Simplified
Chinese Remainder Theorem
-simplified proof:
If gcd(m1,m2) = 1, then for
any a1, a2 Z, the simultaneous congruences
x a1 mod m1
x a2 mod m2
have a solution. Furthermore, if x = x0 is one integer solution, then the complete solution
is
x x0 mod

Fermat's Little Theorem
-If p is a prime and a != 0 is an integer which
is not a multiple of p, then
a^p1 1 mod p
Proof:
Observe that p a.
First we will show that no two of the numbers 0a, 1a, 2a, . . . , (p 1)a are congruent
modulo
p. To see this, suppos

Logical Operators
-We can combine propositions using logical
operators such as and or or.
-For example, 2 + 3 = 5 or 2 + 2 = 7 is
a proposition. Is it true or false?
It is true: in mathematics, the or
(also called the disjunction) of two propositions is t

Linear Congruences
-Thus far, we have done a lot of work on looking at the solutions to congruences of the
form
ax c mod b (9)
which is equivalent to a Seitan Nuggets problem (see Lemma 22). We have seen that the
linear congruence (9) has integer solution

Properties of Equivalence Class
-If [a] is any nonzero element of Zp, where p is prime, then [a] has a unique
multiplicative inverse.
Proof:
if a and b are any relatively prime
integers, then integers x, y such that
ax + by = 1.
Now [a] != [0] in Zp, and

Real Numbers
-A real number x is said to be a square root of a real
number y if x is non-negative and x^2 = y.
-Given x, y non-negative
real numbers, assume x^2 = y^2
. Then x = y.
Proof:
Subtracting y^2
from both sides, we have that x^2 y^2 = 0.
Factorin

Uses of Fermat
-Suppose that we want to know [3]1 in Z7.
At this point we know several ways
to do this:
Try out [1], [2], [3], . . . , [6] to find one that works. It will turn out that
[3] [5] = [15] = [1] in Z7
Use Corollary 27, which tells us that an in

Quantifiers
-A proposition such as For any real number x, 0 x^2 can be seen as a simple propositional
function, 0 x^2
, together with a universal quantier
-For any real number telling us that this
holds regardless of the exact value of x.
-A propositional

Chinese Remainder Theorem
-Solve these simultaneous congruences (note that gcd(9, 7) = 1):
x 2 mod 9
x 3 mod 7
The first congruence tells us that
x cfw_9y + 2 | y Z
Substituting this into the second congruence, we have:
9y + 2 3 mod 7
which can be simplif

Chinese Remainder Theorem
-Consider now that we have 3 congruences:
x 2 mod 9
x 3 mod 7
x 2 mod 5
where gcd(9, 7, 5) = 1.
We already know from the previous example that the complete solution to the first two
congruences is
x 38 mod 63
We are thus reduced