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: 6^{n}-1 is divisible by 5

Example 2.5.3: n^{2} < 2^{n} 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 **Z**_{n} =
{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 5x^{2} ≡ 1 mod 13 and 9x^{2}
≡ 1 mod 13

why is the first the same as
solving x^{2} ≡ 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 a^{p-1} ≡ 1 (mod p)

Z_{13} 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 a^{p-1}
≡ 1 (mod p)

The **order** of a ∈ Z_{p} is the smallest k>0
such that a^{k} ≡ 1 (mod p)

There always is such a k. The set {a, a^{2}, a^{3}, ... }
cannot be infinite, so there are i and j, i<j, so a^{i} ≡ a^{j}.
But then a^{j-i} ≡ 1 mod p.

The next step is to prove that order(a) divides p-1. That's it!

Let A = {a, a^{2}, a^{3}, ... }. 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.

Z_{13} example 3 has order
3, 5 has order 4. A = A_{5} = {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)