Transcript Lec12SeqSum05

Lecture 13
3.2,3.3
Sequences & Summations
Proof by Induction
Cardinality
definition: Two sets have the same cardinality if and only if there is a
one-to-one correspondence between them.
This is simple for finite sets, but what if a set has infinite elements?
definition: A set that is finite or has the same cardinality as the set of
positive integers (Z+) is called countable.
Example: Consider the sequence {an}, an = n^2, n={1,2,3,4...}
Naively speaking, there seem to be much less elements in {an} than in
Z+ (since we skip a lot).
Infinity is weird! Here is the one-to-one mapping:
1 2 3 4 5 6 7 ...
infinity
(intuitively: you can enumerate them)
1 4 9 16 25 36 49 ...
cardinality
Now what about the positive rational numbers: p/q with p,q integer, q not 0 ?
1/1 2 /1 3 /1 4 /1 
1/ 2 2 / 2 3 / 2 4 / 2 
1/ 3 2 / 3 3 / 3 4 / 3 
1/ 4 2 / 4 3 / 4 4 / 4 

1 2
1 ½


3 4 5 6 7 8
9
2 3 1/3 ¼ 2/3 3/2 4
skip

10
5
set of positive rational numbers are countable,
in fact the set of all rational numbers is countable : 4 countable quadrants.
cardinality
That was amazing? Now check this out: a proof that the real numbers are
not countable!
When a prove seems impossible, try to prove a contradiction.
First this: unions of countable sets are countable
intersections of countable sets are countable
a subset of a countable set is countable.
Prove: assume that the reals are countable,
 the reals in (0,1) are countable (since it is a subset),
 Thus there is a sequence that counts all reals as follows:
1: r1 = 0.d11 d12 d12 d14 ...
2: r2 = 0.d21 d22 d23 d24 ...
3: r3 = 0.d31 d32 d33 d34 ...
etc.
construct the number: r = 0.c1 c2 c3 c4 ... with ci=4 if dii not 4
ci = 5 if dii = 4
This is guaranteed to be different than any real in the list, so it isn’t in the list,
so the list is not complete. contradiction!
 real in [0,1] are uncountable  all reals are uncountable.
3.3 Mathematical Induction
If we want to prove propositions P(k) for all positive integers, we may use
inductions.
First we prove: P(1) is true.
Then we prove P(k)  P(k+1).
So, is P(100) true? yes, use “modus ponens” 99 times.
P(1)
P(1)P(2)
____________
P(2)
P(2)P(3)
___________
P(3)
until P(100).
formally:
[ P(1)  k ( P(k )  P(k  1))]  nP( n)
Examples
prove that the sum of the first n odd positive integers is n^2.
n=1: 1 = 1.
assume it’s true for some k.
1+3+5+...+2k-1 = k^2 is true.
add 2k+1 on each side:
1+3+5+...+2k-1 + 2k+1= k^2 + 2k + 1
1+3+5+...+2(k+1)-1 = (k+1)^2

Examples
Prove that n < 2^n for positive integers n.
P(1): 1 < 2
Inductive step: assume P(k) is correct, prove P(k+1) is correct.
k < 2^k
k+1 < 2^k + 1 < 2^k + 2^k = 2^(k+1) 
Note: we could of course also start our induction at another integer b.
Example: Prove that 2^n < n!
for integers n>= 4 .
P(4) = 2^4 = 16 < 4! = 4 3 2 1 = 24.
Induction: assume 2^k < k! 
2 x 2^k < 2 x k!
2^(k+1) < (k+1) x k! (recall k > = 4).
2^(k+1) < (k+1)! 
Strong Induction
induction (I):
[ P(1)  k ( P(k )  P(k  1))]  nP( n)
strong induction (SI): [ P(1)  k ( P(1)  P(2)  ...  P( k )  P( k  1))]  nP(n)
they are equivalent
Strong Induction
Example:
Consider the game where there are 2 piles of n matches. Each player picks
an arbitrary number of matches from one pile. The one who gets the last matches wins.
Proposition: The player who starts second can always win.
P(1): 2 piles with 1 match each. Second player always wins.
Assume player 2 wins when we have 2 piles of k matches.
Can player 2 win when we have 2 piles of k+1 matches?
 Player 1 will have to take between 1 and k+1 matches from 1 pile.
If he picks k+1 matches, player 2 wins by grabbing all matches from the other pile.
If he picks between 1 and k matches, player two takes the same amount from the other
pile reducing the problem to a smaller equivalent problem.
Strong Induction  player 2 always wins 
Examples
Proposition: every positive integer n>2 can be written as the product of primes.
P(2): product of itself (it’s prime).
Assume k can be written as a product of primes.
Can we prove it for k+1?
two cases:
k+1 = prime (thus it is a product of one number – itself).
k+1 = a x b
However since both a and b < k+1 and >= 2, we know that a and b can be written
as the product of primes  a x b is a product of primes. 