# Mathematical proof

A mathematical proof is a way to show that a mathematical theorem is true. To prove a theorem is to show that theorem holds in all cases (where it claims to hold). To prove a statement, one can either use axioms, or theorems which have already been shown to be true. Many techniques for proving a statements exist, and these include proof by induction, proof by contraction and proof by cases.[1][2][3]

## Proof by induction

One type of proof is called proof by induction. This is usually used to prove that a theorem holds for all numbers (or all numbers from some point onwards). There are 4 steps in a proof by induction.

1. State that the proof will be by induction, and state which variable will be used in the induction step.

2. Prove that the statement is true for some beginning case.

3. Assume that for some value n = n0, the statement is true and has all of the properties listed in the statement. This is called the induction step.

4. Show that the statement is true for the next value, n0+1.

Once that is shown, it would mean that for any value of n that is picked, the next one is true. Since it is true for some beginning case (usually n=1), it's true for the next one (n=2). And since it is true for 2, it must be true for 3. And since it is true for 3, it must be true for 4, etc. Induction shows that it is always true, precisely because it is true for whatever comes after any given number.

An example of proof by induction is as follows:

Prove that for all natural numbers n, 2(1+2+3+....+n-1+n)=n(n+1).

Proof: First, the statement can be written as "For all natural numbers n, 2${\displaystyle \sum _{k=1}^{n}k}$=n(n+1)."

By induction on n,

First, for n=1, 2${\displaystyle \sum _{k=1}^{1}k}$=2(1)=1(1+1), so this is true.

Next, assume that for some n=n0 the statement is true. That is, 2${\displaystyle \sum _{k=1}^{n_{0}}k}$ = n0(n0+1).

Then for n=n0+1, 2${\displaystyle \sum _{k=1}^{{n_{0}}+1}k}$ can be rewritten 2(n0+1) + 2${\displaystyle \sum _{k=1}^{n_{0}}k}$.

Since 2${\displaystyle \sum _{k=1}^{n_{0}}k}$ = n0(n0+1), 2n0+1 + 2${\displaystyle \sum _{k=1}^{n_{0}}k}$ = 2(n0+1) + 2n0(n0+1).

So 2(n0+1) + 2n0(n0+1)= 2(n0+1)(n0 + 2), which completes the proof.