Mathematical induction lecture-1
A proof technique
Mathematical Induction is a proof teachnique.
Examples:
1. Sum of first ’n’ natural numbers is $\frac{n(n+1)}{2}$
$1+2+…..+n = \frac{n(n+1)}{2}$
$n^3 -n$ is divisible by 3
Principle of mathematical induction
For given $P(n)$
Goal: $P(n)$ is true for all $ n \in \mathbb{N}$
We can complete this goal in two steps:
We verify that P(1) is true
Assume that P(k) is true for an arbitrary positive integer k. Prove that $p(k+1)$ is true
Remark
Warning: We are not assuming that P(k) is true all k, we will assume for P(k) and then show for P(k+1).
First step - Basis step
Second step - Inductive step
Induction hypothesis
Assumption that P(k) is true is generally refer to as Induction hypothesis.
Question
Why MI is a valid proof technique?
Not inductive reasoning
Example
Show that sum of first n natural numbers is $\frac{n(n+1)}{2}$
$i.e. 1+2+ … +n = \frac{n(n+1)}{2}$
Solution
Use M.I to prove
$1+2+…+n = \frac{n(n+1)}{2}$ $\forall n \in \mathbb{N}$
$P(n) : 1+2+…+n = \frac{n(n+1)}{2}$
Basic step:
L.H.S = 1
= $\frac{1(1+1)}{2}$
i.e $p(1)$ is true
Inductive step
Assume $P(k)$ is true
1+2+….+k = $\frac{k(k+1)}{2}$
To prove that $P(k+1)$ is true
To prove
$= ~ \frac{(k+1)((k+1)+1)}{2}$
$(1+2+3+…+k)+ k+1$
$= ~ \frac{k(k+1)}{2}+ k+1 $ (By Induction Hypothesis)
$\Rightarrow$ $\frac{k(k+1) + 2(k+1)}{2}$
$\Rightarrow$ $\frac{k^2 + {3k} +2}{2}$
$\Rightarrow$ $\frac{(k+2)(k+1)}{2}$
By M.I, P(n) is true for all n
$\therefore$ $1+2+…+n = \frac{n(n+1)}{2} \quad \forall n\in \mathbb{N}$
Conjucture a formula for sum of first n odd positive integers. Verify your conjucture using M.I
$1+3 = 4 = 2^2$
$1+3+5 = 9 = 3^2$
$1+3+5+7 = 16 = 4^2$
.
To P.T.
$1+3+…+2n-1 = n^2 \quad\quad \forall n \in \mathbb{N}$
$P(n): 1+3+…+ 2n-1 = n^2$
We wish $P(n)$ is true for all n
Basis step
To verify $p(1)$
$1 = 1^2$
P(1) is true
Inductive step:
Assume P(k) is true
$1+3+5+…+ 2k-1 $ = $k^2$
We have to P.T. P(k+1) is true
$\Rightarrow$ $1+3+5+…+ 2(k)+1 = (k+1)^2$
$\Rightarrow$ $(1+3+5+… 2k-1) + 2k+1$
$\Rightarrow$ $( k^2 )+ 2k+1$ ( induction hypothesis)
$\Rightarrow$ $(k+1)^2$
Hence by MI
P(n) is true for all n
i.e
Often, we may require to prove P(n) is true for $n = b, b+1, b+2 …..$
b is some integer other than 1
Basic step
We assume P(k)
$k \in \{b, b+1 …\}$
Then we prove P(k+1)
P.T. $1+2+…+2^n= 2^{n+1} -1 \quad \forall$ non negative integer n.
verify P(0) is true
p(n): $1+2+…+2^n = 2^{n+1} - 1$
L.H.S. $2^0 = 1$
R.H.S. $2^{0+1} -1 = 2-1 =1$
$\therefore$ P(0) is true
Assume P(k) is true for k, a non negative integer
To Show that P(k+1) is true
We have to Show that
$1+2+…+2^{k+1} = 2^{(k+1)+1} -1= 2^{k+2} - 1$
$(1+2+…+2^k) + 2^{k+1}$
$\Rightarrow$ $2^{k+1} -1 +2^{k+1}$
$\Rightarrow$ $2.2^{k+1} -1$
$\Rightarrow$ $2^{k+2}-1$
$\therefore$ P(k+1) is true.
Hence by mathematical induction
P(n) is true for all $n \in \mathbb{N} \cup \{0\}$
Basis step:
Induction hypothesis is P(k) is true
i.e. $2^k < k!$
To prove P(k+1) is true
i.e. to prove, $2^{k+1} < (k+1)!$
$2^k < k!$
For an arbitrary, but fixed k, where $k \geq 4$
$2^{k+1} = 2.2^{k}< 2.k!$
$[(k+1)! = (k+1).k!]$
$2^{k+1} < (k+1) k!=(k+1)!$
$(\because 2 < k+1)$
$\Rightarrow 2^{k+1} < (k+1)! $
Hence, by principle of mathematical induction
It follows that
$2^{n} < n!$ fo all $n \geq 4$
Thank you