Proof by Mathematical Induction

=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:


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

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

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.

Example of a proof by summation of series
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Example of a proof by divisibility
Proposition: $$\text{Prove that }4~|~f(n)=5^n+8n+3,\text{ for }n\in\N^+$$

Notice our parameter, $$\text{for }n\in\N^+$$. 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 ($$\N^+$$).

Base case:

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

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

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

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:

$$\begin{align}&4~|~f(k)\implies4~|~f(k+1)\\&\therefore~4~|~f(n),\forall n\in\Z^+\end{align}$$

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

Example of a proof by recurrence relations
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Example of a proof by matrices
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