1 + 2 + 3 + ... + 100
Proof by induction: 1 + 2 + ... + N = N(N+1)/2
Postage in Investigate!
Example 2.5.1: triangle numbers
Example 2.5.2: 6n-1 is divisible by 5
Example 2.5.3: n2 < 2n for n>=5
Warning: Canadians and false induction: The P(1)=>P(2) case fails.
Division algorithm theorem: for integers a and b<a, we can find q and r so a=qb+r and 0<=r<q.
Proof: Induction on a works. If a=qb+r, then either a+1 = qb + (r+1) or else r=q-1 and so a+1 = (q+1)b
Alternative proof: what is the minimum of {a-qb >=0 | q ∈ N}? Let it be r; we claim r<q.
An introduction to divisibility and congruence:
if a,b are in N, then a|b, or "a divides b" if ∃k∈N b=ak
b≡c mod a if a|(b-c)
Not actually done:
Fact: for all a,n∈N there is b so: 0≤b<n and b≡a (mod n). So modular arithmetic comes down to Zn = {0,1,...,n-1}
gcd theorem/algorithm
5x≡1 mod 13, 3x≡1 mod 13
Postage in Investigate!; show 8 cent and 5 cent stamps can make any amount greater than or equal to 28 cents. Proof on p 179
(by the way, there is no solution for 27 cents)
Solving 17x + 80y = 1
Solving 5x2 ≡ 1 mod 13 and 9x2 ≡ 1 mod 13
why is the first the same as solving x2 ≡ 8 mod 13?
Theorem: if p is prime, then for any x not congruent to zero, there exists y so x*y ≡ 1 (mod p).
Fermat's little theorem ap-1 ≡ 1 (mod p)
Z13 example 3 has order 3, 5 has order 4
Binomial-theorem proof
# returns the list a, a*a, a^3, ..., a^n, all mod n
def powerlist(a,n):
lis = [0]*n
lis[0] = a % n
for i in range(1,n):
lis[i] = a*lis[i-1] % n
return lis
a*b ≡ 0 mod 15
a|bc, (a,b) = 1 => a|c
Proof: Note ∃k ka=bc. Also ∃x,y xa+yb=1. At this point c = cxa + cyb = a(cx + ky), so a|c
Group-theoretic proof of Fermat's "little" theorem ∀a≠0 ap-1 ≡ 1 (mod p)
The order of a ∈ Zp is the smallest k>0 such that ak ≡ 1 (mod p)
There always is such a k. The set {a, a2, a3, ... } cannot be infinite, so there are i and j, i<j, so ai ≡ aj. But then aj-i ≡ 1 mod p.
The next step is to prove that order(a) divides p-1. That's it!
Let A = {a, a2, a3, ... }. We show that we can partition {1,...,p-1} into multiple disjoint sets each the same size as A. That proves |A| divides p-1.
Z13 example 3 has order 3, 5 has order 4. A = A5 = {5, 12, 8, 1}.
2A = {10, 11, 3, 2}
3A = {2, 10, 11, 3}
4A = {7, 9, 6, 4}
5A = {12, 8, 1, 5}
6A = {4, 7, 9, 6}
7A = {9, 7, 4, 7}
Maybe:
straightenup(n) sequence. Plot for n=200, n=600, n=1000
fib(n): why is fib(40) slow?
nc(0)=nc(1)=1 For n>1, nc(n) = nc(n-1)
+ nc(n-2) + 1
prove by induction: nc(n) >= fib(n)