# Proof by mathematical induction[1]

Mathematical induction is the process of verifying or proving a mathematical statement is true for all values of ${\displaystyle n}$ within given parameters. For example:

${\displaystyle {\text{Prove that }}f(n)=5^{n}+8n+3{\text{ is divisible by }}4,{\text{ for }}n\in \mathbb {Z} ^{+}}$

We are asked to prove that ${\displaystyle f(n)}$ is divisible by 4. We can test if it's true by giving ${\displaystyle n}$ values.

${\displaystyle n}$ ${\displaystyle f(n)}$ ${\displaystyle {\text{Divisible by }}4}$
${\displaystyle 1}$ ${\displaystyle 16}$ ${\displaystyle 4\cdot 4}$
${\displaystyle 2}$ ${\displaystyle 44}$ ${\displaystyle 4\cdot 11}$
${\displaystyle 3}$ ${\displaystyle 152}$ ${\displaystyle 4\cdot 38}$
${\displaystyle 4}$ ${\displaystyle 660}$ ${\displaystyle 4\cdot 165}$
${\displaystyle 5}$ ${\displaystyle 3168}$ ${\displaystyle 4\cdot 792}$

So, the first 5 values of n are divisible by 4, but what about all cases? That's where mathematical induction comes in.

Mathematical induction is a rigorous process, as such all proofs must have the same general format:

1. Proposition – What are you trying to prove?
2. basis case – Is it true for the first case? This means is it true for the first possible value of ${\displaystyle n}$.
3. Assumption – We assume what we are trying to prove is true for a general number. such as ${\displaystyle k}$
4. Induction – Show that if our assumption is true for the (${\displaystyle k^{th})}$ term, then it must be true for the term after (${\displaystyle k+1^{th})}$ term.
5. Conclusion – Formalise your proof.

There will be four types of mathematical induction you will come across in FP1:

1. Summing series
2. Divisibility
3. Recurrence relations
4. Matrices

## Example of a proof by divisibility

Proposition: ${\displaystyle {\text{Prove that }}4~|~f(n)=5^{n}+8n+3,{\text{ for }}n\in \mathbb {N} ^{+}}$

Note our parameter, ${\displaystyle {\text{for }}n\in \mathbb {N} ^{+}}$ This means it wants us to prove that it's true for all values of ${\displaystyle n}$ which belong to the set (${\displaystyle \in }$) of positive integers (${\displaystyle \mathbb {N} ^{+}}$)

Basis case:

{\displaystyle {\begin{aligned}&{\text{Let }}n=1\\&f(1)=5^{1}+8(1)+3\implies f(1)=16\\&\therefore 4~|~f(1)\end{aligned}}}

Assumption: Now we let ${\displaystyle n=k}$ where ${\displaystyle k}$ is a general positive integer and we assume that ${\displaystyle 4\ |\ f(k)}$

Remember ${\displaystyle f(k)=5^{k}+8k+3}$

Induction: Now we want to prove that the ${\displaystyle k+1^{th}}$ term is also divisible by 4

Hence ${\displaystyle {\text{let }}n=k+1\implies f(k+1)=5^{k+1}+8(k+1)+3}$

This is where our assumption comes in, if ${\displaystyle 4~|~f(k)}$then 4 must also divide ${\displaystyle f(k+1)-f(k)}$

So: ${\displaystyle f(k+1)-f(k)=5^{k+1}+8(k+1)+3-(5^{k}+8k+3)}$

{\displaystyle {\begin{aligned}&f(k+1)-f(k)=5(5^{k})-5^{k}+8\\&f(k+1)-f(k)=4(5^{k})+8\\&\therefore ~f(k+1)-f(k)=4(5^{k}+2)\end{aligned}}}

Now we've shown ${\displaystyle 4~|~{\bigl (}f(k+1)-f(k){\bigr )}}$ and thus ${\displaystyle 4~|~f(k+1)}$ it implies ${\displaystyle 4~|~f(n)}$ because you have successfully shown that 4 divides ${\displaystyle f(n)}$, where ${\displaystyle n}$ is a general, positive integer (${\displaystyle k}$) and also the consecutive term after the general term (${\displaystyle k+1}$)

Conclusion:

{\displaystyle {\begin{aligned}&4~|~f(k)\implies 4~|~f(k+1)\\&\therefore ~4~|~f(n),\forall n\in \mathbb {Z} ^{+}\end{aligned}}}

{\displaystyle {\begin{aligned}&{\text{If }}4{\text{ divides }}f(k){\text{ (as we assumed) then it is implies that }}4{\text{ also divides }}f(k+1)\\&{\text{therefore }}4{\text{ divides }}f(n),{\text{ for all values of }}n{\text{ that belong to the set of positive integers.}}\end{aligned}}}