Induction Principle - Graceland University
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Transcript Induction Principle - Graceland University
Mathematical
Induction
F(1) = 1;
F(n+1) = F(n) + (2n+1) for n≥1
n
1
2
3
4
5
6
7
8
9
10
F(n)
1
4
9
16
25
36
49
64
81
100
F(n) =
n2 for all n ≥ 1
Prove it!
Prove statements of the form:
"p(n) for n ≥ 1"
Verify p(1).
Assume p(n) for some n ≥ 1
"Induction hypothesis"
Show p(n+1) must be true for this value of n.
"Base case"
"Induction step"
Conclude p(n) is true for all n ≥ 1
Prove: F(n) = n2 for n ≥ 1.
Recall
F(1) = 1;
F(n+1) = F(n) + (2n+1) for n>1
Induction
hypothesis
Proof:
Base Case
2
F(1) = 1 = 1
Assume F(n) = n2 for some n ≥ 1.
Then F(n+1) = F(n) + (2n+1), since n+1 > 1.
= n2 + 2n + 1 by hypothesis
= (n+1)2
Therefore F(n) = n2 for all n ≥ 1.
Algebra
Not just for some!
First principle of mathematical
induction
Let S be a set of integers containing a.
Suppose S has the property that whenever
some integer n ≥ a belongs to S, then the
integer n + 1 also belongs to S.
Then S contains every integer greater than or
equal to a.
What does our proof have to
do with S?
Let S = { n ≥ 1 | F(n) = n2 }
F(n) = 1 = 12
Base case shows 1 is in S
F(n) = n2 for some n ≥ 1 => F(n+1) = (n+1)2
Induction step shows that if n is in S, then
n+1 is in S.
By the first principle of induction, S contains all
integers ≥ 1.
That is, F(n) = n2 for all n ≥ 1.
Why does the first principle of
induction work?
Proof: Given S = { n ≥ a | p(n)} has the property that
a is in S and n in S implies n+1 is in S.
It remains to show that S contains all integers ≥ a.
Let T = { n ≥ a | n is not in S}
Assume, towards a contradiction, T is not empty.
By the WOP, T has a smallest element, t
Now t > a, and n = t – 1 belongs to S.
But then n+1 = (t–1)+1 = t belongs to S.
This contradiction shows that T is empty.
Second Principle of
Mathematical Induction
Let S be the set of integers containing a.
Suppose S has the property that n belongs to
S whenever every integer less than n and
greater than or equal to a belongs to S.
Then S contains all integers greater than or
equal to a.
Proof of the 2nd Principle of
Mathematical Induction
Almost identical to the proof of the first.
Try it!
Proof of the Fundamental
Theorem of Arithmetic.
Let S be the set of integers greater than 1 which are
primes or products of primes.
2 belongs to S
Assume that for some integer n ≥ 2, S contains all
integers k with 2 ≤ k < n.
We must show that n is in S.
If n is prime, then n is in S.
If n is composite, n = ab where a,b are in S.
By hypothesis, a and b are primes or products of
primes.
So n is a product of primes.