Proofs:Induction

The beauty of induction is that it allows a theorem to be proven true where an infinite number of cases exist without exploring each case individually. Induction is analogous to an infinite row of dominoes with each domino standing on its end. If you want to make all the dominoes fall, you can either:

1. push on the first one, wait to see what happens, and then check each domino afterwards (which may take a long time if there's an infinite number of dominoes!)
2. or you can prove that if any domino falls, then it will cause the domino after it to fall. (i.e. if the first one falls then the second one will fall, and if the second one falls then the third one will fall, etc.)

Induction, essentially, is the methodology outlined in point 2.

Parts of Induction

Induction is composed of three parts:

1. The Base Case (in the domino analogy, this shows the first domino will fall)
2. The Induction Hypothesis (in the domino analogy, we assume that a particular domino will fall)
3. The Inductive Step (in the domino analogy, we prove that the domino we assume will fall will cause the next domino to fall)

Weak Induction

Weak induction is used to show that a given property holds for all members of a countable inductive set, this usually is used for the set of natural numbers.

Weak induction for proving a statement $P(n)$ (that depends on $n$) relies on two steps:

• $P(n)$ is true for a certain base step. Usually the base case is $n=1$ or $n=0$

• $P(k)\to P(k+1)$. That is, if $P(k)$ is true, then $P(k+1)$ is true.

If these two properties hold, one may induce that the property holds for all elements in the set in question. Returning to the example, if you are sure that you called your neighbor, and you knew that everyone who was called in turn called his/her neighbor, then you would be guaranteed that everyone on the block had been called (assuming you had a linear block, or that it curved around nicely).

Example 1

The first example of a proof by induction is always 'the sum of the first n terms:'

Theorem 2.4.1. For any fixed $n\in \mathbb {N} ,$ $\sum _{i=1}^{n}{i}={\frac {n(n+1)}{2}}$

Proof

• Base step: $1={\frac {1\cdot 2}{2}}$ , therefore the base case holds.

• Inductive step: Assume that $\sum _{i=0}^{n}i={\frac {n(n+1)}{2}}$ . Consider $\sum _{i=1}^{n+1}i$ .

${\begin{aligned}\sum _{i=1}^{n+1}i&=\sum _{i=1}^{n}i+(n+1)={\frac {n(n+1)}{2}}+n+1\\&=\left({\frac {n}{2}}+1\right)(n+1)\\&={\frac {(n+1)(n+2)}{2}}\\&={\frac {(n+1){\big (}(n+1)+1{\big )}}{2}}\end{aligned}}$

So the inductive case holds. Now by induction we see that the theorem is true.

Example 2

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

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

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

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

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 $4~|~f(k+1)$ it implies $4~|~f(n)$ 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:

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

If $4$ divides $f(k)$, then it implies that $4$ also divides $f(k+1)$. Therefore, since $4$ divides $f(1)$, $4$ divides $f(n)$ for all $n\in \mathbb {N}$.