Transcript Induction Principle - Graceland University

Mathematical
Induction
F(1) = 1;
F(n+1) = F(n) + (2n+1) for n≥1
n
1
2
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F(n)
1
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36
49
64
81
100
F(n) =
n2 for all n ≥ 1
Prove it!
Prove statements of the form:
"p(n) for n ≥ 1"
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Verify p(1).
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Assume p(n) for some n ≥ 1
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"Induction hypothesis"
Show p(n+1) must be true for this value of n.
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"Base case"
"Induction step"
Conclude p(n) is true for all n ≥ 1
Prove: F(n) = n2 for n ≥ 1.
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Recall
F(1) = 1;
F(n+1) = F(n) + (2n+1) for n>1
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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
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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?
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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
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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
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Almost identical to the proof of the first.
Try it!
Proof of the Fundamental
Theorem of Arithmetic.
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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.