Download Complete Mathematical Induction and Prime Numbers

Complete Mathematical Induction and Prime Numbers
Original Notes adopted from September 18, 2001 (Week 2)
© P. Rosenthal , MAT246Y1, University of Toronto, Department of Mathematics typed by A. Ku Ong
Complete Mathematical Induction
If S ⊂ N such that:
a) 1 ∈ S
b) (k +1 ) ∈ S whenever { 1,2,3, ..., k } ⊂ S, then S = N.
Example: What is Wrong with this Proof?
Thm: For each n ∈ N, every set of people consists of people the same age.
Proof: True for n = 1.
Use Complete Mathematical Induction
Assume true for n = 1,2,3,...k. Some k.
Consider case n = k +1.
Let S be any set of k +1 people, say S = { x1, x2,... x k, x k+1}
Let S0 = { x1, x2,... xk} k people, so all some age by induction hypothesis.
Let S1 = { x2,x3,x4, ... x k, x k+1}
k people, all same age.
In particular, all same age as x2 etc. for elements of S0.
∴Everyone in S same age as x2
∴By Mathematical Induction (Complete/Ordinary), every set of people consists of people the
same age.
Series with Positive Terms:
a1 + a2 + a3 + .... aj > 0.
Such a series Diverges if for every M, there is a j such that
a1 + a2 + ... aj > M.
Such a series Converges if there is a j such that
aj + a j + 1 + ... + a j + k < 1/2 for every k.
Eg. 1+3+5+ 7... diverges
Eg. 1+ 1/2 +1/4 +1/8 ...converges.
Harmonic Series
1 + 1/2 + 1/3 + 1/4.... diverges
1 + (1/2 + 1/3) + 1/4 + (1/5 +1/6 +1/7 +1/8)
(> 1/4 +1/4)
(1/8 + 1/8 + 1/8 + 1/8)
> 1/2
> 1/2
1+ 1/2 + 1/4 + 1/8 + ... converges.
Suppose {bn} is a sequence of natural numbers with bn+1> bn.
Consider the series
1/b1 + 1/b2 +1/b3 + ...
Case bn = n ⇒ diverges
Case bn = 2n ⇒ converges
Suppose bn is the nth prime number.
1/2 + 1/3 + 1/5 +1/7 + 1/11 + 1/13... Does it converge?
Thm: If Pn is the nth prime number, then 1/p1 + 1/p2 + ... + 1/pj + ... diverges.
Proof: Suppose it converges;
Then there would be some j such that 1/pj+1 + 1/pj+2 + 1/pj+ k <1/2 for all k.
Show that this is impossible: Each natural number n,
(recall: every n ≠ 1 is a product of primes)
Let F(n) be the number of natural numbers ≤ n that are products of primes all of which are in
{ P1, P2, ... Pn}
We will estimate F(n) in Two different ways that will show, for large enough n there is a contradiction.
Fix some n. For each m<= n, write m = s2t where s2 is the largest perfect square in m.
At most √n such s's arise as you go through all m <=n
If m sets combined in F(n) then m = Pi, Pii, Piii
= s 2t
where all i's ≤ j
Each t is a product of primes in {Pi,... Pj}
In t, no two primes are the same, since t is "square free".
Thus there are at most 2 j t's that arise from m's that set counted in F(n).
* F(n) ≤ √n 2j
n – F(n) is the number of natural numbers that are ≤ and have a prime factor among {Pj +1 , Pj + 2... }
1/Pj+1 + 1/Pj+2 + ... 1/ Pj+k <1/2.
At most n/Pj + 1 numbers ≤ have Pj + 1 as a factor
n- F(n) ≤ n/Pj + 1 + n/Pj + 2 + ... ≤ n/2
* F(n) ≤ √n2j
n – F(n) ≤ n/2
n/2 ≤ F(n) ≤ √n2j
√n ≤2 j+1 j, fixed.
True for all n. But false for large n. Contradiction.
Defn: A natural number is composite if it is not 1 & isn't prime.
Can you find 20 consecutive composite numbers?
(21)! + 2, (21) ! + 3, (21)! +4..... (21)! +21 ⇒ 20 consecutive composites.
Thm: For any k, there exist k consecutive composite numbers.
Proof: (k+1)! +2, (k+1)! + 3,..., (k+1)! + k +1 are all composite.
Let p(x) = x2 – x + 41
p(1) = 41 – prime
p(2) = 43 – prime
p(3) = 47 – prime
p(4) = 53 prime
Prime for x = 1,2,3.... 40.
Thm: There is no polynomial (with integer coefficients) whose values are prime for all natural number
values of the variable.
p(x) = anxn + an - 1xn-1 + ... +a1x + a0
Z: set of integers...-3,-2,-1,0,1,2,3 For n ∈N, n+ (-n) =0.
Want distributive law: a(b+c) = ab + ac.
Then (-a)(-b) + (a)(-b)
= (-a +a)(-b)
= 0(-b) =0
ab + a(-b) = a(b + (-b) ) = a*0 = 0
(-a)(-b) + a(-b) = 0 ab + a(-b)
(-a)(-b) =ab
For a,b ∈ Z, a|b if there exists c ∈ Z such that b = ac.
Eg. Is 229 + 3 divisible by 7?
Let m be a fixed natural number.
Def: a ≡b (mod m) "a is congruent to be modulo m"
if a – b is divisible by a (For a,b ∈ Z)
Suppose k,m ∈ N , k = mq +r (divide m into k, get quotient q and remainder r)
k = r(mod m)
ie) a natural number is congruent mod m to its remainder upon division by m.
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