# Proof by mathematical induction

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

${\text{Prove that }}f(n)=5^{n}+8n+3{\text{ is divisible by }}4,{\text{ for }}n\in \mathbb {Z} ^{+}$ We are asked to prove that $f(n)$ is divisible by 4. We can test if it's true by giving $n$ values.

$n$ $f(n)$ ${\text{Divisible by }}4$ $1$ $16$ $4\cdot 4$ $2$ $44$ $4\cdot 11$ $3$ $152$ $4\cdot 38$ $4$ $660$ $4\cdot 165$ $5$ $3168$ $4\cdot 792$ 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. Base case – Is it true for the first case? This means is it true for the first possible value of $n$ .
3. Assumption – We assume what we are trying to prove is true for a general number. such as $k$ , also known as the induction hypothesis.
4. Induction – Show that if our assumption is true for the ($k^{th})$ term, then it must also be true for the ($k+1^{th})$ term.
5. Conclusion – Formalise your proof.

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

1. Summation of series;
2. Divisibility;
3. Recurrence relations;
4. Matrices.

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## Example of a proof by divisibility

Proposition: ${\text{Prove that }}4~|~f(n)=5^{n}+8n+3,{\text{ for }}n\in \mathbb {N} ^{+}$ Notice our parameter, ${\text{for }}n\in \mathbb {N} ^{+}$ . This means that what we want to prove must be true for all values of $n$ which belong to the set (denoted by $\in$ ) of positive integers ($\mathbb {N} ^{+}$ ).

Base case:

{\begin{aligned}&{\text{Let }}n=1\\&f(1)=5^{1}+8(1)+3\implies f(1)=16\\&\therefore 4~|~f(1)\end{aligned}} Assumption (Induction Hypothesis): Now we let $n=k$ where $k$ is a general positive integer, and we assume that $4\ |\ f(k)$ .

Remember that $f(k)=5^{k}+8k+3$ .

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

Hence:${\text{let }}n=k+1\implies f(k+1)=5^{k+1}+8(k+1)+3$ This is where our assumption comes in, if $4~|~f(k)$ then 4 must also divide $f(k+1)-f(k)$ .

So: $f(k+1)-f(k)=5^{k+1}+8(k+1)+3-(5^{k}+8k+3)$ {\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 $4~|~{\bigl (}f(k+1)-f(k){\bigr )}$ and thus $4~|~f(k+1)$ . This implies that $4~|~f(n)$ because you have successfully shown that 4 divides $f(n)$ , where $n$ is a general, positive integer ($k$ ) and also the consecutive term after the general term ($k+1$ )

Conclusion:

{\begin{aligned}&4~|~f(k)\implies 4~|~f(k+1)\\&\therefore ~4~|~f(n),\forall n\in \mathbb {Z} ^{+}\end{aligned}} {\begin{aligned}&{\text{If }}4{\text{ divides }}f(k){\text{ (as we assumed) then it 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}} (Empty)

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