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```Some Proof Techniques
• Induction
• Pigeonhole Principle
• Diagonalization
Induction
Let P (n) be a proposition involving integer n. The Principle of Mathematical Induction states that P (n) is true
for n0 ≤ n, if the following are true:
• P (n0), and
• for n ≥ n0, P (k), n0 ≤ k ≤ n implies
P (n + 1)
Problems
1. The number of subsets of a set of
size n is 2n.
2.
n
X
i = n(n + 1)/2
i=0
3.
n
X
i=0
i2 = n(n + 1)(2n + 1)/6
4.
n
X
3
i =(
i=0
5.
n
X
n
X
i)2
i=0
2i = 2n+1 − 1
i=0
6.
n
X
1/(i(i + 1)) = n/(n + 1)
i=1
7. For n ≥ 4, n! > 2n
Pigeonhole Principle
If A and B are finite sets and |A| >
|B|, then there is no one-to-one function from A to B.
Example: In any group of at least two
people there are at least two persons
that have the same number of acquaintances within the group.
Example: How many shoes must be
drawn from a box containing 10 pairs
to ensure a match?
Example: In NYC there are at least
two people with the same number of
Example: If A is a set of 10 numbers
between 1 and 100, there are two distinct disjoint subsets of A whose elements sum to the same number.
Diagonalization Proofs
• Theorem (Georg Cantor). The set
of all subsets of natural numbers is
uncountable.
– Suppose there is a one-to-one function f from N onto pow(N).
– pow(N) = {S0,S1,S2,...}
where Si = f(i)
– D = {n in N | n is not in Sn}.
D is the diagonal set for N.
– D is a set of natural numbers,
hence D = Sk for some k in N
– if k is in Sk, then k is not in D.
But Sk = D.
– if k is not in Sk, then k is in D.
But Sk = D.
– So pow(N) is not countable.
```
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