Download 4. Properties of Integers: Mathematical Induction

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4. Properties of Integers:
Mathematical Induction
Z+ : set of all positive integers
The Well Order Principle: any non-empty subset of Z+ contains a smallest element
Principle of Mathematical Induction:
S(n): an open mathematical statement of n
(a) If S(1) is true
(b) If S(k) is true for some k ∈ Z+ then
S(k + 1) is true
then S(n) is true for all n ∈ Z+
or
[S(1) Λ [ V- k ≥ 1 [S(k) → S(k + 1)]]]
=> V- n ≥ 1 S(n)
-59Proof:
Let F = { n ∈ Z+ | S(n) is false}
If F ≠ Φ then F has a least element t.
t > 1 (... S(1) is true)
... t − 1 ∈ Z+ and S(t − 1) is true.
But by (b), S(t) must be true too.
This is a contradiction.
... we must have F = Φ.
Mathematical induction does not have to start
with 1, it can start with any integer n0 .
The above statement can be put in a more general form, as follows:
[S(n0 ) Λ [ V- k ≥ n0 [S(k) → S(k + 1)]]]
=> V- n ≥ n0 S(n)
-60Example: Prove that P(n) = 22n+1 + 1 is divisible by 3 for all n ≥ 0
Proof. By induction.
P(0) = 21 + 1 = 3
(n0 = 0)
divisible by 3
Assume P(k) is divisible by 3 (i.e.,
P(k) = 3 * l for some l)
The we have
P(k + 1) = 22(k+1)+1 + 1 = 4(22k+1 ) + 1
= 4(3l − 1) + 1 = 3(4k − 1) ,
divisible by 3
Hence, by mathematical induction, P(n) is
divisible by 3 for all n ≥ 0.
-61Example: Consider the following four equations:
1=1
2+3+4=1+8
5 + 6 + 7 + 8 + 9 = 8 + 27
10 + 11 + 12 + 13 + 14 + 15 + 16 = 27 + 64
Conjecture the general formula suggested by
these four equations, and prove your conjecture.
Proof. General formula:
( j+1)2
Σ
2
j +1
i = j 3 + ( j + 1)3
Prove by induction:
Key:
( j+2)2
( j+2)2
( j+1)2
( j+1) +1
j +1
j +1
i− Σ i
Σ2 i = Σ
2
2
-62Example: Let S 1 and S 2 be two sets where
|S 1 | = m, |S 2 | = r, for m, r ∈ Z+ , and the elements in each of S 1 , S 2 are in ascending order.
It can be shown that the elments in S 1 and S 2
can be merged into ascending order by making
no more than m + r − 1 comparisons. Use this
result to establish the following.
For n ≥ 0, let S be a set with |S| = 2n . Prove that
the number of comparisons needed to place the
elements of S in ascending order is bounded
above by n ⋅ 2n .
Proof. By induction on n.
n = 0, |S| = 20 , 0 ⋅ 20 = 0 comparison
n = k, |S| = 2k , k ⋅ 2k = 0 comparisons
When n = k + 1, |S| = 2k+1 .
Let S = A ∪ B, | A| = 2k , |B| = 2k .
k ⋅ 2k for A,
k ⋅ 2k for B
Then merge A and B.
-63Finite Induction Principle:
- alternative form
S(n): open math statement of n
n0 ≤ n1 ∈ Z+
(a) If S(n0 ), S(n0 + 1), ..., S(n1 ), are true, and
(b) If S(n0 ), ..., S(k − 1), S(k) true for some
k ≥ n1 then S(k + 1) is true, then S(n) is true for
all n ≥ n0
-644.2 Recursive Definition
When a concept (term) is defined in terms of
similar prior results, we say that the concept is
defined recursively, using the concept of
recursion.
You need a basis step and a recursive step.
Example:
a0 = 1, a1 = 2, a2 = 3
a n = a n−1 + a n−2 + a n−3
for n ≥ 3, n ∈ Z+
Example:
0! = 1
V- n ∈ Z+ , (n + 1)! = n!(n + 1)
-65Example:
1 ∈ X,
V- a ∈ X, a + 2 ∈ X
=> X = the set of all positive odd integers.
S(n) : 2n + 1 ∈ X
Example: to prove something is well-defined
Let n ∈ Z+ where n ≥ 2, and let A1 , A2 , ...,
An ⊆ U for each 1 ≤ i ≤ n. Then
A1 ∩ A2 ∩ . . . ∩ A n = A1 ∪ A2 ∪ . . . ∪ A n
-664.3 The Division Algorithm: Prime Numbers
__
Definition: b|a (b ≠ 0) if _| n s.t. a = bn
a: multiple of b, b: divisior of a
Theorem: V- a, b, c ∈ Z,
(a) 1|a, a|0
(b) [( a|b ) Λ ( b|a )] => a = ±b
(c) [( a|b ) Λ ( b|c )] => a|c
(d) a|b => a|bx for all x ∈ Z
(e) if x = y + z and if a divides two of x, y,
z then a divides the third.
(f) [( a|b ) Λ ( a|c )] => a |(bx + cy)
for all x, y ∈ Z
(g) if a|c i , 1 ≤ i ≤ n =>
a |(c 1 x 1 + . . . + c n y n ), x i ∈ Z
-67Lemma 4.1: If n ∈ Z+ and n is composite, then
there exist a prime p such that p|n.
Proof Let P be the set of all positive divisors of
n. P has a smallest element. This element is a
prime.
Theorem 4.4: (Euclid)
There are infinitely many primes.
Proof Let P be the set of all primes. If P is
finite, say P = { p1 , p2 , . . . , p n }. Consider
q = p1 p2 . . . p n + 1.
q is a composite. So there exists p j such that
p j |q, i.e., there exists m ∈ Z such that q = p j m.
But then
p j |1 = p j m − p1 p2 . . . p n ,
a contradiction.
-68Theorem 4.5: (Division Algorithm)
If a, b ∈ Z and b > 0 then there exist unique
q, r ∈ Z with a = qb + r, 0 ≤ r < b.
Proof Consider {ib | i ∈ Z}. There exist qb,
(q + 1)b such that
qb ≤ a < (q + 1)b
Let r = a − qb. If
a = qb + r = q′b + r′
we have
b(q − q′) = r′ − r
This implies that
b|(r′ − r) => r′ − r = 0
=> r′ = r
(why?)
But then since b(q − q′) = 0, we must have
q − q′ = 0 or q = q′
Hence q and r are unique.