand side until you get what you want. As is, it feels a little backwards. I hope this makes some sense.

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